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The Earth’s Gravitational Field and Near Sea Level Atmospheric Temperatures – Slaying the Sky Dragon Excerpt

Written by Charles R. Anderson

The primary source of warmth for the Earth is radiant energy provided by our Sun. Just as the Sun radiates energy into space, the Earth, being warmer than space, also radiates energy out into space. Most of the energy it radiates into space is in the form of infra-red radiation, though light is a contributor as well. The Earth will radiate about the same amount of energy into space as it receives from the Sun on average. The Earth is often called a black body radiator, though this is not technically correct.

psi 2

The energy equilibrium with the Sun ignores some heat from the Earth’s core, energy due to the Earth’s magnetic field interactions with the magnetic field of the Sun, gravitational tide effects due to the moon, or energy due to material from space entering our atmosphere at high speeds. It is commonly claimed by those who advocate catastrophic global warming due to man’s emissions of carbon dioxide that the total greenhouse gas effect is a warming of the Earth’s surface by about 33ºC.

They say this warming is caused by the infra-red (IR) radiation absorbing gases of water vapor, carbon dioxide, and methane in the atmosphere. These gases are said to cause the Earth to retain and even multiply the energy it receives from the Sun, so that the Earth’s surface is warmer than it would otherwise be. Of these gases, water vapor is much the most important, but carbon dioxide is said to have a large enough effect that man’s additions to the concentrations in the atmosphere will do serious harm to the Earth’s flora and fauna, as well as man himself.

That CO2 might have such a big effect is due to a gross exaggeration of the total effect of the atmosphere’s IRabsorbing gases. The exaggerated scale of the effect comes from the observation that the temperature of the Earth as seen from space is about 33ºC cooler than the average temperature of the Earth’s surface. That entire temperature difference is attributed by them to the greenhouse gas effect, or the effect of IR-absorbing gases. Exaggerating that effect allows them to exaggerate the effect of man’s emissions of carbon dioxide and methane. This chapter will discuss the real sources of the 33ºC difference between the average temperature of the Earth’s surface and its radiation temperature as seen from space, with primary focus on the contribution of the Earth’s gravitational field.

Back in 1976, before the politicization of the Catastrophic Greenhouse Gas Warming hypothesis, the National Oceanic and Atmospheric Administration, the National Aeronautics and Space Administration, and the United States Air Force collaborated on a project to calculate the properties for a U.S. Standard Atmosphere as a function of altitude. They calculated the temperature, pressure, density, and molecular speeds, among other parameters for an ideal gas in the Earth’s gravitational field as a function of altitude. “The equations used are those adopted 15 October 1976 by the United States Committee on Extension to the Standard Atmosphere (COESA), representing 29 U.S. scientific and engineering organizations. The values selected in 1976 are slight modifications of those adopted in 1962. The equations and parameters used are documented in a book entitled U.S. Standard Atmosphere, 1976 published by the U.S. Government Printing Office, Washington, D.C.”

The tables are based on rocket and satellite data and perfect gas theory to provide atmospheric densities and temperatures from -5,000 meters below sea level to 1000 kilometers altitude. Below 32 kilometers altitude, the U.S. Standard Atmosphere is identical with the Standard Atmosphere of the International Civil Aviation Organization.

My copy of these tables is in the 71st Edition of the Handbook of Chemistry and Physics for 1990 – 1991. The preface to the table says: “The U.S. Standard Atmosphere, 1976 is an idealized, steady-state representation of the earth’s atmosphere from the surface to 1000 km, as it is assumed to exist in a period of moderate solar activity. The air is assumed to be dry, and at heights sufficiently below 86 km, the atmosphere is assumed to be homogeneously mixed with a relative-volume composition leading to a mean molecular weight M. The molecular weights and assumed fractional-volume composition of sea-level dry air were” and a list of the gas molecules followed. I will leave out the molecular weights and just give the gas molecules, their fractional volumes, and the translation of the chemical symbol.

N2, 0.78084, nitrogen

O2, 0.209476, oxygen

Ar, 0.00934, argon

CO2, 0.000314, carbon dioxide [Note this is less than the usual fraction used now]

Ne, 0.00001818, neon

He, 0.00000524, helium

Kr, 0.00000114, krypton

Xe, 0.000000087, xenon

CH4, 0.000002, methane

H2, 0.00000005, hydrogen

For many purposes, the molecules of the atmosphere are well represented by an ideal gas model. Note again that the primary greenhouse gas is water vapor and it is left out entirely. Now let us list some values for the temperature, pressure, density, and speed of the average gas molecule of molecular weight M for various altitudes as calculated from the volume fractions in this U.S. Standard Atmosphere. They are given in the table below:

psi 1

The temperature at sea level here is 288.15 Kelvin, which is 15.00ºC, and a change of 1K, equals a change of 1ºC. This temperature will vary with the time of year and the time of day. The change of temperature will then shift the values of this table as the sea level temperature changes from the near average value of 15ºC.

The main source of energy affecting this sea level temperature is solar radiation, which is affected by the time of day, the time of year, cloud cover, the humidity, volcanic ash and aerosols, and industrial and transportation aerosols. Energy also comes from the Earth’s hot interior, the interaction of the Earth’s magnetic field with the sun’s magnetic field, the solar wind, debris from space, the decay of radioactive elements, and the gravitational effects of the moon. These sources of energy can shift the sea level temperature relative to that given in this table, but then other values will shift with it. The very fluid atmosphere will seek out an equilibrium temperature distribution as a function of altitude for each such sea level temperature.

It will do this by means of gas molecule collisions, convection, radiation, and evaporation or sublimation of water. With a sea level temperature of 288.15K or 15ºC, this table gives the values of other parameters which represent the equilibrium condition.

Now let us note that the black body temperature of the Earth as seen from space is 255K. This is almost identical to the temperature of the Standard Atmosphere at an altitude of 5000 m. Thus, the effective altitude at which radiative cooling of the atmosphere takes place with respect to space is 5000 m. Above about 4000 m, radiative cooling becomes the dominant mechanism, so the effective altitude should be higher than that as this implies. Now, the Greenhouse Gas hypothesis claims that the reason the temperature at sea level is 288K, rather than 255K, is because greenhouse gases such as water vapor, CO2, and methane absorb the long wavelength IR radiation from the sunlight heated surface of the Earth and re-emit half of it back to the ground. That “extra” radiation heats the Earth’s surface to a temperature 33K higher than it would otherwise be.

Let us now take a look at the below sea level temperatures of this ideal gas with no water vapor. The lowest open area on the surface of the Earth’s land masses is the Dead Sea shore. 

The altitude there is about -413 m, or 413 meters below sea level. At -500 m, the temperature would be 3.25K higher than at sea level according to this table. Interpolating from the table, the temperature at -413 m would be about 2.68 C or K warmer than it would be at sea level. Indeed, the Dead Sea area is very often warmer than the surrounding sea level areas. This tends to be true of other areas below sea level also. Yet, there is no reason to think that IR emissions from the surface and their capture by water vapor and CO2 should be very different, so this suggests that the mechanism of IR-absorbing gases is not the primary reason for any temperature difference between the Dead Sea and the surrounding sea level areas.

Why does the table give values down to -5000 m altitude? Because the table is of value to deep mine engineers who have to deal with air ventilation in deep mines. Deep mines become warm primarily because the surrounding rock becomes warm due to the Earth’s high core temperature, which also means that some of the energy at the Earth’s surface causing it to be warm is due to heat from the core. Deep mines also become warm due to the effect of gravity on air molecules! That effect is an important effect and has to be taken into account. Let us see how important the U.S. Standard Atmosphere tables imply it is. At -5000 m, the temperature of the standard atmosphere is given in the table as 320.676K, which is 32.526 K or C higher than the temperature at sea level and at atmospheric pressure. Now note that this is almost the same temperature increase that the surface of the Earth has relative to the altitude with the same temperature as its effective ‘black body’ radiation temperature, that is, 5000 m.

Now the bottom of a mine shaft 5000 m deep does not have any sunlight being absorbed and it is not emitting IR because sunlight has warmed it only to have much of that energy radiated as IR which is reflected back onto the bottom of the mine. No, the higher temperature of the atmosphere at the bottom of the mine shaft is due to the higher kinetic energy of gas molecules that comes from them having about as much less potential energy in the Earth’s gravitational field compared to sea level as those at sea level have compared to the potential energy at 5000 meters altitude. An ideal gas molecule at a lower altitude has as much more kinetic energy compared to a molecule at higher altitude as that higher altitude molecular has more potential energy.

Recalling high school physics, the total energy of a mass M in the Earth’s gravitational field near the surface is E = ½ Mv2 + Mgh, where g is 9.8 m/s2 and h is an altitude small compared to the radius of the Earth. This is an approximation of course. The U.S. Standard Atmosphere tables include the variation of g with altitude. At sea level it is 9.8066 m/s2, while at 5000 m altitude it is 9.7912 m/s2, for instance. This instance of conservation of energy is very important because the temperature of an ideal gas is proportional to the square of its particle velocity, which is proportional to its kinetic energy.

The ideal gas law is PV = nRT. P is the pressure, V is the volume, n is the number of moles of the gas in the volume, R is a constant, and T is the temperature in Kelvin. Let’s go down our mine shaft. Let’s say it is 5000 m deep and we will calculate the average T for each horizontal slab 1 meter tall as we go down the shaft. Assuming the air density is increasing as the pressure increases, we know that n is proportional to the air density and increasing. It is not constant as in your simple high school science class calculations. The tables show us that this is the case. Now if the pressure, P, were proportional to this density, then T would remain constant. It is not, however. 

The pressure P is proportional to the product of the density and the kinetic energy of the molecule. This is why T for the ideal gas is proportional to the kinetic energy of the molecules.

So, the temperature change upon going 5000 m below sea level is about a 33K increase, while the temperature change on going from 5000 m altitude to sea level is about a 33K increase. Now if the temperature at sea level is due to greenhouse gas effects smuggled into the table by the use of some inputs from rocket and satellite data, then the temperature increase values in the table for below sea level should be nonsense.

Yet similar tables were generated in 1958, 1962, 1966, and 1976. This table is still widely used. Over those years, deep mine engineers would surely have had input to correct the values of temperature for the below sea level entries if they did not hold up in their calculations for mine heat management and temperature, pressure, and air density measurements. Or would they? We will discuss this more.

Let us illustrate this mining application problem. The Tau Tona gold mine in South Africa is 11,760 feet deep or 3,584.4 meters deep. The air in the mine reaches a temperature of 130 F. The gravitationally determined air temperature at that depth is 311.46K or 38.31ºC or 100.96ºF as interpolated from the full table of the U.S. Standard Atmosphere given in 500 meter altitude increments. Clearly the rock wall temperature is higher than the temperature of air in gravitational equilibrium in this deep mine. The air at sea level from the table is at 288.15K or 59ºF. The mine engineer would love to bring this air at 59ºF down to the bottom of the mine and be able to keep it at 59ºF as he does so. Air at 59ºF would be much more effective in cooling the bottom of the mine than is air at 101ºF.

But, even if the mine engineer flows this air down to the bottom of the mine in well-insulated pipes, it is still in the Earth’s gravitational field. So, he can keep it from being warmed much by the very warm rock walls of the mine, but he cannot keep it from rising to a temperature of 101ºF, according to the tables! When he calculates the necessary air flow to achieve a given cooler temperature in the bottom of the deep mine, he should be very aware of the temperature increase in the air due to bringing it down to that depth.

So, if the increased gas molecule kinetic energy in the atmosphere at -5000 m creates an increase in temperature of about 33º C, then the increased gas density and kinetic energy in the atmosphere at sea level compared to that at 5000 m, is also responsible for the 33ºC increase in temperature at the surface of the Earth compared to its black body radiation temperature of 255K and the temperature of air at 5000 meters altitude according to the logic of the tables.

That the temperature of a gas in a gravitational field increases as the strength of the field increases is further confirmed by the temperature and pressure relationships to altitude of the other planets with gaseous atmospheres or compositions. The actual gases vary widely as does the amount of radiation incident upon them from the sun. The reflectivities of those gases also differ greatly. Despite that, as the pressure increases as one moves deeper into the gas atmosphere of each planet, the temperature increases. When the gravitational field is extremely high, as in the case of Jupiter, the temperatures become extremely high in the interior. Yet, Jupiter receives a pitiful 50.5 watts/sq. m of solar radiation compared to the Earth’s incident solar radiation of 1368 W/sq. m.

The examples of the gaseous planets being heated by gravitational field effects have often been presented by critics of the greenhouse gas hypothesis with no effective reply. Somehow, the Earth is different. On the Earth, only manmade effects are claimed to be important.

Now we can bring this back closer to home again. The materials at the core of the Earth are extremely hot due to the very high pressures caused by the very high gravitational field strength there. The molten core liquids heat up due to high pressure, but the gases of the Earth’s atmosphere do not? We should be asking the Greenhouse Gas hypothesis advocates why liquids under pressure become very hot, but a gas under increased pressure in a gravitational field does not?

You may want to ask where the energy is coming from that is heating the gas deep in the mine shaft. The gas at the surface of the Earth has a potential energy due to gravity. The gas at the bottom of the mine shaft has a lower potential energy, with the difference in energy converted into a higher kinetic energy because energy is conserved. The gas molecules therefore have a higher mean velocity. The gas temperature is proportional to the kinetic energy of the gas molecules, so it also goes up. The effect on the pressure is that it goes up as the number density of the molecules goes up and it also goes up in proportion to the increase in kinetic energy.

The pressure is increasing faster than either n or T. In fact it is increasing as the product of these two increasing properties, as stated in the ideal gas law, PV = nRT. Note that the temperature does not depend on any flow of gases, according to the tables. It only depends upon the kinetic energy of the air molecules, which is a function of the strength of the Earth’s gravitational field at a given altitude.

The pressure has a more complex functionality, because it depends both on the strength of the gravitational field and the weight of the atmosphere above a given altitude.

It is gravity which accounts for the increased pressure and density of the atmosphere at the bottom of the mine shaft. This is also where most of the energy comes from to heat the Earth’s core or the atmosphere of Jupiter. But the pressure is proportional to the product of the density and the temperature and both are increasing, so the pressure is increasing faster than the density does.

Alternatively, the temperature is proportional to the pressure divided by the density. Since the pressure is increasing faster than the density is, the temperature goes up as we go down the mine shaft.

For the same reason, the temperature goes up as we go 5000 meters downward from the effective black body radiation altitude of about 5000 meters to the surface of the Earth. There is simply no need to posit a complex and unproven theory of greenhouse gas warming to explain why the surface of the Earth is 33ºC warmer than the black body temperature of the Earth as seen from space and as in equilibrium with the incident energy from the sun. The tables seem to imply the cause of this difference is the Earth’s gravitational field. However, despite the fact that one would think that the calculations of mine engineers would have confirmed or invalidated the table for below sea level air temperatures, we will find that the tables do imply a stronger gravitational field effect on the atmosphere than makes sense.

In my article called Do IR-Absorbing Gases Warm or Cool the Earth’s Surface?, I pointed out that:
“If we assume that the sphere [at an altitude of 5000m] with the temperature of 255K is in equilibrium with a slightly smaller black body sphere of the radius of the Earth at sea level, we can calculate the temperature of that surface given that it must radiate a power equal to the power of the surrounding sphere which is in equilibrium with space. The temperature will be higher, since the surface area of the sphere is smaller. In fact, the temperature of the Earth’s surface as a black body would be 255.100K or 0.1ºC warmer than the sphere at the altitude of 5000 meters above sea level which is in equilibrium with space. But the Earth’s surface is not really a black body, so the Stefan-Boltzmann equation has to have an emissivity factor multiplied times the temperature side of the equation. For the Earth’s surface this emissivity factor is about 0.7 on average. This causes the Earth’s surface to have to be at the more elevated temperature of 278.89K to be in equilibrium. This is only about 9K or 9ºC below its usual temperature of 288K. Anything otherwise violates the Law of Energy Conservation.”

So, the increased kinetic energy of air molecules at sea level when compared to that at the effective black body temperature of the Earth altitude of 5000m does not account for the 33K temperature difference despite the tables seeming implication. The effect is real, but its size must be exaggerated. Unless there are counteracting cooling effects, the gravitational temperature increase would be limited to about 9K when changing altitude from 5000m to sea level. There are countervailing cooling effects, so the gravitational effect may be larger than 9K, but it is unlikely to be as large as 33K.

Apparently, the tables did smuggle in effects due to other heating effects and then improperly projected those to the below sea level altitudes.

Yet, mine engineers do attest to such an effect, but they have not apparently been effective in getting the U. S. Standard Atmosphere tables adjusted to any good scientific measurements of the effect.

This is another of many examples of contributing warming effects which have not been properly taken into account by the climate models and which appears to have an inadequate basis in experimental measurements by physicists.

There are still other sources of warming that can claim to contribute to the 9K remaining temperature increase at sea level. As mentioned above, one is the conduction of heat from the Earth’s core. Another is the retention of heat by the oceans with their very high heat capacity. The land surface also retains heat and releases it slowly as the brightest part of the day passes and is replaced with night.

The atmosphere itself, especially when laden with water vapor, holds and retains considerable heat. These heat retention materials with significant heat capacities, even out the temperatures between day and night and when clouds momentarily block strong sunlight. In doing this, they raise the average temperature somewhat through the course of the day. At this time, the size of the average increased temperature contribution of each effect is not known, but they are each making a contribution.

The major difference between the sea level temperature and the effective altitude at which radiation into space balances the energy fluxes into and away from the Earth, a temperature difference of about 33ºC, is not properly attributed to greenhouse gases. These gases are gases that absorb IR radiation effectively. The main greenhouse gas is water and CO2 and methane gas are minor greenhouse gases. Those who promote the idea that man’s emissions of CO2 and methane are likely to cause catastrophic global warming, claim that these greenhouse gases are responsible for this substantial heating of the Earth’s surface by an additional 33ºC. As we have seen here, this is not so.

This argument does not mean that it is not possible for greenhouse gases to shift the sea level temperature a degree or two from the present average temperature or to influence the temperature through the course of a short time, such as between night and day. Whether there are such effects caused by IR-absorbing gases needs to be examined in other ways. What has been established here is other effects, including warming due to the Earth’s gravitational field on its atmosphere, make significant contributions to the warming usually attributed to greenhouse gas theory. Most of the warming is simply a radiative equilibrium having nothing to do with greenhouse gases. Some is due to gravitational effects on the gases of the atmosphere or on the Earth’s core. Some is due to heat retention by the materials near the Earth’s surface which have significant heat capacities.

As I discussed in Do IR-Absorbing Gases Warm or Cool the Earth’s Surface? the effect of carbon dioxide and methane is a net cooling effect, not a warming effect. Water has a much more complex role because it can retain absorbed heat in ways that CO2 and methane cannot.

It is the main greenhouse gas, but it has many properties and many roles not well-described by that name. In many of those roles, it also acts to cool the surface by evaporation and sublimation. In fact, its cooling power is probably presently underestimated in climate models. Water vapor as a cloud former is not yet well-understood, as the nucleation of clouds by cosmic rays makes clear. Water and the IR-absorbing gases keep the Earth’s surface from becoming much warmer during the day when there is direct sunlight. Their role at night in net heat retention is very complex and is not yet understood at all well.

Instead of creating evermore fanciful computer climate models, there are many real warming and cooling effects which will only be well-understood when good scientific measurements from well-designed experiments are performed.

After a huge expenditure of monies and effort on global climate models it is clear that the basic science needed as input into such models is sadly lacking. The effort has been badly misplaced.
Climate modeling without an adequate basis in the understanding of the physics of radiation and heat transfer and transport is pointless.

It is very telling when well-known climate modelers make fun of physicists, chemists, geologists and engineers who take a look at the very effects which are essential inputs to their climate models.

Comments (62)

  • Avatar

    Rosco

    |

    I’m calling this quits – the sequence of comment and reply has become ridiculously convoluted.

    I do get your arguments – we simply disagree.

    I wrote :-

    “Object at 37 C – call it 310 K… I’ll make an assumption the object is irradiated by a
    constant ~523 W/m2 (or body metabolism if you like) and it absorbs all of it hence a
    temperature of ~37 degrees C or ~310 K.”

    “Case 1 – “cool surroundings” – lets say 20 C – call it 293 K.
    Q(net) = 1.sigma(310^4 – 293^4) – ~105.

    Case 2 – “colder surroundings” – lets say minus 20 C – call it 253 K
    Q(net) = 1.sigma(310^4 – 253^4) – ~291.”

    And yes when there is nothing but empty space the Q(net) is 523 W/m2.

    In response you wrote –

    “If there is no other power source, yes that is true. But if there is a separate power
    source – the sun for the earth, metabolic processes for the human body — the replacement
    of cool surroundings for colder surroundings reduce the Q(net) between the object and its
    surroundings, resulting in an increase in temperature of the object until the Q(net) to
    the surroundings matches the separate Q(in). For the life of me, I can’t understand why
    people like you have so much trouble with this concept!”

    You have also written:-
    “The SB equation gives you the radiative flux OUTPUT of an object as function of its
    temperature and emissivity. It does NOT give you the temperature of an object (transient
    or steady state) as a function of its radiative FLUX input! You should have learned this
    early in your first thermo class.”

    Am I right in believing that at a certain temperature an object will emit radiation in
    proportion to its temperature and this radiation is given by the equation :-

    P = A*e*sigma*T^4 ?

    And isn’t emissivity a kind of afterthought added by later experimentation because the
    very premise of cavity radiation is it has emissivity of 1 ?

    The texts always present the SB equation as H = A*e*sigma*T^4 (Young & Freedman). They
    then go on to introduce the “net” form BUT each term is exactly the original SB equation.

    We can assume unit area and emissivity and still discuss by simply ignoring these as they
    remain the same ? Obviously one must be consistent.

    So I know that you choose to use the emission from the 310 K object from the initial
    equation as your Q(net) or Q(in) ? and you hold it constant as the necessary output from:-

    Q(net) = 523 = 1 * sigma (Tobj^4 – 253^4)

    Whereas I have said that an object emitting 523 W/m2 is at temperature of 310 K and I use
    this in the
    Q(net) = 1*e*sigma(310^4 – 253^4)

    And I say Q(net) does not need to be held constant because all objects emit radiation in
    proportion to their temperature.

    If I accept your viewpoint that because the 523 net output must be preserved there still
    remains the fact that the only way the object can increase in temperature is by absorbing
    an extra 232 W/m2 equivalent to the output from an environment at 253 K ?

    Even if you say this 232 W/m2 comes from Q(net) Q(in) ? – which you don’t by the way you add it to Q(net) – there is still the fact that before you play this “bait and switch” the object is
    still emitting 523 W/m2 equivalent to the 523 W/m2 Q(net).

    Thus the only way the extra 232 W/m2 can figure is to come from the cold environment.

    That is your argument isn’t it – the ambient environment reduces the radiation output from
    the object by 232 W/m2 (253^4) – from ~523 to ~291 – and hence the object MUST increase to
    340 K to overcome this “radiation resistance” ?

    But an object radiates in proportion to its temperature – ALL of the texts state this !

    And heat does not transfer from cold to hot – ALL of the texts state this !

    Thus if the object is prevented from radiating 232 W/m2 of its 523 W/m2 emitted when it is
    at 310 K this can only occur through the effective transfer of heat from the cold
    environment to the hotter object.

    You also show this explicitly in what you write

    Q(net) = 523 = 1 * sigma (Tobj^4 – 253^4)

    Tobj = ((523/sigma) + 253^4) ^ (1/4) = 340K”.

    I don’t see any way that anyone can claim this is not the transfer of 232 W/m2 or
    ~sigma*253^4 from cold to hot.

    You object that my equations do not balance energy output BUT both terms in the equations
    I wrote are correct.

    The object emits 523 W/m2 in mine, the surroundings emit 232 in mine and Q(net) is 291
    W/m2 in mine.

    The object does not heat up because heat does not transfer from cold to hot.. I believe the
    transfer from cold to hot is absolutely necessary to increase the output from ~523 to ~757
    as your equations show.

    Experiments for the 17th and 18th century established heat transfer is a one way process and are the basis of Thermodynamics.

    So lets leave it there.

    I do know where you are coming from ! We simply don’t agree and probably never will.

    I still say :-

    http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node37.html

    The Clausius statement:-
    “No process is possible whose sole result is the transfer of heat from a cooler to a
    hotter body.”

    • Avatar

      Rosco

      |

      I meant to mention the Washington University lecture

      http://www.atmos.washington.edu/2002Q4/211/notes_greenhouse.html

      It states the combination of solar radiation at 239 W/m2 combines with 239 W/m2 atmospheric back radiation to force the Earth’s surfaces from 254 K to 303 K.

      This explicitly claims the “atmospheric back radiation” has equivalent heating capacity to the solar radiation. This is universally claimed to be the basis of the greenhouse effect.

      I do not believe the “atmospheric back radiation” is capable of inducing the same effects as the solar radiation or has even a tiny proportion of the heating power and I reject this concept out of hand no matter how many insults or obfuscations I am subjected to !

    • Avatar

      Quokka

      |

      Maybe this is the crux of the misunderstanding:

      That is your argument isn’t it – the ambient environment reduces the radiation output from the object by 232 W/m2 (253^4) – from ~523 to ~291 – and hence the object MUST increase to 340 K to overcome this “radiation resistance” ?

      No. If the object is at 310K, has a 532W internal power source, and is in an ambient environment of 253K, then:

      It is radiating 532W. (i.e. radiation output is not reduced at all)
      It is absorbing 232W from the 253K environment.
      The net energy loss from the object is 532 – 232 = 291W.
      But it also has an internal energy source of 532W, so it is gaining actually 232W of energy, which means it has to get hotter.
      It will keep getting hotter until it reaches equilibrium at 340K. At that point:
      The energy inputs to the object are 523W (internal), 232W (external), total of 755W.
      At 340K, it also radiates 755W, inputs and outputs are equal so temperature will not change any further.

      But an object radiates in proportion to its temperature – ALL of the texts state this !

      And I’m pretty sure Ed Bo has never stated otherwise. The object above always radiates in proportion to its temperature.

      And heat does not transfer from cold to hot – ALL of the texts state this !

      True. You may be thinking that the 232W going from the environment into the object is heat transferring from cold to hot in the above equilibrium. It is not. Energy is allowed to flow in both directions. Heat cannot, since it is the net energy flow.

    • Avatar

      Ed Bo

      |

      Rosco:

      One last shot. You say: “Am I right in believing that at a certain temperature an object will emit radiation in proportion to its temperature and this radiation is given by the equation :

      P = A*e*sigma*T^4 ?”

      Yes.

      You say: “And isn’t emissivity a kind of afterthought added by later experimentation because the very premise of cavity radiation is it has emissivity of 1 ?”

      No, but not really relevant to our discussion. (A cavity with a small hole is just a way of demonstrating how you can get virtually complete absorption out of real-world materials whose absorptivity is less than 1.0.)

      You continue: “The texts always present the SB equation as H = A*e*sigma*T^4 (Young & Freedman). They then go on to introduce the “net” form BUT each term is exactly the original SB equation.”

      Yes. The equation you cite is for (gross) radiative power from an object. The net power transfer between two objects is just the difference in the two gross flows.

      Next you say: “We can assume unit area and emissivity and still discuss by simply ignoring these as they remain the same ? Obviously one must be consistent.”

      Yes, we both have been doing this to keep the discussion simple and focus on the key topics.

      But now you claim: “So I know that you choose to use the emission from the 310 K object from the initial equation as your Q(net) or Q(in) ? and you hold it constant as the necessary output from:- Q(net) = 523 = 1 * sigma (Tobj^4 – 253^4)

      Whereas I have said that an object emitting 523 W/m2 is at temperature of 310 K and I use
      this in the
      Q(net) = 1*e*sigma(310^4 – 253^4)”

      NO!!! Here’s where you miss the point entirely and go completely wrong. The object at 310K (which we will take at least as the initial condition) has a Q(gross) [object to ambient] of 523W, as in the Y&F equation you cited. But you must also account for the Q(gross) [ambient to object] AND the Q(in) from our power source of 523W.

      So in the case of an ambient of 253K producing a Q(gross) [ambient to object] of 232W, the object has 523W input from the power source and 232W input from ambient, and a 523W Q(gross) [output to ambient]. This results in a +232W power imbalance for the object.

      The 1st Law DEMANDS that this imbalance result in an increase in internal energy of the object. (All the texts state this!!) Since we are sticking to the simple case of no phase change, chemical reactions, etc., this means that the object’s temperature must increase.

      You seem to be arguing that despite the power imbalance in and out of the object, its temperature will not increase. How do you square that with the 1st Law???

      We both agree that the object will increase its emitted radiation as its temperature increases. But you miss the simple chain of causality that the power imbalance leads to temperature increase leads to increased radiative output until the object’s power balance i restored. I showed you how to solve for the object temperature at which this happens.

      You say: “If I accept your viewpoint that because the 523 net output must be preserved there still remains the fact that the only way the object can increase in temperature is by absorbing
      an extra 232 W/m2 equivalent to the output from an environment at 253 K ?”

      You should accept it because that’s the way it works!

      Next you say: “Even if you say this 232 W/m2 comes from Q(net) Q(in) ? – which you don’t by the way you add it to Q(net) – there is still the fact that before you play this “bait and switch” the object is still emitting 523 W/m2 equivalent to the 523 W/m2 Q(net). Thus the only way the extra 232 W/m2 can figure is to come from the cold environment.”

      Once again you are getting yourself horribly confused between gross and net flows. In the steady state, the Q(net) between the object and ambient must match the Q(in) from the power source. At 310K, the object’s Q(gross) {to ambient} is 523W. But with the Q(gross) {from ambient} of 232W, there is not enough (gross) power out to match the power in.

      You continue: “That is your argument isn’t it – the ambient environment reduces the radiation output from the object by 232 W/m2 (253^4) – from ~523 to ~291 – and hence the object MUST increase to 340 K to overcome this “radiation resistance” ? But an object radiates in proportion to its temperature – ALL of the texts state this !”

      Again, you are completely confusing gross and net flows. The object’s GROSS radiative output is a function of its temperature. But the NET radiative exchange is also a function of the ambient temperature.

      High-school-dropout cashiers making change understand the difference between gross and net flows. Why can’t you???

      Then you say: “And heat does not transfer from cold to hot – ALL of the texts state this !”

      The heat transfer between the object and ambient — Q(net) — is positive in all our examples, as you yourself have calculated, which means it is from hot to cold. Every heat transfer textbook I have ever seen discusses radiative heat transfer as an exchange, with the hot-to-cold gross transfer exceeding the cold-to-hot gross transfer.

      You seem to believe that because replacing a 0K ambient with a 253K ambient results in an increase in object temperature, (net) heat must be transferred from cold to hot. NO!!! That is why I brought up the example of the partial clog in the sink drain, which results in an increase in the level of water in the sink, even though no water is moving up. Of course, the passive clog cannot push water up against gravity. But we’ve all seen the increase in water level.

      Finally, you quote Clausius: “The Clausius statement:-
      “No process is possible whose sole result is the transfer of heat from a cooler to a
      hotter body.” ”

      Two comments: In the radiative exchange between our object and its ambient, there is also a transfer from a hotter to a cooler body, so “the transfer of heat from a cooler to a hotter body is not the SOLE result.

      Second, Clausius anticipated confusions like yours over 150 years ago, so he wrote and published this clarification:

      “In the first place, the principle implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it. The principle holds, however, not only for processes of this kind, but for all other by which a transmission of heat can be brought about between two bodies of different temperatures, amongst which processes must be particularly noticed those wherein the interchange of heat is produced by means of one or more bodies which, on changing their condition, either receive heat from a body, or impart heat to other bodies.
      On considering the results of such processes more closely, we find that in one and the same process heat may be carried from a colder to a warmer body and another quantity of heat transferred from a warmer to a colder body without any other permanent change occurring. In this case we have not a simple transmission of heat from a colder to a warmer body, or an ascending transmission of heat, as it may be called, but two connected transmissions of opposite characters, one ascending and the other descending, which compensate each other.”

      So Clausius explicitly talks about an “ascending transmission of heat” (cold to hot), which is real, but must be less than the “descending transmission of heat”.

  • Avatar

    Quokka

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    An interesting discussion. Everyone apparently agrees that if you have a sphere with 1m^2 surface area in 0K space, and an energy source of 532 W inside it, then it will stabilize at a temperature of 310K. But Ed Bo maintains that the sphere will get hotter when placed in an environment with background radiation > 0, while Rosco maintains that it will stay at the same temperature of 310 K (for ambient temperatures of <= 310K, at least).

    Consider the case where the sphere has no power source inside. If the ambient temperature is 310K, then the sphere will reach a temperature of 310K as it stabilizes with its environment. It will sit there, radiating 532W of energy, and absorbing 532W of energy to maintain the same temperature. It must be absorbing 532W from the environment, because otherwise the net loss of 532W would cause it to cool down.

    If we now start a 532W heating element inside the 310K sphere, what will happen? Surely it will warm up. How could it not? The energy from the heat source inside it can't radiate away (if the sphere is opaque), can't convect away (if the sphere is solid), and can't quickly conduct away (especially if the substance of the sphere is a good insulator). Presumably it will heat up until the temperature differential between the sphere's surface and the environment is enough to allow the excess energy to flow outwards.

    If you think this doesn't happen – why not? What does happen to the heat generated inside the sphere?

    • Avatar

      Rosco

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      I am unsure about your question because I think you assume I am talking about something different – which I am not.

      In this discussion I have ALWAYS said the object IS “heated” to ~310 K by an external or internal heat source and not by its surroundings.

      I have ALWAYS said that at ~310 K it radiates ~523 W/m2 – this is what the SB equation as derived initially explicitly states – the 523 W/m2 radiation will occur at this 310 K temperature no matter what the external temperature is.

      What I have not claimed is the object heats up in contradiction to the laws of thermodynamics!

      The Second Law of Thermodynamics –

      http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node37.html – The Clausius statement:-

      “No process is possible whose sole result is the transfer of heat from a cooler to a hotter body.”

      I wrote :-

      “Object at 37 C – call it 310 K… I’ll make an assumption the object is irradiated by a constant ~523 W/m2 (or body metabolism if you like) and it absorbs all of it hence a temperature of ~37 degrees C or ~310 K.”

      I explicitly show that in each equation I wrote – for example: –

      “Case 1 – “cool surroundings” – lets say 20 C – call it 293 K.
      Q(net) = 1.sigma(310^4 – 293^4) – ~105.” There it is – 310 as the object.

      All through this I have been trying to show what ALL physics texts explicitly state –

      The equation – Q(net) = A*e*sigma (T^4 – T(surroundings)^4), assuming unity for e and A and the usual conventions – indicates that “heat” transfers exclusively from hot to cold.

      If the result is positive net energy transfers from the object to the surroundings and its temperature will decrease.

      Conversely, If the result is negative net energy transfers from the surroundings to the object and its temperature will increase.

      Every Physics text will support this.

      The topic of thermodynamics is complex and there are no simplistic solutions despite what lots of people claim.

      While we all know about thermal equilibrium it is surely self evident that a “sphere” with some form of energy supply capable of maintaining 310 K in “zero K” space will not heat to 340 K when placed in a freezer – I mean come on !

      Yet this is exactly what has been vigorously and offensively asserted continually throughout this exchange !

      • Avatar

        Ed Bo

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        Rosco:

        You say: “While we all know about thermal equilibrium it is surely self evident that a “sphere” with some form of energy supply capable of maintaining 310 K in “zero K” space will not heat to 340 K when placed in a freezer – I mean come on !”

        It is most certainly NOT self evident, and if you had ANY practical experience in the field, you would not be saying that!

        I have been involved in thermal testing of electronics for over three decades now, and these are real cases of objects with constant power inputs at varying ambient temperatures. We, like all electronic manufacturers, invest heavily in thermal test chambers that permit us to precisely vary and control the ambient temperature around the object under test.

        In every single case I’ve seen in over 30 years, when ambient temperature is increased, the steady-state temperature of the constantly power object increases. The engineers and technicians in our test labs would be rolling on the floor laughing if they saw your assertion!

        Let’s consider the following scenario. I, the evil warmist, have captured you and tied you down on a very large block of ice at 0C. All of your body’s homeostatic mechanisms cannot compensate for the heat loss, and your core body temperature drops to 35C – you have hypothermia.

        But I relent, and offer to move you to an equally large block of slate (which has the same thermal conductivity as ice) at 25C. Do you turn down the offer because it is still below your body temperature, and therefore by your logic cannot result in an increase in body temperature?

      • Avatar

        Quokka

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        Replying to Rosco:

        In this discussion I have ALWAYS said the object IS “heated” to ~310 K by an external or internal heat source and not by its surroundings.

        Yes, I get that if it is heated by 532W, it will reach 310K, whether the heat source is internal or external. But I don’t understand any distinction between ‘surroundings’ and ‘external’. A 532K background is ‘the surroundings’, but is also an ‘external’ heat source, and will heat the object to the same temperature as, say, an external 532W laser would.

        While we all know about thermal equilibrium it is surely self evident that a “sphere” with some form of energy supply capable of maintaining 310 K in “zero K” space will not heat to 340 K when placed in a freezer – I mean come on !

        Actually I don’t find that self evident. I infer that you also believe that the 310K sphere in zero K space, powered say by internal radioactive decay, will likewise not warm up if placed in space with 310K background radiation (since I see no qualitative difference between a 310K background and a 269K freezer). My thought experiment added the heat sources in the other order: let the inert sphere come to equilibrium at 310K in space with 532W background radiation. Now, if you start up a 532K internal power source inside a 310K object, how can that extra energy *not* further warm the sphere? Where else could it possibly go?

        By the way, stating the 2nd law of thermodynamics does not answer the question.

        I know I’m repeating my question; that’s because you didn’t answer it before. I don’t now what you didn’t understand, since you didn’t say, but this is a very simple scenario.

        Whether you add the internal source then the external background source, or the external source then the internal source, or both at once, should make no difference to the eventual equilibrium state, so effectively this experiment is the same as placing the powered 310K sphere into a 310K environment.

    • Avatar

      Ed Bo

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      Quokka:

      You are absolutely right that if you have an unpowered object at 310K in equilibrium with its ambient also of 310K, and then you add a separate power source to the object, it will increase in internal energy, and therefore temperature (absent phase changes, etc.) The 1st Law demands it. (Also, it happens every time you turn on an electrical appliance in your house!) This is not (or rather, should not be) a difficult concept to get.

      One of the things you learn early in a real thermodynamics course is the “path independence” of these state variables like internal energy. That is, for a given set of conditions, you get the same result regardless of the starting conditions and the “path” taken to get to the final state.

      So it does not matter if we start at 310K equilibrium and add power, or start powered in a lower temperature ambient and increase the ambient temperature — you get to the same result.

      Rosco seems completely unfamiliar with this basic concept.

    • Avatar

      Jeff Greenwell

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      This is so easily testable it is ridiculous. Simply run out to Amazon and purchase yourself a Spectrometer. Crank up your oven to its highest temperature (usually self-cleaning mode) with the light bulb inside turned on. Using the spectrometer, measure the temperature of the light bulb. Now, let the oven and light bulb cool down. Take the light bulb out to your deep freezer. So the same experiment.

      I will absolutely and unequivocally guarantee that the temperature will be EXACTLY the same in both cases.

      • Avatar

        Quokka

        |

        My quick back-of-the-envelope calculation is that, if you take the temperature of the bulb filament to be 2550C (according to a quick web search) then you would expect the temperature to increase by 0.8 of a degree, i.e. about 1/3500th of the temperature. I very much doubt cheap Amazon spectrometers can detect this difference.

        • Avatar

          Ed Bo

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          You’re right once again, Quokka. Since the absolute temperature of the filament is about 10 times that of our ambient, its dQ/dT is about 1000 times greater, so its temperature only needs to increase by 1/1000 the amount of an object at our ambient to emit the same extra amount of power. So Jeff’s experiment might be challenged to find an effect.

          Still, there is an experiment both better and simpler. Get a standard incandescent bulb and a halogen incandescent bulb of the same power rating, say 40W. Turn on one, and then the other, sequentially in a dark room. Your eyes are good enough spectrometers to see the substantial difference in spectrum between the two – with the halogen bulb producing a much “whiter” light.

          Why the difference? The halogen bulb, being much smaller, has less surface area to reject energy by conductive/convective means, so the glass is a much higher temperature than that of the standard bulb.

          This in turn means that the gas inside the bulb is also hotter than in the standard bulb, but still colder than the filament. So the filament, exposed to a warmer (conductive) ambient than the standard bulb, runs significantly hotter and emits a spectrum of considerably shorter wavelengths with a much higher percentage in the visible range, and appearing whiter than the standard incandescent.

          You can get an even further improvement if you buy a halogen incandescent with the inner glass surface having a coating that reflects the shortwave infrared back to the filament, increasing its temperature further, and shifting the spectrum even more toward shorter wavelengths, with an even greater percentage in the visible range.

  • Avatar

    Rosco

    |

    Ed – I think you are confused.

    You wrote:-

    “I will introduce a Case 3, with an ambient of 0K (“cold” space). Using the SB equation, we see that the Q(net) from the object of 310K to 0K ambient is 523 W/m2. This balances the power input to the object, so 310K is the steady state temperature of the the object IN THIS CASE (and only in this case). You CANNOT assume, as you did, that this is the general case.”

    I’ll put your words into an equation

    Q(net) = 1*sigma(310^4 – 0^4) = ~523.

    “So let’s compare Case 3 and Case 2, replacing the very cold ambient of 0K in Case 3 with an increased ambient (but still cooler than the object) of 253K. What is the object’s steady state temperature in this case? It is simply the temperature at which the Q(net) from the object to ambient is 523 W/m2 so the object’s power output matches its input. So we have:

    Q(net) = 523 = 1 * sigma (Tobj^4 – 253^4)

    Tobj = ((523/sigma) + 253^4) ^ (1/4) = 340K”

    What you have written in your equations is you accept that “heat” transfers from a colder “object” – your ambient environment at 253 K – to an object you have already assigned a higher temperature of 310 K by your first proposition.

    This is not in accordance with the established laws of thermodynamics which clearly state that heat flow in a natural process is exclusively from hot to cold – we all know one can perform work to overcome this natural state just to anticipate another red herring from you.

    No matter what you say you have explicitly stated that heat transfers from a cold “object” to an already hotter object.

    This transfer of heat is explicitly stated in your equation and your text – your words not mine.

    I say for this and a whole host of other reasons you are wrong.

    • Avatar

      Ed Bo

      |

      Rosco:

      You write: “What you have written in your equations is you accept that “heat” transfers from a colder “object” – your ambient environment at 253 K – to an object you have already assigned a higher temperature of 310 K by your first proposition.”

      No, not at all! I got the same result for heat transfer from a 310K object to 253K ambient that you did: Q(net) = +291 W/m2. That is, a 291 W/m2 heat transfer FROM the 310K object TO the 253K ambient. This is completely in accordance with the 2nd Law. You just can’t handle the implications of this.

      You seem to be completely confused as to how the “warm” object’s temperature can increase when the ambient is changed from “cold” to “cool” (to use qualitative terms) even when the heat transfer is still FROM the warm object TO the cool ambient. But you have to remember the separate power source – that is what is really “heating” the object.

      A simple analogy may make it clearer. You are running water from the tap into a sink with an open drain. The water in the sink reaches a certain steady-state height at which the outflow through the drain matches the inflow from the tap.

      But now, there is a partial clog in the drain that reduces the outflow for that height of water. With the inflow from the tap unchanged, the height of the water in the sink increases. So is this clump of hair in the drain actually pushing water uphill against gravity? This is the equivalent of what you are arguing!

      No! The water is actually always moving downward, first from tap to sink, and then from sink through the drain. This is true EVEN AS THE LEVEL OF WATER IN THE SINK IS INCREASING!!!

      The same is true in your heat transfer example. The power transfer is always from high temperature to low temperature, first from the power source to the object, then from the object to ambient EVEN AS THE TEMPERATURE OF THE OBJECT IS INCREASING!!! The Q(net) from warm object to cold or cool ambient is always positive, in accordance with the 2nd Law.

      You have not even begun to deal with the problem of how the object can have 523W in and 291W out, and not increase in internal energy and temperature. Do you understand what the 1st Law says?

  • Avatar

    Ed Bo

    |

    Rosco:

    I’m glad you finally figured out your error. But it took you long enough to figure out one of your most trivial errors, and it’s just symptomatic of your profound ignorance of thermodynamics and heat transfer. I will go into depth here.

    Let’s look at your comment of June 23 at 10:21pm, where you actually try to do some quantitative analysis. You start out with:

    “Object at 37 C – call it 310 K… I’ll make an assumption the object is irradiated by a constant ~523 W/m2 (or body metabolism if you like) and it absorbs all of it hence a temperature of ~37 degrees C or ~310 K.”

    You’ve gone off the rails before you’ve even started! This is a far more profound misuse of the SB equation than your effective mixing up of temperature and flux density. You are using the SB equation exactly “backwards” here, showing that you have no clue as to the underlying meaning of the equation.

    The SB equation gives you the radiative flux OUTPUT of an object as function of its temperature and emissivity. It does NOT give you the temperature of an object (transient or steady state) as a function of its radiative FLUX input! You should have learned this early in your first thermo class.

    Even to do the simple computation of a steady state temperature requires you to find the temperature at which the constant power input is matched by the temperature-dependent outputs. This is what the 1st Law of Thermodynamics demands — it is not a difficult concept.

    So let’s take your Case 1, with an object temperature of 310K and an ambient of 293K. You correctly use the SB equation here to compute a Q(net) of 105 W/m2. But you completely fail to understand the implications of this calculations. This Q(net) is far less than the input of 523 W/m2, so the object will increase in temperature.

    The same is true in your Case 2, with an ambient of 253K. Once again you plug numbers into the SB equation and get a Q(net) of 291K. But this is still less than the 523 W/m2 input, so the internal energy of the object, along with its temperature, will increase.

    I will introduce a Case 3, with an ambient of 0K (“cold” space). Using the SB equation, we see that the Q(net) from the object of 310K to 0K ambient is 523 W/m2. This balances the power input to the object, so 310K is the steady state temperature of the the object IN THIS CASE (and only in this case). You CANNOT assume, as you did, that this is the general case.

    So let’s compare Case 3 and Case 2, replacing the very cold ambient of 0K in Case 3 with an increased ambient (but still cooler than the object) of 253K. What is the object’s steady state temperature in this case? It is simply the temperature at which the Q(net) from the object to ambient is 523 W/m2 so the object’s power output matches its input. So we have:

    Q(net) = 523 = 1 * sigma (Tobj^4 – 253^4)

    Tobj = ((523/sigma) + 253^4) ^ (1/4) = 340K

    These are very, very basic thermodynamic calculations typical of an introductory class. This is also the fundamental principle of the so-called greenhouse effect — that the object’s (earth surface) temperature is higher when it is radiating to an ambient of about ~253K than when it is radiating to an ambient of ~0K, as it would with a transparent atmosphere.

    • Avatar

      Rosco

      |

      I wrote :-
      “Object at 37 C – call it 310 K… I’ll make an assumption the object is irradiated by a constant ~523 W/m2 (or body metabolism if you like) and it absorbs all of it hence a temperature of ~37 degrees C or ~310 K.”

      How is this using the SB equation exactly backwards ?

      “Even to do the simple computation of a steady state temperature requires you to find the temperature at which the constant power input is matched by the temperature-dependent outputs.”

      I have the object at 310 K consistent with 523 W/m2 !

      You have agreed that my use has been right in each instance ! – “You correctly use the SB equation here to compute a Q(net) of 105 W/m2.”

      I have the object at 310 K emitting 532 W/m2 in each example – If you claim this is not true then you are simply not telling the truth.

      I do not have Q(net) at 523 – that is true. Nor do I believe it needs to be – the temperature of 310 K means the object is emitting 523 W/m2 – this is absolutely correct !

      Here is exactly what I wrote :-

      “Case 1 – “cool surroundings” – lets say 20 C – call it 293 K.
      Q(net) = 1.sigma(310^4 – 293^4) – ~105.

      Case 2 – “colder surroundings” – lets say minus 20 C – call it 253 K
      Q(net) = 1.sigma(310^4 – 253^4) – ~291.”

      And yes when there is nothing but empty space the Q(net) is 523 W/m2.

      But seriously I have the object at 310 K balancing the power input in each case !

      The Stefan-Boltzmann equation is P = sigmaT^4. 523 = sigma.310^4.

      An object radiates in proportion to its temperature regardless of the surroundings. It will only heat up when it receives net energy.

      You have simply switched the value of Q(net) to give the value you want and completely changed the scenario yet again.

      What happens when we bring the object into an environment of 310 K ?

      Q(net) = 1.sigma(310^4 – 310^4) = 0.

      Two objects at the same temperature exchange no “net” energy – this is a fundamental reality.

      But you claim:-

      Q(net) = 523 = 1.sigma(Tobj^4 – 310^4) and thus you claim the object must have a temperature of

      (523/sigma + 310^4)^(1/4) = ~369 K. – 369 K means the power radiated is ~1051 W/m2

      Thus 523 W/m2 completely absorbed by an object causes temperatures of

      310 K when radiating to empty space. – 310 K means the power radiated is ~523 W/m2.

      340 K when radiating to air at 253 K. – 340 K means the power radiated is ~758 W/m2

      358 K when radiating to air at 293 K. – 358 K means the power radiated is ~931 W/m2

      369 K when radiating to air at the same temperature the object was initially with no extra energy input.

      All of this occurs with a constant input of 523 W/m2 ?

      You have created energy out of nothing – doubling it when the two objects are at the same temperature despite the fact that there is no net exchange of energy.

      As I have said continually you just make stuff up.

      • Avatar

        Ed Bo

        |

        Rosco:

        You are really too much! You continually display your absolute ignorance of the most fundamental concepts of thermodynamics. You see an equation and you just plug numbers in (often in the wrong place) without any conceptual understanding of the meaning of the equation!

        You need to read the following, which I am just repeating from my last comment, very carefully, if you want any chance of getting the most basic problems correct:

        “The SB equation gives you the radiative flux OUTPUT of an object as function of its temperature and emissivity. It does NOT give you the temperature of an object (transient or steady state) as a function of its radiative FLUX input! You should have learned this early in your first thermo class.”

        You continue to use it in the second sense – temperature as a function of flux input – which is absolutely incorrect! This is what I mean by using the equation “backwards”. I defy you to find a single engineering or science thermodynamics text that uses the equation as you do.

        Now let’s look at the 1st Law energy balance analysis – I doubt you have ever done one correctly in your entire life, because you are doing them completely wrong here. Using the 1st Law for our object, we have:

        DeltaU = Sum(EnergyIn) – Sum(EnergyOut)

        where U is the internal energy of the object. Without phase changes, chemical reactions, or the like, the object temperature varies directly with the internal energy. In differential form, we have:

        dU/dt = Sum(PowerIn) – Sum(PowerOut)

        where dU/dt is the rate of change of internal energy of the object. In the steady state, dU/dt is zero, so Sum(PowerIn) must equal Sum(PowerOut).

        So let’s look at your example. To keep it simple, but without losing generality, we’ll say that your object has a surface area of 1 m2.

        Your object has a separate constant power source or 523W, and it is thermally interacting with ambients of different temperatures. We will stipulate an initial temperature in each case of 310K.

        In Case 3 of 0K ambient, we have the separate power input of 523W, a Q(gross) from the object to ambient of 523W, and a Q(gross) from ambient to the object of 0W – so Q(net) between object and ambient is 523W. Sum(PowerIn) = 523 + 0 = 523W, and Sum(PowerOut) = 523W, so dU/dt is 0 at 310K, so 310K is the steady state temperature in this case.

        In Case 2 of 253K ambient, we have the separate power input of 523W, a Q(gross) from the object to ambient of 523W, and a Q(gross) from ambient to the object of 232W – so Q(net) between object and ambient is 291W. Using gross transfers, Sum(PowerIn) = 523 + 232 = 755W, and Sum(PowerOut) = 523W, so dU/dt = 232W, and the temperature will increase.

        Using net transfers, Sum(PowerIn) = 523W, and Sum(PowerOut) = 291@, so dU/dt = 232W. It doesn’t matter whether you use gross or net transfers, AS LONG AS YOU ARE CONSISTENT (which you are not).

        Following up on Case 2, the object temperature will continue to increase as long as dU/dt is positive. When the temperature reaches 340K, we have Sum(PowerIn) = 523 + 232 = 755W and Sum(PowerOut) = 755W, so dU/dt = 0, and 340K is the steady state temperature in this case.

        You claim that I “have created energy out of nothing”, when in fact I have scrupulously used the conservation of energy in all my calculations. Your very fundamental mistake here is to confuse power and energy, a very common mistake for beginners. Watts are units of power, not energy, and we are talking about conservation of energy, not conservation of power.

        Conservation of energy (1st Law) DOES NOT constrain power flows! This is a vital point!

        Let’s use a financial analogy with “conservation of money”. I am your banker, and each week you deposit a $523 paycheck in my bank. Each week you also spend $523 from the account, so at the end of the week your bank balance is the same as that of the previous week. So far, so good.

        But now, you start getting rebates on your $523 of purchases of $232 per week, which you deposit back in the bank account. By your logic, this would be “creating money out of nothing” and this would not increase your bank balance each week. I would love to be your banker!!!

        Please, please, take a real thermodynamics course so you can stop making a complete fool of yourself!

  • Avatar

    Ed Bo

    |

    Rosco:

    Once again, let’s go back to the beginning. You are so horribly confused, you don’t even understand what you are saying or what the issues are.

    You keep saying things like: ” if Ed had the intelligence to comprehend he would realise he has absolutely agreed with my proposal – that any area of space where the minimum radiation flus is 586 W/m2 cannot be described as cold.”

    And I keep explaining to you why your proposal is completely … what’s the word I’m looking for… oh, yes: WRONG!!!! And I’ve given you specific examples, too. Here is another one: Haven’t you ever been out on a sunny winter day, where the air temperature is far below zero even though there is over 1000 W/m2 passing through the air? By your logic I should not need heavy clothing to protect me from this “can’t be considered cold” environment.

    You obviously have no concept of what temperature really is, so I guess I have to explain it to you. Thermodynamic temperature is directly proportional to the average kinetic energy of the molecules of the substance (no matter how sparse). It has NOTHING to do with the flux passing through the space unabsorbed.

    This is such a basic concept that is introduced early in a first thermodynamics course — so I realize you could not have taken and understood such a course.

    You object to my statement that: ” Rosco claimed that it was absurd to consider space “cold” for the purposes of radiative heat transfer with the earth.” So let’s review what was said:

    In the original post, Charles said: ” the Earth, being warmer than space, also radiates energy out into space.”

    You responded that: “Earth is “warmer” than space is the sort of ludicrous statement climate science advocates say all the time and it appears time and again in all sorts of references.”

    So if you don’t understand that my characterization of your claim is ABSOLUTELY CORRECT, then your only defense is that you are simply too confused to understand what you are saying.

    You say: “The lowest temperatures measured appear to be of the order of ~100 K – on the Moon and Mercury. There may be some crater areas where the Sun never shines that may be ~95 K.”

    Wrong! Polar lunar temperatures during the “winter” for that hemisphere get down to 35K, so they must be radiating to an ambient of lower temperature. Why not all the way down to 3K? Because at 35K, the highest thermal radiation possible is 0.08 W/m2, which is virtually nothing.

    When I pressed you to say what temperature you would use for space in radiative calculations, you said 0K, which last I checked is pretty damn cold!

    You say: ” I have always understood the point about no radiation from empty space but it is irrelevant because the solar radiation is always present.”

    This just shows your absolute confusion on the subject. The [virtually] “no radiation from empty space” is absolutely relevant and necessary to understand to do the correct heat transfer calculations. Nobody is claiming the sun is not shining, and when that radiation is absorbed, it increases the energy of the absorbing body.

    I could go on and on, but I will just respond again to your final line: ” the inner solar system space is not cold by any definition.” As I showed above, it is cold by the THERMODYNAMIC definition.

    We determine a lot of temperatures by measuring the frequency and magnitude of their spectrums. It’s how we know the surface of the sun is at 5770K; it’s how we know the equatorial lunar temperature at dawn is at 90K; it’s how we know the polar lunar winter temperature is at 35K. By the same technique, we calculate the (effective, if you want) temperature of space in all other directions as 3K. But it is this temperature that we must use for proper heat transfer calculations (at least you realize that 0K is close enough). At most you have an irrelevant semantic quibble, but even there you are wrong!

    Oh, and I have to cite this equation of yours for solar/earth radiative exchange:

    Hence Q(net) = e.sigma.(Te^4 – 1361).

    I laughed out loud when I saw this one, as it shows you have no idea what these equations mean — you just plug numbers into an equation you have seen somewhere!

    • Avatar

      Rosco

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      Are you for real ?

      “Haven’t you ever been out on a sunny winter day, where the air temperature is far below zero even though there is over 1000 W/m2 passing through the air? By your logic I should not need heavy clothing to protect me from this “can’t be considered cold” environment.”

      Of course it is a cold environment – the temperature of the air is below zero. And on a windy day where forced convection is induced it will feel even colder.

      Typical bullshit red herring from you – what does this have to do with space ?

      And no – I’ve never been out “on a sunny winter day where the air temperature is far below zero…” – air temperature rarely drops below 10 degrees C where I live.

      If indeed there is 1000 W/m2 normal to the surface then it must be reflected away or not absorbed for some reason. Sorry I can’t accurately describe your hypothetical situation with the limited information provided by you.

      You state – “Polar lunar temperatures during the “winter” for that hemisphere get down to 35K, so they must be radiating to an ambient of lower temperature”.

      I will search for the reference and check it out but I do not disbelieve you on the 35 K – I just haven’t seen it.

      However the only reason this occurs is because for significant amounts of time these areas are not irradiated by the Sun at all – possibly never. This is indisputable fact.

      In the volume of the inner solar system such areas are extremely rare indeed and this is simply what I have been saying all along !

      This statement is as accurate as anything you have said and absolutely supports my claim that any area of the inner solar system where the solar radiation is present should not be described as cold.

      At any point in space any object placed there will heat up because of the solar radiation – this does not happen in a cold “environment”.

      If these craters are irradiated they would not be as cold.

      Surely any object radiates in proportion to its temperature as expressed by the SB equation – P = e.sigma.T^4 per unit area regardless of the external environment. This is what they derived empirically.

      The Earth does not have to be warmer than space to radiate – it will radiate no matter what.

      It will only heat up if the radiation in represents a “hotter radiation”. It will radiate towards the Sun as well as all other directions. As everyone knows the majority of the Sun facing hemisphere is being heated whilst the other is not. (If you take 51% of the solar constant incident on the Earth’s surfaces as realistic then the area of the hemisphere irradiated by more than 239 W/m2 extends from ~70 N to ~ 70 S.)

      You will probably mock this “hotter radiation” description but it is simply shorthand and spectral features of radiation are important and the 1361 W/m2 solar constant is entirely different in character to the radiation from some object with near unity emissivity at ~394 K.

      If you remember correctly YOU introduced the cooling radiation fins on satellites and the vacuum chambers chilled by liquid nitrogen into the conversation.

      I simply said that satellites must be designed to cope with being able to radiate accumulated heat away as they are either permanently or temporarily exposed to the solar radiation.

      It is more difficult to shed heat in space where conduction or convection are not present.

      YOU claimed the cold of space is the prime design factor – despite the obvious fact that something radiating to something as cold as you insist it is probably doesn’t need ANY radiation fins at all !

      The impact of the solar radiation is the crucial design parameter for near Earth orbit satellites.

      I should have realised that I need to write a thesis and explicitly state everything when I post a comment because you twist things into something to attack even if it is actually realistic.

      ACTUALLY – most of the time you simply make up lies about what the other person says.

      Here is another quote from you –

      “If there is no other power source, yes that is true. But if there is a separate power source – the sun for the earth, metabolic processes for the human body — the replacement of cool surroundings for colder surroundings reduce the Q(net) between the object and its surroundings, resulting in an increase in temperature of the object until the Q(net) to the surroundings matches the separate Q(in). For the life of me, I can’t understand why people like you have so much trouble with this concept!”

      So in accordance with your instructions I wrote:-

      Q(net) = e.sigma.(Te^4 – 1361) –

      I’m uncertain about your real problem with this – if the atmosphere is transparent, cloud free then albedo is low – the 1361 may be a bit high but I wrote it because you have also been banging on about satellites.

      1361 is Q(in). Even if you find this laughable it is a correct statement as you wrote and represents heating because the equation is negative.

      What would you suggest I put instead of the accepted solar constant ?

      Satellites measure radiation to space and they are designed to account for the solar radiation.

      The cosmic microwave background radiation is ignored.

      As for “the replacement of cool surroundings for colder surroundings reduce the Q(net) between the object and its surroundings, resulting in an increase in temperature of the object until the Q(net) to the surroundings matches the separate Q(in).”

      How about you write what you are stating in plain simple English with equations and numbers to show what you mean to us plebs ?

      What is Q(in) ?

      And what does introducing an atmosphere into the discussion have to do with space ?

    • Avatar

      Rosco

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      OK – I’ll concede Q(net) = e.sigma.(Te^4 – 1361) is not right nor is it what I meant – obviously I should have removed the brackets. This is the type of mistake people like you seek out.

      What I meant was Q(net) = e.sigma.Te^4 – 1361 which is a valid statement for the net flux as a result of Te minus the solar constant. This is negative indicating the Sun heats the Earth.

      You are patronizing in your manner thinking you are clever for enlightening us with stuff we already knew – you have not added one new thing to the discussion.

      Your Polar lunar example absolutely proves my original point – if such areas were irradiated by the Sun they would not be cold and most of the inner solar system is subject to the solar radiation !

      I still think you’re an idiot incapable of having a discussion without introducing loads of red herrings or outright lies.

      I don’t give a damn about your opinion or pontifications. I think I’ll take Jerry’s advice and simply ignore your BS from now – you have offered absolutely zero new to any discussion on this site.

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    Rosco

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    Jerry – I never wrote this –

    “But there are no objects here, just empty space! The solar energy just passes through this space on the way to the interstellar regions. If there was anything significant there to absorb this radiation, we could measure what it re-radiates to earth. BUT WE DON’T!!!”

    So what ?

    In fact if Ed had the intelligence to comprehend he would realise he has absolutely agreed with my proposal – that any area of space where the minimum radiation flus is 586 W/m2 cannot be described as cold.

    This is Ed Bo’s deliberate lying about what I said.

    I will say it explicitly – Ed Bo is an outright liar !

    Because that is the only way to describe someone who claims you say different things you never said time and again.

    Here is another way he just makes things up to discredit others in response to your comment :-

    “Jerry: Rosco claimed that it was absurd to consider space “cold” for the purposes of radiative heat transfer with the earth. I was pointing out that measurements showed that space radiates towards the earth exactly as a blackbody of ~3K would.”

    I have never mentioned almost all of the things he says I said.

    I never said anything about the radiative heat transfer with the Earth – Ed Bo is a liar !

    All I have ever said is it is ABSURD BEYOND BELIEF to claim space with a minimum radiation flux of 586 W/m2 can be considered cold.

    I always said the solar radiation is directional and just because it is not intercepting any object does not mean it is not always there.

    It IS absurd and Ed never admits this simple point but continually shifts the goal posts.

    He HAS EXPLICITLY claimed it is appropriate to consider a vacuum with a continuous radiation flux of 586 W/m2 can be considered cold and that is just plain bullshit.

    Here’s another example of his lies:-

    “You say that you “never said the the directional Solar radiation that isn’t intercepted by the Earth by has any impact on our energy budget” but by insisting that it changes the temperature of space around earth, which would affect the earth’s energy budget.”

    “by insisting that it changes the temperature of space around earth” !

    Find one statement that I wrote where I claim that the solar radiation “changes the temperature of space around earth”.

    Ed Bo is an outright liar !

    Here is another example:-

    “Regardless, detailed measurements by astronomers (nothing to do with climate science) show that the radiation received by earth from space is equivalent to that of a perfect blackbody (as close as they can measure) at 2.725K — within +/-0.001K in all directions.”

    “all directions” ?

    This is absolute bullshit !

    The reference to the NASA mission to explore and measure the “the cosmic microwave background” – http://science.nasa.gov/missions/cobe/ shows Ed is deliberately misleading with the statement he made as quoted.

    The argument has NEVER been about the radiation “received” by Earth from space and even that is certainly absolutely NOT “equivalent to that of a perfect blackbody (as close as they can measure) at 2.725K — within +/-0.001K in all directions.”

    The solar radiation impacts the Earth and heats it up as anyone knows – as it would heat up anything anywhere in Ed Bo’s fanciful cold space of the inner solar system.

    Seriously, who accepts that an area of space where the minimum radiation flux is 586 W/m2 and significantly higher than that over the vast majority of the inner solar system is accurately described as cold ?

    All the rest of Ed Bo’s ravings fail to challenge that simple reality by deception and lies about what I said.

    I set out a scenario and some equations in answer to another of his claims.

    Ed needs to do the same to show how he arrives at his conclusion that simply moving an object from a cold area to a less cold area causes it to heat up.

    But, returning to the simple proposition I said which lead to all of this pseudoscience on Ed Bo’s part based on lies and deception –

    Seriously, who accepts that an area of space where the minimum radiation flux is 586 W/m2 and significantly higher than that over the vast majority of the inner solar system is accurately described as cold ?

    • Avatar

      Jerry L Krause

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      Hi Rosco,

      I believe the solution to Ed Bo is to ignore him. As I wrote to Charles: “I have been a fool for allowing Ed Bo distract me from considering the real system to which you drew to my and others’ attentions.”

      Have a good day, Jerry

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        Rosco

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        Jerry I am always prepared to have a reasonable discussion with anyone and make my point.

        Unfortunately I have learnt from experience that this is impossible when people involved in the discussion deliberately lie about what one has said and then use this invented statement for ridicule.

        I commenced with a simple statement that space has no physical properties and hence applying terms such as hot or cold is misleading. I supported that statement with the observation that the inner solar system for example has a continuous radiation flux of at least 586 W/m2.

        But some people see such a simple statement as a challenge to their perceived knowledge and ability and start using lies and distortion by claiming things one never said at all.

        I would like Ed Bo to make his case with equations as I did.

        But he won’t do this – rather he makes confusing undefined statements combined with belittling arguments he ascribes to his “opponent” – the vast majority of these statements are entirely his.

        No-one can have a discussion in those circumstances but I welcome his putting his case succinctly with clear definition so we could all understand whether he can enlighten us with something we do not understand and discuss the proposition honestly.

        Enough on this for now.

    • Avatar

      Jerry L Krause

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      Hi Rosco,

      Because Pat Obar commented: “hhhh” to Carl Brehmer’s recent article, I researched who Pat Obar might be. And a search led me to: http://principia-scientific.org/arctic-explorers-or-buccaneers.html/#comment-11224. You responded to this article and then Hans Schreuder responded to it and to a comment you had made. Then, according to Hans, Jim McGinn (using an alias) shows up and begins to criticize Hans’ general understanding. Hans replies with an empirical case of clothes drying to support his understanding. To which Jim responds: “Evaporation involves miniature clumps/droplets of H2O. It does not involve steam or gaseous H2O.”

      Rosco, I know I have just reviewed that I had written to Charles: “I have been a fool for allowing Ed Bo distract me from considering the real system to which you drew to my and others’ attentions.” But I consider anyone who attempts to have a ‘conversation or debate’ with anyone who writes: “Evaporation involves miniature clumps/droplets of H2O. It does not involve steam or gaseous H2O.”, is a fool. And evidently, Hans considered there is little future (he does make one short additional comment) in having a dialogue with Jim.

      But knowing what has been written, Pat Obar engages Jim. Was Pat being a fool? I do not know. For I can see some interesting issues being discussed which I had not read about being considered before. For while the words condensation nuclei were never used, I could see the “colloids” could be used a different term for condensation nuclei. And I believe we need to consciously consider what we know about these critical components of the earth’s atmosphere.

      But I still do not think it profitable to have such a discussion (dialogue) with such individuals such as Ed Bo and Jim McGinn (if this is his actual name) who attempt to rewrite our most basic knowledge of matter. We need to limit our efforts to those who at least do have a fundamental understanding of basic modern science; even though we have still have differences of understanding about issues which have not yet been considered for near as long as our basic understanding of how clothes dry.

      R. C. Sutcliffe, a somewhat esteemed meteorologist, was invite by Norton & Company to write a book for their Advancement of Science Series and his book was published in 1966. In his preface he began: “This is not a textbook on meteorology, neither a general introduction nor a formal course, but it has a serious purpose and that is to explain to the general reader what it is that meteorologists are doing and trying to do.” Then in his second chapter he reviewed: “When it became firmly established from observations on mountains and in manned and free balloons that the air became steadily colder as the altitude increased, scientists were very ready to generalize and to assume that the cooling went on indefinitely to the limit of the atmosphere. This was the general belief until in 1899 the Frenchman Teisserenc de Bort, announced to an astonished and even incredulous world that his sounding balloons had reached heights above which the temperature decreased no further.”

      Hence, the historical truth is that meteorology is an infant science. Jet streams became an established fact during WWII. Atmospheric soundings at scattered land locations became somewhat regular shortly before WWII. And I know of no such sounding being routinely made in the midst of oceans. Yes, they are made from islands in the midst of oceans without an obvious recognition that even small islands in the midst of tropical oceans have the potential to create localized weather because the temperatures of land surfaces diurnally oscillate over a significantly greater range than the surface temperatures of the oceans which surround them.

      But what astounds me is how many continue to write as if the temperature at the base of the atmosphere always cools with increasing altitude. Balloon soundings made shortly before sunrise, disclose, given cloudless nighttime skies, the formation of a temperature inversions. And at the poles and at lower latitudes, during their dark winter season, I believe temperature inversions are an almost permanent feature of their atmospheres. I admit I do not know how meteorologists, forecasters, and atmospheric modelers are dealing with these observed temperature inversions about which I seldom read.

      I write this to you simply to get your take on what I have written.

      Have a good day, Jerry

  • Avatar

    Rosco

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    Jerry – are you serious ?

    “We must recognize the position in which Ed Bo finds himself. He is an engineering professor ” ???

    Surely any professor cannot be so confused to make the claims this guy has.

    He suffers from JMSU syndrome when someone challenges him (JMSU is just make stuff up by the way).

    No-one mentioned Earth’s energy budget except him and he then uses this straw man argument to attack his opponents ?

    I hope you are wrong about him being an Engineering professor – I have engineering qualifications myself and none of the professors I knew ever talked the gobbledygook this guy has sprouted.

    How can an educated person argue that the inner solar system space has the equivalent of the “cosmic background radiation” when the rest of the world knows the solar radiation is continually present ?

    Heaven help us !

    • Avatar

      Ed Bo

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      Rosco:

      Let’s recap. You stated: “Earth is ‘warmer’ than space is the sort of ludicrous statement climate science advocates say all the time and it appears time and again in all sorts of references.”

      I pointed out that the measured radiation received by the earth from all angles of space except for the half-degree diameter subtended by the sun is the same as would be generated by an object at about 3K.

      Now you insist that because solar radiation is passing through this space, even though it is not directed at the earth, affects the effective temperature of this space, and therefore the energy transfer from the earth.

      You say: “Effectively any non reflective object randomly placed inside this sphere will absorb this radiation and heat up”.

      But there are no objects here, just empty space! The solar energy just passes through this space on the way to the interstellar regions. If there was anything significant there to absorb this radiation, we could measure what it re-radiates to earth. BUT WE DON’T!!!

      You cite lunar temperatures, which even at the equator get below 100K in the lunar night and are still decreasing. Do you at least agree that the effective temperature of the space it is radiating to must be below 100K?

      You wouldn’t answer my question as to what temperature should be used for space in radiative heat transfer calculations between earth and space. Since you claim to be an engineer, I’ll make it a very basic engineering problem:

      You are working on the thermal system of a satellite for low earth orbit. The satellite has a long flat fin to radiate away the energy generated by its internal power supply. It is black with virtually 100% absorptivity/emissivity. When it is on the “night” side of the earth, one side of the fin is directly facing the earth, and the other side is facing the outer solar system (of which only a tiny, tiny fraction is in the earth’s shadow). The fin is at a temperature of 400K.

      In the classic heat transfer equation:

      Q (W/m2) = sigma * (T1^4 – T2^4)

      where T1 is the temperature of the fin (400K here), what would you use for T2 for the side facing the earth? What would you use for T2 for the side of the fin facing away from the earth? What is the heat transfer flux density from each side?

      This is a trivial problem for an introductory undergraduate heat transfer course. Even a mediocre student would be expected to answer it in a couple of minutes. Can you?

      • Avatar

        Jerry L Krause

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        Hi Rosco and Ed Bo,

        Why both of you? Because your and my comments produces thoughts which each of us might not have considered but for the comments of others. Now as I come to submit my composition I see I am confused. For I wrote the following assuming what immediately follows was directly written by Rosco. But now I see what I have copied is from Ed’s comment. Although Ed has stated that he is reviewing what you, Rosco, has stated, I am not certain where Rosco’s comments end and Ed’s begin. With this qualifier, I submit what I had composedl

        Rosco you wrote: “But there are no objects here, just empty space! The solar energy just passes through this space on the way to the interstellar regions. If there was anything significant there to absorb this radiation, we could measure what it re-radiates to earth. BUT WE DON’T!!!”

        This comment caused me to reflect upon what has been seen with the Hubble telescope that (images) is so far from the earth (telescope). What we (the telescope) see is radiation that does interact with matter way out there but then cannot interact with matter in its long trip to the telescope. So, Rosco, you make an excellent point and it will be interesting how Ed Bo reacts to it.

        But your comment also caused to think of something closer to earth—comet tails. Newton wrote extensively about these tails in the 3rd Book of The Principia. Again, they support your consideration that space is quite empty of matter. For as soon as there is matter we can see it presence. Now I do not agree with you that the matter of the comet tail is absorbing radiation, as this is what you seem to have implied. I agree with Feynman that the interaction of radiation with minute particles of matter (but much larger than the atoms and molecules of our atmosphere) of which I believe we all understand the comet tails are composed, is not an absorption phenomenon but a scattering phenomenon. For Feynman has explained how minute particles of matter, much larger than atoms and molecules, are capable of intensely scattering the solar radiation. (The Feynman Lectures on Physics, Vol I, pp 32-8,9) Yes, I have given this reference before and before and as yet no one has responded as to what Feynman taught.

        And I remember that you have previously replied to Ed that you will trust Feynman, as I do. This even though I am aware that he (Feynman) made a very serious mistake during his first lecture to his class at Caltech. But this does not mean he must always be wrong. As Anthony Bright-Paul has written: “All human beings have faults.”

        Have a good day, Jerry

  • Avatar

    Rosco

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    I probably had too much coffee yesterday !

    Simply put my position is this.

    Imagine a sphere with a radius of 228 million kilometres – roughly Mars orbit.

    At the centre is the Sun.

    Every bit of space inside that sphere has a continuous radiation from the Sun with at least 586 W/m2 value – significantly larger values as one approaches the Sun.

    The only portions of this space not subject to this continuous radiation are the areas where the 3 planets and the moon block the radiation.

    These areas are tiny compared to the total volume of this sphere – less than 1/300 billion th of the total volume.

    Effectively any non reflective object randomly placed inside this sphere will absorb this radiation and heat up – with the above mentioned exception noted.

    We can make an estimate of the power such a randomly placed object will be irradiated by (happy about my use of “irradiated” Ed Bo ?).

    We can do this by considering the Sun as a point source and applying the inverse square law and arrive at figures for the various orbits of the planets. NASA provides this information.

    This is standard science as taught universally.

    Accordingly only an idiot could ignore this simple analysis and claim the inner solar system space can be considered cold because they read something about the “cosmic microwave background”.

    Only Ed Bo mentioned Earth’s energy budget. It is irrelevant to the FACT that the Solar radiation is always present as is the directionality of the radiation.

    I was simply trying to make the point that there is absolutely space in the inner solar system where the only radiation is the “cosmic microwave background”.

    Clearly there is no space in the inner solar system that can be considered cold or that could possibly be considered to “act” “like a blackbody at 2.725 K.”

    The lowest temperatures found in the inner solar system confirm this reality and also the nonsense about instantaneous radiative equilibrium.

    The lowest temperatures measured appear to be of the order of ~100 K – on the Moon and Mercury. There may be some crater areas where the Sun never shines that may be ~95 K.

    That is still a huge difference to the 2.725 K Ed Bo talks about existing throughout the whole of the inner solar system space.

    The Moon’s night is about a fortnight on Earth – 14 + days- whilst Mercury’s night is roughly 3 months on Earth – ~88 days.

    Exposed to effectively zero radiation from space – I’m not arguing that – neither cools to anywhere near the levels claimed as the “temperature” of space despite all of the bull-dust associated with the claim that radiative cooling occurs “at the speed of light”.

    But just because there exists a small volume of space where the planets block the solar radiation does not mean that the 3 hundred billion times this area can be considered cold.

    That is simply stupid beyond belief – and Ed YES I have always understood the point about no radiation from empty space but it is irrelevant because the solar radiation is always present.

    Your arguments are spurious nonsense – the inner solar system space is not cold by any definition.

  • Avatar

    Jerry L Krause

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    Hi Charles,

    I have been a fool for allowing Ed Bo distract me from considering the real system to which you drew to my and others’ attentions.

    I am sure you know but just didn’t state that the ‘atmosphere’ of the mine, because of its temperature gradient from it bottom to the surface, is absolutely unstable. Yet, vertical convection does not occur to ‘correct’ this unstable condition.

    You stated, correctly I believe, that if the surface atmosphere, at its surface temperature, could adiabatically subside to the bottom of the mine its temperature would only be warmed to about 100oC. The mining engineer could cause this to happen by adding an insulated pipe through which this surface atmosphere could subside without being heated by the warmer environment of the insulted pipe. For the reason that the unstable condition exists is there is no cooler, more dense, atmosphere to lift the warmer ‘atmosphere’ from the mine shaft.

    It is obvious that the temperature of the atmosphere in the mine shaft is due to the temperature of its environment (the walls of the mine shaft). And this is because there is no natural circulation of the atmosphere in the mine shaft. We might conclude that this mine shaft has created an artificial (man-created) systems which would never occur ‘naturally’.

    However, C. Donald Ahrens, in his popular meteorology textbook—Meteorology Today 3rd Ed.—wrote: “As the sun rises in the morning, sunlight warms the ground, and the ground warms the air in contact with it by conduction. However, air is such a poor heat conductor that this process only takes place within a few centimeters of the ground. As the sun rises higher in the sky, the air in contact with the ground becomes even warmer, and there exists a thermal boundary separating the hot surface air from the slightly cooler air above. Given their random motion, some air molecules will cross this boundary. The “hot” molecules below bring greater kinetic energy to the cooler air; the “cool” molecules above bring a deficit of energy to the hot, surface air.” Ahrens has just described the thermal conduction process by air, but I am not sure that he knows this. Which he has ignored, which you (Charles) do not, is that the “hot” molecules from below lose kinetic energy as they move upward and gain potential energy. And that the “cool” molecules from above gain kinetic as they move downward as they lose some potential energy as they change their position relative to the earth’s center of mass.

    But Ahrens continued: “However, on a windless day, this form of heat exchange is slow, and a substantial temperature difference usually exists just above the ground. This explains why joggers on a clear, windless, summer afternoon may experience air temperatures of over 50oC (122oF) at their feet and only 32oC (90oF) at their waist.”

    And he continued: “Near the surface, convection begins, and rising air bubbles (thermals) help redistribute heat. In calm weather, these thermals are small and do not effectively mix the air near the surface. Thus, large vertical temperature gradients are able to exist.”

    Ahrens is doing a tremendous job of helping the reader to image a certain, real, atmospheric condition that naturally occurs. However, he leaves out the reason of how (why) the convection begins and therefore why “these thermals are small”. He does not describe what convection is (that is at least at this point). Convection is the organized movement of the air molecules in a certain direction instead of their purely random motion when the “atmosphere” is calm. Consider the fact we cannot ‘feel’ these molecule commonly striking our skin at 600 or 700 miles per hour when there is no convection (calm atmosphere), but our skin can detect the presence of a slight draft of a mile or two per hour. Work must done to create convection. And we know it is a natural heat engine that does the work necessary to cause atmospheric convection. And we know that a heat engine, natural or artificial, requires there be a temperature difference.

    If the surface upon which the jogger ran were perfectly homogeneous, it seems unlikely its surface temperature would locally become different as the surface is heated by the solar radiation. So no temperature difference, no convection. However, if the surface is heterogeneous in one way or another, surface temperature differences can form and small, localized, convection can begin as described by Ahrens.

    But I have by passed the purpose I began quoting Ahrens. Which was the incidental observation that “air temperatures of over 50oC (122oF) at their feet and only 32oC (90oF) at their waist” naturally existed. We need to consider the consequences of these sometimes natural conditions.
    However, that would be again going away from the system which Charles brought to our attention. So, Charles, to what temperature were mining engineers able to cool the atmosphere at the bottom if the mine so miners could do a shift’s work at the bottom of the mine. And it would be interesting to learn how they did this.

    Have a good day, Jerry

  • Avatar

    Jerry L Krause

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    Hi Ed Bo,

    Practical—of pertaining to, manifested in practice or action—opposed to theoretical.

    Newton in his preface wrote: “The ancients considered mechanics in a twofold respect; as rational, which proceeds accurately by demonstration; and practical. To practical mechanics all the manual arts belong, from which mechanics took its name. But as artificers do not work with perfect accuracy, it comes to pass that mechanics is so distinguished from geometry, that what is perfectly accurate is called geometrical; what is less so, is called mechanical.”

    I live in a practical mechanical world in which I cannot measure with perfect accuracy, as I have already stated. I have now read a bit about CMB: “The anisotropy of the cosmic microwave background (CMB) consists of the small temperature fluctuations in the blackbody radiation left over from the Big Bang. The average temperature of this radiation is 2.725 K as measured by the FIRAS instrument on the COBE satellite. Without any contrast enhancement the CMB sky looks like the upper left panel of the figure below. But there are small temperature fluctuations superimposed on this average. One pattern is a plus or minus 0.00335 K variation with one hot pole and one cold pole: a dipole pattern. A velocity of the observer with respect to the Universe produces a dipole pattern with dT/T = v/c by the Doppler shift. The observed dipole indicates that the Solar System is moving at 368+/-2 km/sec relative to the observable Universe in the direction galactic longitude l=263.85o and latitude b=48.25o with an uncertainty slightly smaller than 0.1o. If we subtract the average temperature and expand the contrast by a factor of 400, we get the upper right panel below. This shows the dipole pattern and the emission from the Milky Way which dominates the red color in the picture, which represents the longest wavelength data. After the average temperature and the dipole pattern are removed, there are intrinsic fluctuations in the CMB which can be seen faintly away from the Milky Way in the lower left panel below, which has constrast enhanced by 2000X. Finally we can combine the multiple frequencies in a way that eliminates the Milky Way, giving the CMB map in the lower right with a 30,000X contrast enhancement. (http://www.astro.ucla.edu/~wright/CMB-DT.html)

    “The anisotropy of the cosmic microwave background (CMB) consists of the small temperature fluctuations in the blackbody radiation left over from the Big Bang.” It might surprise you Ed, but I claim to be able to read. But I have a problem with this first sentence. I have to question: Is it the small temperature fluctuations which are left over from the Big Bang? Or is it the blackbody radiation that is left over from the Big Bang? And then I read: “If we subtract the average temperature … .” Which was 2.725 K leaving a temperature of 0.000 K, if my math is correct. However, we are still seeing emission from the Milky Way. So it seems the ‘radiation’ temperature?? cannot actually be 0.000 K unless it is 0.00499 K or less and it is being rounded to 0.000 K.

    Now I would like to have you practically explain how a temperature variation of 0.00335 K is so terribly critical when the issue can be a temperature difference of about 33 K.

    So if you cannot be practical, please have conversations with your theoretical astronomers about topics which interest you. For it is not polite to criticize handicapped people who have to measure temperatures of systems where the temperatures can fluctuate a degree or so within say 15 minutes, to be conservative.

    Have a good day, Jerry

    • Avatar

      Ed Bo

      |

      Jerry: Rosco claimed that it was absurd to consider space “cold” for the purposes of radiative heat transfer with the earth. I was pointing out that measurements showed that space radiates towards the earth exactly as a blackbody of ~3K would.

      The only reason I quoted the +/-0.001K uncertainty was to show how precisely this is known.

  • Avatar

    Rosco

    |

    Ed – you seem determined to misinterpret what is said and then you resort to insult.

    The space between the Sun and Mars is always subject to a continuous radiative power which varies from a huge 69 million W/m2 to ~586 W/m2 at Mars.

    Apparently you don’t believe this which makes you amazingly gullible !

    To call that cold is simply absurd beyond belief !

    I never said the the directional Solar radiation that isn’t intercepted by the Earth by has any impact on our energy budget – that is YOUR absurdly irrelevant statement not mine !

    It doesn’t change the FACT that the Solar radiation is always there at the power levels quoted by NASA anyway – your arguments in support of your absurd hypothesis are wrong.

    I also have read about the measurement of the “cosmic microwave background” and I have no reason to challenge the story but it does not change the reality that the Solar radiation is there and only a fool could claim space with a continuous power of the order of 1361 W/m2 passing through it is cold.

    You keep making this assertion and frankly it is a lie – I have never said what you say I have !

    Your statements –

    “You are in effect arguing that if I shine a laser beam away from you, it can still burn you. It’s just ridiculous!”; and,

    “YOU emphasize that space is essentially empty, but yet you insist that this nothingness is “irradiated”. That’s just meaningless! That radiation is just passing through your empty space, and it does not do a U-turn and head back to the earth.” are puerile beyond belief !

    I used the word “irradiated” as a form of shorthand – BUT any object in the location would be irradiated.

    Your puerile statements do not make your argument right.

    You are typical of someone who has little intelligence – resort to making stupid arguments which you blame on someone who NEVER said them because you are totally incapable of sustaining a credible intelligible argument.

    Why not just admit you are wrong ? I would if I were wrong.

    How anyone can be stupid enough to claim an area of space subject to radiation of at least 586 W/m2 has any characteristics of a “blackbody” at 2.725 K is simply astounding !

    Let me make it plain in language even a fool can understand – there exists an area of space at Mars’ orbit where a rectangular plane object normal to the Sun will be irradiated by 586 W/m2.

    That is reality and no matter what BS argument you make about about the “cosmic microwave background” changes that reality.

    How can you ignore reality ?

    You also say “Rosco’s fundamental mistake in asserting the “space has no temperature” mantra is not understanding the difference in mechanisms between conductive/convective heat transfer and radiative.”

    You ignorant insulting simpleton !

    You showed that you do not understand the correct method of applying the Stefan-Boltzmann equation when you tried to be you were clever in a previous posting.

    All you did was demonstrate that you have little understanding and you are excelling in that endeavour again.

    What level of radiation is there in the portion of space that the Earth will occupy in 2 months time ?

    Is it 1361 W/m2 according to NASA or ~3microwatts ?

    The answer is it is ALWAYS 1361 W/m2 – who gives a damn about its direction ?

    Any object intercepting that radiation will heat up just as Skylab was uninhabitable when its heat shields were damage during blast off and remained so until a shielding mechanism allowed astronauts to enter to repair the damage.

    To call that cold is absurd beyond belief ! I know I’ve said that before but I just can’t get over how absurd your claims are !

    It is always there and any vacuum that has 1361 W/m2 of radiation cannot be considered cold – except by idiots who think that because they read something about the “cosmic microwave background” makes them clever.

    Seriously – are you so stupid you don’t realise electromagnetic radiation is always a vector quantity and thus has direction and magnitude ?

    It has to be – it is the very definition of something “emitted” !

    Your claim that the inner Solar system space is cold because there is only the “cosmic microwave background” except for the half a degree by half a degree from the Sun simply demonstrates you are completely incompetent and lack comprehension of even the simplest statements.

    I get what you are saying about the “cosmic microwave background” – it is just totally irrelevant and your assertions that this means one can consider space cold is incomprehensible.

    I suppose you also believe objects only radiate towards cooler regions ?

    Every bit of the inner solar system space has powerful Solar radiation ALWAYS present and only someone incapable of even basic comprehension could not appreciate that to claim this represents an area of space where only the “cosmic microwave background” is relevant and hence conclude it is cold is simply mind boggling in its stupidity.

    Stop making up BS statements you think others have made.

    • Avatar

      Ed Bo

      |

      Rosco:

      In all my years of teaching undergraduate and graduate engineering students, I have (fortunately) never had a student as profoundly confused as you.

      I use lasers to melt steel. The incredibly high flux density of the laser beam does not superheat the air it passes through, because the air is transparent to the laser light.

      You say that you “never said the the directional Solar radiation that isn’t intercepted by the Earth by has any impact on our energy budget” but by insisting that it changes the temperature of space around earth, which would affect the earth’s energy budget.

      I continually point out that ACTUAL MEASUREMENTS of radiation hitting the earth show that this radiation is equivalent to that produced by a blackbody at about 3K, and that this has less than one-billionth the intensity of the 1361 W/m2 you claim (yes, you do) is important because it is passing through this space unimpeded.

      I am the one properly considering the directionality of radiation, not you!

      So Rosco, in the standard radiative heat transfer equation:

      Q = epsilon * sigma * (T1^4 – T2^4)

      applied to earth’s transfer to space (let’s limit it to the nighttime sky, but not just in the earth’s shadow), with T1 to be the earth’s effect radiating temperature, and T2 to be the effective radiating temperature of space, what value would you use for T2?

      Space agencies have giant vacuum chambers to test their spacecraft before launch. These chambers have “shrouds” that can be supercooled with liquid nitrogen and sometimes helium to get their temperature below 100K and even down to about 20K. These cost millions of dollars. Why do they bother?

      • Avatar

        Rosco

        |

        Ed – I never said the solar radiation changes the temperature of the space around Earth. You just make stuff up because you seem to be incapable of understanding a simple concept – the Sun emits radiation over 4pi steradians and about ~228 million kilometres away from the Sun that radiation has a power of ~586 W/m2.

        To state as you do this area of space has only the cosmic background radiation passing through it is simply stupid beyond belief.

        I have never encountered anyone as confused or unable to understand a simple argument as you !

        As for your BS about the measurement of the cosmic background radiation you are also completely wrong on that as well.

        From http://science.nasa.gov/missions/cobe/

        “The experiment module contained the instruments and a dewar filled with 650 liters of 1.6 K liquid helium, with a conical Sun shade. The base module contained the attitude control, communications and power systems. The satellite rotated at 1 rpm about the axis of symmetry to control systematic errors in the anisotropy measurements and to allow observations of the zodiacal light at various solar elongation angles. The orientation of the spin axis was maintained anti-Earth and at 94 degrees to the Sun-Earth line. The operational orbit was dawn-dusk Sun-synchronous so that the Sun was always to the side and thus was shielded from the instruments. With this orbit and spin-axis orientation, the instruments performed a complete scan of the celestial sphere every six months.”

        As is obvious to anyone reading this NASA acknowledge the fact that the inner solar system space is NOT a place where the radiation level is equivalent to the cosmic background radiation.

        Note “conical Sun shade” and “The orientation of the spin axis was maintained anti-Earth and at 94 degrees to the Sun-Earth line.”

        And most tellingly “The operational orbit was dawn-dusk Sun-synchronous so that the Sun was always to the side and thus was shielded from the instruments”

        The solar radiation is there no matter what you claim as is obviously acknowledged by NASA so what is your point in claiming space is cold ?

        As to your question

        “Q = epsilon * sigma * (T1^4 – T2^4)

        applied to earth’s transfer to space (let’s limit it to the nighttime sky, but not just in the earth’s shadow), with T1 to be the earth’s effect radiating temperature, and T2 to be the effective radiating temperature of space, what value would you use for T2?”

        I would probably say T2 was zero ! However this in no way implies that that the space surrounding Earth is cold – the radiation aimed at Earth from space is low and hence it simply just doesn’t apply in the equation – except for the solar radiation of course. Using ~3 microwatts in the calculation is simply not relevant.

        If we have 2 objects radiating against each other with similar properties and there is no other source of radiation then we have

        Q(net) = e.sigma(T1^4 – T2^4). The hotter one will cool and the cooler one will warm until both reach T at which time Q(net) becomes zero.

        Everything radiates energy with a power proportional to the 4th power of its temperature – you’d agree with that surely ?

        The temperature of the surroundings do not reduce the radiation loss without changing the object’s temperature – you’d agree with that surely ?

        Cooler surroundings cannot make the hotter object increase in temperature because the value of Q(net) is positive indicating “net” energy transfer from 1 to 2 – you’d agree with that surely ?

        The energy radiating from 2 to 1 cannot even replace the energy radiating from 1 to 2 and we always have the same result – 1 cools, 2 warms and Q(net) reduces to zero when both are at the same temperature – you’d agree with that surely ?

        I began this series of exchanges a few months ago with a claim that heat loss from satellites in space requires significantly larger heat radiators than would be necessary in an atmosphere.

        You took exception to that and challenged me with some nonsense involving the SB equation similar to what you have written again above.

        I answered it correctly then and I have done so again.

        What is the point you are trying to make with this ? What is your point about the cosmic background radiation ?

        NASA clearly acknowledge the reality I have said exists – that the solar radiation absolutely means that the inner solar system space cannot be considered cold by any rational definition !

        “Space agencies have giant vacuum chambers to test their spacecraft before launch. These chambers have “shrouds” that can be supercooled with liquid nitrogen and sometimes helium to get their temperature below 100K and even down to about 20K. These cost millions of dollars. Why do they bother?”

        My assumption is that there may be material testing reasons for this – extreme cold adversely affects all metals.

        Why would they bother testing the effectiveness of radiation in extreme low levels of radiation ?

        Surely any satellite is going to be subject to the solar radiation some of the time and it is going to be much, much more difficult to radiate heat away from something exposed to the solar radiation than it is in an area of space where the satellite is shielded ?

        • Avatar

          Ed Bo

          |

          Rosco:

          You started this by objecting to the concept of the earth being warmer than space and so radiates away power to space. Now you admit you would use a temperature value for space of absolute zero in SB calculations — which is certainly much colder than earth temperatures of 200 – 300K.

          We’re finally getting through to you! You finally agree with the point I have been making all along!

          You would use 0K temperature even though this space has ~1000 W/m2 coursing through it. What do you think the actual temperature of the very low density space near earth really is?

          Why do you continue to insist, against all reason, that radiative energy passing unimpeded through a space with nothing absorbed in that space affects the temperature of that space? I used the example of a powerful laser beam shot through the air to melt steel. The intensity of that beam is incredibly high. Does it superheat the air that is transparent to the laser light? According to your analysis, it must. According to experimental reality, it does not. Maybe you have to re-examine your analysis!

          You say: “I began this series of exchanges a few months ago with a claim that heat loss from satellites in space requires significantly larger heat radiators than would be necessary in an atmosphere.”

          It’s not so simple! Yes, you lose the conductive/convective cooling that is possible at 1 atmosphere. I’ve had to deal with motors overheating in the vacuum chambers of X-ray synchrotrons for this reason.

          But even in a vacuum chamber, if the walls are at room temperature, they provide an ambient radiation level of over 400 W/m2. This is why the space agencies supercool the walls of the chamber — to eliminate this “back radiation”. When this ambient radiation is reduced almost to zero, and the net radiative output from the fins is increased enough that it more than compensates for the loss of conductive/convective cooling. And in space, in most directions, the ambient radiation level is virtually zero.

          So the reason they supercool the walls of the vacuum chamber is to properly simulate the thermal exchanges the craft will actually encounter in space.

          Of course, if the sun is shining on it, whether in air or in a vacuum, the added power input makes cooling harder. The vacuum chambers also have banks of incandescent lights on one wall to simulate the power from the sun.

          You say: “Cooler surroundings cannot make the hotter object increase in temperature because the value of Q(net) is positive indicating “net” energy transfer from 1 to 2 – you’d agree with that surely ?”

          If there is no other power source, yes that is true. But if there is a separate power source – the sun for the earth, metabolic processes for the human body — the replacement of cool surroundings for colder surroundings reduce the Q(net) between the object and its surroundings, resulting in an increase in temperature of the object until the Q(net) to the surroundings matches the separate Q(in). For the life of me, I can’t understand why people like you have so much trouble with this concept!

          • Avatar

            Rosco

            |

            I knew you would think that what I wrote was some justification for the BS you have been spruiking !

            Again you are just making shit up.

            I have not made the claims your twisted imagination thinks I have.

            Seriously your ability to completely twist facts and invent things borders on the pathological.

            I never said anything about a laser – YOU did – and I certainly never said the majority of the things you claim I said !

            All I ever said was that it is just absurd to claim space (which has no physical properties) and which is continuously subject to the solar radiation of at least 586 w/m2 can be in any way be considered COLD !

            YOU said it is COLD quoting the “cosmic microwave background”.

            The NASA site I referenced proves that you have been talking nonsense !

            I NEVER said “radiative energy passing unimpeded through a space with nothing absorbed in that space affects the temperature of that space?”

            I did say that any object subject to the solar radiation will certainly not be cold.

            I continually said space has NO temperature ! Neither hot nor cold !

            By your ridiculous arguments it would be more appropriate to say the inner solar system space is actually hot !

            But I don’t.

            You insist space can be considered cold despite the continuous presence of levels of radiation of at least 586 w/m2.

            I plainly said that objects in the inner solar system will be subject to that radiation and heat up.

            I stated it is difficult to avoid the solar radiation and that the volume where the planets block it is tiny.

            Your example of the stefan-boltzmann equation is absolutely meaningless – a typical example of misdirection practised by deceitful people like you.

            For the Earth radiating to space there are really 2 scenarios – actually many more because of the spherical shape of Earth but lets keep it to two.

            In one I would put Ts^4 to zero where Ts^4 is the temperature of the surroundings. The reason I put that to zero is because there are effectively NO surroundings and hence obviously NO temperature.

            Hence Q(net) = e.sigma.(Te^4 – 0) as I said before.

            There is also obviously no significant radiation flux from these areas of space.

            Note I have never argued that is not the case – I simply said that is no justification for claiming space is cold in an area where there is a large radiation flux present. You have continually said that it is cold !

            I always said all radiation is directional – it was your JMSU claims that I said something I never said !

            The other case is the one people like you never mention !

            In this case the Q(net) equation includes not a temperature but a powerful radiation flux because half of the Earth is continuously irradiated by the Sun.

            Hence Q(net) = e.sigma.(Te^4 – 1361).

            Obviously this is strongly negative indicating a net heating effect.

            But you also misinterpret this equation !

            It is not a case of “We’re finally getting through to you!” applying to me – I have always understood the position you have put – I just know you are wrong.

            You make patronising statements to bolster your argument which we ALL get – really we get your stuff – you are not teaching us anything – we all get your position.

            Everyone understands a vacuum chamber is subject to the radiation fluxes of its surroundings yet you pronounce such an obvious reality as if you are sharing your superior knowledge. Obviously a vacuum chamber will have radiation levels from the walls of the chamber equivalent to the temperature of the walls.

            Actually you have agreed with me when I said a positive Q(net) indicates a transfer of net energy from hot to cold – “If there is no other power source, yes that is true.”

            But as usual you JMSU and ignore my clear and precise statement –

            “If we have 2 objects radiating against each other with similar properties and there is no other source of radiation then we have …”

            I clearly said “no other source of radiation” and yet AGAIN you choose to imply I have said something I have not said.

            Seriously you are either incapable of understanding simple written English are you are deceitful.

            Why do I say this hurtful thing ?

            Because time after time you claim I have said absurd things I have never uttered and this clear example is proof of that !

            You continue on to prove you are deceitful with the statement –

            “But if there is a separate power source – the sun for the earth, metabolic processes for the human body — the replacement of cool surroundings for colder surroundings reduce the Q(net) between the object and its surroundings, resulting in an increase in temperature of the object until the Q(net) to the surroundings matches the separate Q(in).”

            Changing the goalposts to deceive is the tactic of someone with a psychological problem – they are often called pathological liars.

            You have done this again – let us be plain here – that is what you have done by introducing “a separate power source”.

            Also your statement is convuluted and I think designed to mislead – something I think you do all the time as a result of our little discussions.

            Really, who says “the replacement of cool surroundings for colder surroundings” ?

            The statement is poor english and could be taken to imply that you replace the cool surroundings. If that is what you intended then your argument is entirely wrong as shown by a few simple calculations.

            I’m assuming you mean if we move the object from cold surroundings to cool surroundings – ie to warmer surroundings.

            But lets examine your statement. We’ll use unit emissivity – thats 1 if you don’t quite get it.

            Object at 37 C – call it 310 K.

            Now how do we account for your “separate power source” ? I’ll make an assumption the object is irradiated by a constant ~523 W/m2 (or body metabolism if you like) and it absorbs all of it hence a temperature of ~37 degrees C or ~310 K.

            I firmly believe you will change goalposts again but at least I am trying to be honest and consistent.

            Case 1 – “cool surroundings” – lets say 20 C – call it 293 K.

            Q(net) = 1.sigma(310^4 – 293^4) – ~105.

            Case 2 – “colder surroundings” – lets say minus 20 C – call it 253 K

            Q(net) = 1.sigma(310^4 – 253^4) – ~291.

            So Q(net) in case 1 is obviously a smaller value than in case 2 – we all knew that anyway – we did not need you to “enlighten” us.

            Now just how does this justify increasing the temperature of the object beyond 37 degrees C ?

            As I said you need to clarify your statement. I’ve assumed what you meant.

            If you really meant to say that Q(net) is less when the surroundings are warmer then we all agree but that was always obvious.

            It is unnecessarily obtuse – state your case simply and concisely.

            The only way the “surroundings” increase the temperature of the object is if the surroundings are hotter than the object and “hotter” than the external radiative flux heating the object and the Q(net) calculation is negative using all the USUAL conventions applied to the equation – the first T is the object and the second T is the surroundings.

            You have failed to qualify what you mean by Q(in) so I see no logic in your statement “reduce the Q(net) between the object and its surroundings, resulting in an increase in temperature of the object until the Q(net) to the surroundings matches the separate Q(in).”

            I commence from the position where the object is at a temperature consistent with the input radiation I chose – Q(in) at 30 degrees C.

            The Stefan-Boltzmann equation was derived from cavity radiation experiments and resulted in the equation

            Q = sigma T^4.

            Everything emits radiation in proportion to its temperature – obviously real objects also have emissivity. These values are derived by experiment.

            I am consistently amazed that people think that Q(net) must be preserved when it obviously changes as circumstances change.

            The temperature of the object will not change simply by moving it from a minus 20 to a plus 20 environment.

            All that will change is Q(net).

            The object will increase in temperature when Q(net) becomes negative which obviously means the surrounding environment is hotter than the object and its “external or internal” power source.

            The equation explicitly says the object is radiating in proportion to its temperature – it is right there in the equation and I’ll requote it if necessary.

            Q(net) = e.sigma( T1^4 – T2^4) – there it is T1^4.

            Q(net) IS NOT and (DOES NOT have to be) preserved.

            In fact Q(net) is not even defined unless you explicitly know T1 and T2.

            When the object is at 37 degrees C in a minus 20 C environment we have already calculated that Q(net) is ~291.

            What your statement –

            “the replacement of cool surroundings for colder surroundings reduce the Q(net) between the object and its surroundings, resulting in an increase in temperature of the object until the Q(net) to the surroundings matches the separate Q(in).” –

            is claiming is that Q(net) = e.sigma(T1^4 – T2^4) means that T1 must increase to 61 degrees C or 334 K.

            Here it is in equation form.

            At minus 20 degrees C – 253 K

            Q(net) = 1.sigma(310^4 – 253^4) – ~291. We did this above.

            In the “cooler” surroundings of 20 degrees C or ~293 K you need T1 of the object to be 61 degrees C or slightly more than 334 K to preserve Q(net).

            Q(net) = 1.sigma(334^4 – 253^4) – ~287 – the small difference is due to rounding.

            I find it astounding that you can assert such an absurdity reflects reality !

  • Avatar

    Greg House

    |

    Rosco, climate professors and others can be rightfully called stupid if they claim stupid things. I hope we can agree on that. Of course, they are not completely stupid. But this claim about -18°C is a stupid one. The attempts to “explain” the 33°C difference are logically also stupid, since they are based on this false value in the first place, sorry. Even double stupid, because the explanation in fact explains, how to get energy out of nothing, so as such it must be physically wrong. MUST, Rosco. It is not my fault that they keep posting that. And it is not my task to find out who is what exactly, a stupid person a liar or whatever.

  • Avatar

    Rosco

    |

    Again Greg is up to his old standard – insult without any actual evidence to back up his claims.

    Here is the challenge I’ve posted twice before –

    Tell us all what happens to a Planck curve when you change variable from wavelength to wavenumber.

    If you can’t do this simple task you have no actual ability or knowledge at all. All of the people you describe as “do not know what they are talking about” or “stupid” or “liars” could explain this in a few sentences off the top of their head without any research.

    Bet you can’t !!!!

  • Avatar

    Ed Bo

    |

    Charles:

    The earth absorbs from the sun an amount of power equal to 240 W/m2 times the area of the earth. The surface of the earth emits an amount of power equal to 490 W/m2 times the area of the earth. These numbers are from actual measurements, easily known to within +/-10 W/m2.

    So there is an imbalance of 250 W/m2 times the area of the earth, much much larger than any errors in the measurements. No one, alarmist or skeptic alike, thinks the earth’s power balance is off by more than 1 W/m2 times the area of the earth.

    What do you propose to make up the difference? You seem to think that gravitational effects do. But for a gravitational field to transfer power, something must fall in that field. Hydroelectric power generation depends on the water ending up lower than it started. But the atmosphere cannot fall — it is already “fallen”. (Total downdraft has to be matched by equivalent updraft.)

    So I ask, where does the extra 250 W/m2 down to the surface come from in your analysis?

    • Avatar

      Greg House

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      Yeah. 2+2=5. So I ask, where does the extra 1 comes from? Let the stupidity go on.

      • Avatar

        Ed Bo

        |

        What extra “1” are you talking about? I am talking strictly about conservation of energy. What in God’s name are you referring to?

    • Avatar

      Greg House

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      But you are good, no doubt. Keep trolling them, Ed Bo, they deserve it.

    • Avatar

      Jerry L Krause

      |

      Hi Ed Bo,

      You wrote: “The earth absorbs from the sun an amount of power equal to 240 W/m2 times the area of the earth. The surface of the earth emits an amount of power equal to 490 W/m2 times the area of the earth. These numbers are from actual measurements, easily known to within +/-10 W/m2.”

      “These numbers are from actual measurements.” 490 W/m2 could be an actual measurement of the upward emission from the surface but it varies during a 24 hour period, given cloudless skies, over a range significantly greater than +/- 10 W/m2 and during a 24 hour period at many locations the solar radiation that reaches the surface varies between 0 W/m2 to maybe as high as about 1200 W/m2.

      Your ignorance of this topic is showing.

      Have a good day, Jerry

      • Avatar

        Ed Bo

        |

        Jerry:

        You are apparently unfamiliar with the shorthand commonly used in this field to talk about energy transfers in a convenient fashion. So I will repeat my point more pedantically for you.

        Over a year, the earth absorbs from the sun an amount of energy equal to:

        Eabs = 240 (W/m2) * Aearth (m2) * SecondsPerYear (sec)

        Over this same year, the earth’s surface emits an amount of energy equal to:

        Esurf = 490 (W/m2) * Aearth (m2) * SecondsPerYear (sec)

        Considering just these two factors, there is an energy imbalance at the surface equal:

        Ediff = 250 (W/m2) * Aearth (m2) * SecondsPerYear (sec)

        Now obviously, the power densities are not constant over time or area. Values are higher in the tropics than the polar regions, higher in winter than summer, and higher during the day than the night. But integrated over the earth’s surface and over a year, we know these total values from real measurement to within a few percent.

        • Avatar

          Jerry L Krause

          |

          Hi Ed Bo,

          I am very aware that averaged anything cannot be an actual measurement. In quantitative chemical analysis we did average our results because we always ran three independent analysis on a portion of the same sample. We did this to determine how reproducible the results were. And we calculated the average of the three so we could apply a statistical test if an odd ball could be reasonably disregarded. Of course, today machines can do much of the analysis we did quicker and with more precision. But I believe the accuracy of the machine still needs to be established by running standards. For we learned there was a difference between precision and accuracy also.

          Have a good day, Jerry

  • Avatar

    Jerry L Krause

    |

    Hi Charles,

    I much enjoyed reading your article. I had never considered the atmosphere of a mine. It seems obvious you have brought a very controlled ‘natural’ experiment to our attention. But I have some comments that I would like you to consider and comment about.

    “Just as the Sun radiates energy into space, the Earth, being warmer than space, also radiates energy out into space.”

    Would the Earth stop radiating energy out to space if space were warmer than the earth? I would merely omit “being warmer than space” as it seems to create an issue that need not be considered. Also, I have recently read that space has no temperature because it has no matter. And later in your article you make the point that temperature is proportional to the average kinetic energy of matter. Which is easiest understood by a consideration of gaseous matter.

    “The energy equilibrium with the Sun ignores some heat from the Earth’s core …”

    Later, in the depths of mines, you seem to ignore what this somewhat continuous source of heat from the interior of the Earth is generally understood to be. We generally understand the source of the energy which the sun is somewhat continuously radiating to space is natural nuclear fusion occurring in its interior and that the source of energy which the earth would continuously radiate to space is natural nuclear fission occurring in its solid (and possibly liquid) interior.

    “Deep mines become warm primarily because the surrounding rock becomes warm due to the Earth’s high core temperature, which also means that some of the energy at the Earth’s surface causing it to be warm is due to heat from the core. Deep mines also become warm due to the effect of gravity on air molecules! That effect is an important effect and has to be taken into account.”

    “Let us illustrate this mining application problem. The Tau Tona gold mine in South Africa is 11,760 feet deep or 3,584.4 meters deep. The air in the mine reaches a temperature of 130 F. The gravitationally determined air temperature at that depth is 311.46K or 38.31ºC or 100.96ºF as interpolated from the full table of the U.S. Standard Atmosphere given in 500 meter altitude increments. Clearly the rock wall temperature is higher than the temperature of air in gravitational equilibrium in this deep mine. The air at sea level from the table is at 288.15K or 59ºF. The mine engineer would love to bring this air at 59ºF down to the bottom of the mine and be able to keep it at 59ºF as he does so. Air at 59ºF would be much more effective in cooling the bottom of the mine than is air at 101ºF.”

    I am not sure you have considered that there are two temperature gradients involved in this situation which, I consider, you accurately describe (define). The temperature gradient which exists in the solid mine walls due to the thermal conductivity of energy through solid matter from hotter interior to the cooler surface. The atmosphere temperature gradient, which is your principal focus, is a fiction until the mining engineer tries to cool the atmosphere at the bottom of the mine by pumping the surface atmosphere to the bottom of the mine to ‘lift’ the warmer, less dense, atmosphere to the surface.

    However, my comment is not meant to be critical of what you have brought to our attentions. It is just to add that the conclusion of what you ‘analyze’ is that the mining engineer can never cool, no matter the rate of pumping, the atmosphere at the bottom of the mine below 101ºF by pumping surface atmosphere to the bottom of the mine.

    More important, I consider, is that my purpose to draw your (Charles) and the readers’ attentions, to that the Earth’s atmosphere is seldom, if ever, a natural equilibrium system. That it is a natural non-equilibrium dynamic system which is always striving toward some equilibrium state but never reaching it because the Earth continuously revolves about an axis with a period of about 24 hours.

    But Charles, what you wrote is so good, even if in detail it is not absolutely accurate. Richard Feynman wrote (relative to a story of how his father taught him): “Finally he says, “So you see, everywhere there’s a source of food, there’s some form of life that finds it.”

    “Now, I knew that it may not have been exactly a louse, that it might not be exactly true that the louse’s legs have mites. The story was probably incorrect in detail, but what he was telling me was right in principle.”

    Have a good day, Jerry

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      Rosco

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      ““Just as the Sun radiates energy into space, the Earth, being warmer than space, also radiates energy out into space.”

      Would the Earth stop radiating energy out to space if space were warmer than the earth?”

      Good points.

      Vacuum Space has no temperature because temperature is a physical property and vacuum space is by definition the absence of physical materials.

      Earth is “warmer” than space is the sort of ludicrous statement climate science advocates say all the time and it appears time and again in all sorts of references.

      Charles has simply stated an often quoted statement – probably should get rid of it though.

      The only thing one can say about the “space” at near Earth orbit is that it is continually “awash” with powerful solar radiation with values varying from ~1320 to ~1415 W/m2 over one orbit.

      To claim that is cold is absurd.

      That aside though, nothing has to be “warmer” than its surroundings to radiate provided it has some temperature. This is a common mistake climate science advocates get wrong all the time.

      I will still radiate in proportion to my internal temperature near 37 C even when the air temperature exceeds 40 C.

      Any attempt by my body to regulate my skin temperature is through evaporation not by an attempt to lower my core temperature – it has nothing to do with radiation. It is also why a hot humid climate feels uncomfortable – our body loses its most potent cooling mechanism because the humid air cannot readily accept more water vapour – again nothing to do with “greenhouse gas” radiation.

      Still a great article.

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        Greg House

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        They calculate the -18°C by wrongly putting the area of a sphere into the Stefan-Boltzmann formula. They also wrongly handle averages. Very stupid. The story about satellites conforming the nonsense is another falsehood.

        WHAT CHEMICAL REACTIONS? I do not think you know what you are talking about.

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          Greg House

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          Sorry, it was the wrong place. Please delete it.

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        Ed Bo

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        Rosco, you say: “Vacuum Space has no temperature because temperature is a physical property and vacuum space is by definition the absence of physical materials.”

        So, at what density does matter cease to have a temperature? The density of matter in space is extremely low, but it is not zero. And its extent is extremely vast, so there is a lot of “physical materials” out there in all directions.

        Regardless, detailed measurements by astronomers (nothing to do with climate science) show that the radiation received by earth from space is equivalent to that of a perfect blackbody (as close as they can measure) at 2.725K — within +/-0.001K in all directions.

        So even if you refuse to believe that space “has a temperature”, it is still valid that for the purposes of radiative energy transfer, it behaves exactly like a blackbody at 2.725K, providing only 3 microwatts per square meter to the earth. Call this the “effective blackbody temperature” if you like, it is still vitally important to heat transfer calculations.

        Consider the radiative fins of a satellite on the night side of the earth. The side facing the earth is receiving about 240 W/m2; the other side is receiving 3uW/m2. The engineers must account for this properly.

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          Jerry L Krause

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          Hi Ed Bo,

          Thank you! Thank you! For being so wrong. You wrote: “Regardless, detailed measurements by astronomers (nothing to do with climate science) show that the radiation received by earth from space is equivalent to that of a perfect blackbody (as close as they can measure) at 2.725K — within +/-0.001K in all directions.”

          You did this to support your argument that space is not void of matter as stated by Rosco. The implication of your reference to the observations of astronomers seems to be that this diffuse matter of space must be the source of this radiation observed by the astronomers. Now I have no idea what the astronomers consider to be the source of this radiation but I know I ‘see’ stars. And it seems generally agreed that the sun is a star. The difference being the sun is much closer to the earth than the other stars I see. And because I have had the opportunity to be in very isolated locations of the earth at night, I have observed what has been termed the Milky Way. Which has been explained, I believe, by astronomers to be the result of many, many, many distant stars so far away that individual stars cannot be seen by the naked eye but the light from so many of these very distant stars blend together to make a continuum of very dim light which we term the Milky Way. So, while I cannot claim that space has no matter, I can claim that the source of most of the radiation that the astronomers, here on earth, observe and measure has stars as its origin.

          And this is not the only thing your mistake called (brought) to my attention. In a comment to the article (http://principia-scientific.org/how-can-uranus-have-storms-hot-enough-to-melt-steel-a-runaway-greenhouse-effect/) I had drawn attention to a seldom considered scattering theory taught by Richard Feynman to students at Caltech. And at the time I wrote my comment I thought its ‘validity?’ hinged on Feynman’s scattering hypothesis. What I had overlooked was the importance of the common observations, non-debatable, to which Feynman had directed his student’s attention as an introduction to his hypothesis. Which observations, I did not quote in an effort to make my comment brief. Which I now do.

          “There are several points to be made about the above results. One interesting question is, why do we ever see the clouds? Where do the clouds come from? Everyone knows it is the condensation of water vapor. But, of course, the water vapor is already in the atmosphere before it condenses, so why don’t we see it then? After is condenses it is perfectly obvious. It wasn’t there, now it is there. So the mystery of where the clouds come from is not really such a childish mystery …” (The Feynman Lectures On Physics, Vol I pp 32-8)

          But your mistake not only drew my attention to the importance of our observations of clouds, it drew my attention to a quote (wisdom) of Galileo. “Nature is relentless and unchangeable, and it is indifferent as to whether its hidden reasons and actions are understandable to man or not.” And it drew my attention to Newton’s second rule of reasoning in philosophy. “Therefore to the same natural effects we must, as far as possible, assign the same causes.” To which he added the comment: “As to respiration in a man and in a beast; the descent of stones in Europe and in America; the light of our culinary fire and of the sun; the reflection of light in the earth, and in the planets.”

          Clearly, if clouds ‘reflect’ or ‘scatter’ solar radiation does not matter, what matters is what we observe. And according to Newton’s last comment, we must reason that what we observe here at earth about cloud and solar radiation must be observed about cloud and solar radiation in the Uranus atmosphere. And Galileo understood it does not matter how we reason what we observe, it is only the observation which matters.

          So thank you again, for enlightening me by your mistake.

          Have a good day, Jerry

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            Ed Bo

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            Jerry:

            I’m afraid the mistakes are all yours, and just demonstrate that you don’t understand the most basic concepts involved here.

            I was referring to what is commonly known as the “cosmic microwave background” (CMB) radiation, which has a spectrum completely different from that of starlight. Its peak wavelength is about 1.1 millimeters, compared to the peak of about 500 nanometers for most stars.

            As I said, this CMB radiation has a spectrum profile and magnitude that is exactly the same as what an ideal blackbody at 2.725K would have, so for the purposes of radiative heat transfer calculations, it can be treated as an ambient of 2.725K with an emissivity of 1.0.

            Interestingly, the total flux density from stars (excluding the sun, of course) has about the same (tiny) magnitude of about 3 microwatts per square meter, but as I said, with a spectrum of much shorter wavelengths than the CMB. Both sources can safely be ignored in any quantitative calculations of earthly heat transfer.

            Rosco’s fundamental mistake in asserting the “space has no temperature” mantra is not understanding the difference in mechanisms between conductive/convective heat transfer and radiative. With the density of space so low, it does not matter what the temperature is for any conductive or convective transfers, because there is not anything significant to conduct to or convect with.

            But for purposes of radiative transfer, it does matter. The earth can and does radiate energy away to (through?) space, even when there is very low density. And the fact that space radiates virtually no energy back to earth means that it behaves identically to an ambient of ~0 K.

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            Jerry L Krause

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            Hi Ed Bo,

            Are you saying that stars emit different photons of drastically different wavelengths than our Sun?

            If you are, then I am wrong. However, I am an experimentalist and am quite sure it is quite difficult (maybe impossible) to measure a temperature of 2.725K. And before you tell me I do not know anything about such a low temperature be advised that I made magnetic measurements down to minimum pumped He temperatures.

            And are you really saying that stars do not emit inivisble IR radiation even though we know they emit visible radiation? Do you have an explanation how it is we can observe the visible and that the IR is not tagging alone?

            You can write about any other wavelengths you like, but you do have to explain how the sun heats the earth but the star light which see does not also heat the earth because it certainly being emitted by something whose temperature is definitely greater than that of the earth. You are right I do not know much of what you wrote. But can you admit that you do not know what I am writing?

            Have a good day, Jerry

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            Ed Bo

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            Jerry, you ask: “Are you saying that stars emit different photons of drastically different wavelengths than our Sun?”

            No, not at all (although there is some variation between stars). I am saying that the “cosmic microwave background” radiation is of “drastically different wavelengths” than that of any stars, including our sun. It does NOT come from stars.

            99% of our sun’s radiation has shorter wavelengths than 4 micrometers, as is true of similar stars. About 45% has longer wavelengths than 700 nanometers, which we cannot see and so consider “infrared”.

            We infer the sun’s surface temperature (~5770K) from these spectral measurements. We do not stick a thermometer in the sun!

            Similarly, we take measurements of CMB radiation, which is in the 1 millimeter wavelength range — a thousand times longer wavelengths than typical stellar radiation. From these spectral measurements — done with radio telescopes — we infer the background temperature of about 3K.

            You write: ” you do have to explain how the sun heats the earth but the star light which see does not also heat the earth because it certainly being emitted by something whose temperature is definitely greater than that of the earth.”

            All radiation absorbed by the earth “heats the earth”. But the sun’s radiation hitting the earth has a flux density of 1365 W/m2, whereas all of the starlight hitting the earth combined has a flux density of 0.000003 W/m2. It can safely be ignored.

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          Rosco

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          Ed – you seem to enjoy just disagreeing.

          “So, at what density does matter cease to have a temperature? ”

          I’d say that at the density of the Thermosphere temperature is an irrelevant concept – molecular “temperatures” of 2500 C as reported have little meaning in the near vacuum.

          All of the comments I have made about space not being cold have been made in the context of near Earth orbit space or out to Mars orbit.

          In this region of space the radiation from the Sun is always present.

          Why do you ignore this reality ? The only area of space between Mars’ orbit and the Sun not continuously irradiated by powerful Solar radiation is the volume of a small cone “shadow” cast by a planet.

          The only reason that the measurements you like to cite could be made is because of Earth’s shielding the Solar radiation. By similar triangles the height of this cone of “shadow” is of the order of 1.4 million kilometres. The Moon’s shadow on Earth is a mere few hundred kilometres wide during a total eclipse.

          Just because the Solar radiation is directional and there is little radiation coming from all other directions does not mean you can conclude space acts like a blackbody at 2.725 K.

          Such a claim is not only wrong it is irrelevant because any object outside the tiny shadows cast by planets is irradiated by the Sun – it is unavoidable.

          It is just wrong to make the claim you make – the Solar radiation is always present and simple geometry of similar triangles shows the “shadow” cast by Earth is insignificant.

          A cursory glance at any photo of Venus transiting the sun shows this is indisputably true.

          The volume of a sphere defined by Mars’ orbit is ~ 5 x 10^25 cubic kilometres.

          The cone “shadow” cast by Earth has a volume of ~6 x 10^13 cubic kilometres – 6371 radius and 1.4 million kilometre height.

          The volume of space continuously irradiated by the Sun is ~ 9 x 10^11 times larger than the “shadow” cast by Earth. Divide this by 3 to include Venus and Mercury and it is still so insignificant as to be meaningless – it is at least 300 billion times larger.

          Seriously I think satellite engineers can and do ignore the few milliwatts – any Earth orbiting satellite is designed to cater for the Solar radiation.

          And I think I pointed out in a previous comment that the space station is either irradiated by the Solar “constant” or by whatever IR is emitted by Earth – which we are told is ~239 W/m^2.

          Why does the fact that Earth receives a few milliwatts from space even have any relevance when half of Earth is continuously irradiated by the Sun ?

          Is there something wrong with the maths I used ? I can’t see any problem with my calculations.

          Effectively- ALL of the space in a sphere defined by Mars’ orbit as radius is permanently irradiated by radiation with at least 586.2 W/m2 (NASA Mars Fact Sheet) power and growing significantly more powerful as one proceeds towards the Sun.

          Any other claim is simply irrelevant.

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            Ed Bo

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            Rosco: You can’t be serious! You say: “Just because the Solar radiation is directional and there is little radiation coming from all other directions does not mean you can conclude space acts like a blackbody at 2.725 K.”

            Of course you can! The highly directional solar radiation that is not headed directly for the earth does not affect the earth’s energy balance AT ALL. You are in effect arguing that if I shine a laser beam away from you, it can still burn you. It’s just ridiculous!

            YOU emphasize that space is essentially empty, but yet you insist that this nothingness is “irradiated”. That’s just meaningless! That radiation is just passing through your empty space, and it does not do a U-turn and head back to the earth.

            I’m citing real measurements of the radiative power that is directed toward the earth and reaching the earth at every angle but the 1/2-degree by 1/2-degree of the sun. And that radiative power is only 3 microwatts per square meter in the microwave spectrum (and another 3 uW/m2 of starlight centered in the visible spectrum). This is equivalent to what a body at about 3K would radiate.

            So for the purposes of radiative heat transfer, it is absolutely appropriate to consider space to be very “cold”. (For the purposes of conductive or convective heat transfer, the incredibly low mass density of space means that it does not matter what the temperature of what little there is out there, there is essentially no heat transfer by these means, but radiative power passes unimpeded through space.)

            You are simply showing that your conceptual framework for these matters is completely wrong, so you have gone off the rails before you even start your math.

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            Jerry L Krause

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            Hi Rosco,

            We must recognize the position in which Ed Bo finds himself. He is an engineering professor who seemingly has stepped outside his profession of engineering into what is termed science. It seems Ed Bo’s position on some scientific issues is like that of the academics at the time of Galileo who subscribed to the popular, but wrong, ideas of Aristotle who had reasoned without the aid of observations. How you are going back to your students admitting you were wrong?

            Have a good day, Jerry

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    Greg House

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    Charles, you have explained something that is physically impossible, hence your explanation must be wrong. In absence of a more powerful source of heat there can not be any difference between the real temperature and what the Sun can possibly induce. Neither 33°C nor 0.000001°C. Because the warmer surface would radiate away more energy than it gets, which is physically impossible.

    The -18°C thing is the result of a wrong calculation.

    • Avatar

      Rosco

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      “In absence of a more powerful source of heat there can not be any difference between the real temperature and what the Sun can possibly induce.”

      So work, chemical reactions and all of the other mechanical factors cannot possibly create any temperature increase ?

      As for “The -18°C thing is the result of a wrong calculation.” – the only calculations involved in this are the calculations used in the satellites to convert the measured radiation fluxes emitted by Earth into an “ensemble temperature”.

      Most of the educated people on Earth disagree with you on this point – most accept that the radiation emitted by Earth and measured by satellites is similar to that emitted by an object at ~255 K calculated by the SB equation.

      If such calculations were the result of a wrong calculation then infra-red thermometers are an impossibility – tell that to blast furnace or power station engineers.

      Show your depth of understanding by telling us what happens to the shape of a Planck curve when one changes variable from say wavelength to wavenumber.

      • Avatar

        Greg House

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        They calculate the -18°C by wrongly putting the area of a sphere into the Stefan-Boltzmann formula. They also wrongly handle averages. Very stupid. The story about satellites conforming the nonsense is another falsehood.

        WHAT CHEMICAL REACTIONS? I do not think you know what you are talking about.

        • Avatar

          Rosco

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          We all know you don’t know what you’re talking about !

          You apparently do not believe that satellites can measure radiation emitted to space or you believe that all personnel associated are liars.

          As I asked and you chose to ignore (because I know you can’t even comprehend the question ) – Tell us all what happens to a Planck curve when you change variable from wavelength to wavenumber.

          Do you even know what the question means ?

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            Greg House

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            Again, your or whoever claim that satellites measured what it is in fact a false value, a result of a false calculation, is sheer nonsense.

            As for the people who claim it, they either do not know what they are talking about (the best option for you) or they are stupid or they are liars or a mixture of different options. Some of them, I do not want to mention the names, have understood, in my opinion, that they are wrong, but can not admit it for some psychological reasons. So they keep posting the same crap.

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