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The greenhouse effect and the 2nd law of thermodynamics

What the science says...

Select a level... Basic Intermediate

The 2nd law of thermodynamics is consistent with the greenhouse effect which is directly observed.

Climate Myth...

2nd law of thermodynamics contradicts greenhouse theory

 

"The atmospheric greenhouse effect, an idea that many authors trace back to the traditional works of Fourier 1824, Tyndall 1861, and Arrhenius 1896, and which is still supported in global climatology, essentially describes a fictitious mechanism, in which a planetary atmosphere acts as a heat pump driven by an environment that is radiatively interacting with but radiatively equilibrated to the atmospheric system. According to the second law of thermodynamics such a planetary machine can never exist." (Gerhard Gerlich)

 

Skeptics sometimes claim that the explanation for global warming contradicts the second law of thermodynamics. But does it? To answer that, first, we need to know how global warming works. Then, we need to know what the second law of thermodynamics is, and how it applies to global warming. Global warming, in a nutshell, works like this:

The sun warms the Earth. The Earth and its atmosphere radiate heat away into space. They radiate most of the heat that is received from the sun, so the average temperature of the Earth stays more or less constant. Greenhouse gases trap some of the escaping heat closer to the Earth's surface, making it harder for it to shed that heat, so the Earth warms up in order to radiate the heat more effectively. So the greenhouse gases make the Earth warmer - like a blanket conserving body heat - and voila, you have global warming. See What is Global Warming and the Greenhouse Effect for a more detailed explanation.

The second law of thermodynamics has been stated in many ways. For us, Rudolf Clausius said it best:

"Heat generally cannot flow spontaneously from a material at lower temperature to a material at higher temperature."

So if you put something hot next to something cold, the hot thing won't get hotter, and the cold thing won't get colder. That's so obvious that it hardly needs a scientist to say it, we know this from our daily lives. If you put an ice-cube into your drink, the drink doesn't boil!

The skeptic tells us that, because the air, including the greenhouse gasses, is cooler than the surface of the Earth, it cannot warm the Earth. If it did, they say, that means heat would have to flow from cold to hot, in apparent violation of the second law of thermodynamics.

So have climate scientists made an elementary mistake? Of course not! The skeptic is ignoring the fact that the Earth is being warmed by the sun, which makes all the difference.

To see why, consider that blanket that keeps you warm. If your skin feels cold, wrapping yourself in a blanket can make you warmer. Why? Because your body is generating heat, and that heat is escaping from your body into the environment. When you wrap yourself in a blanket, the loss of heat is reduced, some is retained at the surface of your body, and you warm up. You get warmer because the heat that your body is generating cannot escape as fast as before.

If you put the blanket on a tailors dummy, which does not generate heat, it will have no effect. The dummy will not spontaneously get warmer. That's obvious too!

Is using a blanket an accurate model for global warming by greenhouse gases? Certainly there are differences in how the heat is created and lost, and our body can produce varying amounts of heat, unlike the near-constant heat we receive from the sun. But as far as the second law of thermodynamics goes, where we are only talking about the flow of heat, the comparison is good. The second law says nothing about how the heat is produced, only about how it flows between things.

To summarise: Heat from the sun warms the Earth, as heat from your body keeps you warm. The Earth loses heat to space, and your body loses heat to the environment. Greenhouse gases slow down the rate of heat-loss from the surface of the Earth, like a blanket that slows down the rate at which your body loses heat. The result is the same in both cases, the surface of the Earth, or of your body, gets warmer.

So global warming does not violate the second law of thermodynamics. And if someone tells you otherwise, just remember that you're a warm human being, and certainly nobody's dummy.

Basic rebuttal written by Tony Wildish


Update July 2015:

Here is the relevant lecture-video from Denial101x - Making Sense of Climate Science Denial

 


Update October 2017:

Here is a walk-through explanation of the Greenhouse Effect for bunnies, by none other than Eli, over at Rabbit Run.

Last updated on 7 October 2017 by skeptickev. View Archives

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Further reading

  • Most textbooks on climate or atmospheric physics describe the greenhouse effect, and you can easily find these in a university library. Some examples include:
  • The Greenhouse Effect, part of a module on "Cycles of the Earth and Atmosphere" provided for teachers by the University Corporation for Atmospheric Research (UCAR).
  • What is the greenhouse effect?, part of a FAQ provided by the European Environment Agency.

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Comments 351 to 375 out of 882:

  1. “Higher is Colder”, KR,337 is not “part of the greenhouse effect”. It is the only plausible way of explaining how increasing atmospheric absorption and emission can increase the surface temperature. Incidentally, it is a mechanism which G and T did not discuss, although it was current from 1900 onwards. Think about an atmosphere without a lapse rate – an isothermal atmosphere where higher is not colder. Add greenhouse gasses, increase absorption, and you suggest that the atmospheric temperature will increase. What would happen if it did? Apply the Stefan-Bolzmann equation to the radiation to space, and energy emission will also increase (proportional to the fourth power of the atmospheric temperature). But the incoming energy, from the sun, will not change. So the atmospheric temperature will fall back to its original value. My simple model, is designed to make the same point. With a lapse rate, you can suggest that the effective emission level moves up to a colder region, reducing energy emission. All the temperatures must then increase to restore the balance. The only snag with that argument is that the evidence from the last 30 years shows that it does not happen to any detectable extent. Neither G and T (nor I) claim that AGW contradicts the second law. It is just that some of the sillier explanations of AGW do. Most of them confuse heat and energy, which is where entropy comes in. The silliest explanation, which you can still find in modern text-books, (Houghton for example) is the original greenhouse radiative effect. Consider a greenhouse made of non-absorbing material, such as rock salt. It will absorb heat from the sun, the interior will heat up, and, with convective cooling eliminated, the internal temperature will be higher than the surroundings (G and T’s car interiors, for example). The greenhouse will radiate W watts per square meter, proportional to the fourth power of its temperature. Now replace the rock salt cover with glass, which absorbs infra-red radiation. Half of the outgoing radiation will return to the interior, which, so the story goes, will heat up until it radiates 2W. The original W will then be radiated to the atmosphere, and W will be returned to the interior. The ratio of the glass interior temperature to the rock salt interior temperature will be the fourth root of 2, or 1.19. An increase of 19% of the rock-salt interior absolute temperature, or about 60 degrees C. Does that argument sound familiar? You will find it in part 1 of the Rabett paper to which SOD contributed. It is, of course, nonsense. Back radiation from the cooler glass cannot heat the warmer interior. It would breach the second law if it did. To check this R W Woods built two greenhouses – one rock salt, one glass – so that their convective warming would be identical. Any back-radiative effect would heat the glass green-house preferentially. Their temperatures were the same.
  2. damorbel - I begin to see some of your issues, and quite frankly shudder to consider where to even begin. First - Do you think thermal emission is monochromatic? It's not! It covers a broad band of photon energies, due to a large number of possible electron band transitions of different levels. Second - Absorptivity describes the spectral efficiency of absorbing a photon at any particular energy/wavelength. It turns out to be equal to emissivity when the object is at thermal equilibrium. The ground, for example, has about a 95% probability of absorbing a photon at 6 micron wavelength. The thing is, photons do not carry ID cards - a 6 micron photon may be coming from a superheated plasma or a cold atmosphere - there's still a 95% probability of the ground absorbing it, and hence receiving energy from it. The recipient of a photon cannot know and does not care what the source of the photon is. But that photon still adds to the energy of the receiving object. So your statement "The temperatures are the same because the energies of the photons from both sources are the same" is incorrect. The sun provides a bunch of photons at various energies, the atmosphere emits (downward) a smaller bunch of photons at various lower energies, and these sum up to the total energy received by the ground. Which then emits it's own photons of thermal radiation. The heat flow, the net/summed power, is from sun -> ground -> atmosphere -> space, but even a cold atmosphere adds a tiny bit of energy to the warmer ground. If you cannot understand these basics, well, I can't help you, and I can't see spending my time banging my head on the wall.
  3. Fred Staples - "“Higher is Colder”, KR,337 is not “part of the greenhouse effect”. It is the only plausible way of explaining how increasing atmospheric absorption and emission can increase the surface temperature." Quite frankly, no. It's part of the story, but certainly not the entire thing. You might find the simple Excel models I posted here and here of interest. The first is a simple iterative single-layer atmosphere model (no convection/evaporation), so the numbers won't be accurate. But it starts with the surface of the Earth emitting exactly what it receives from the sun (240 W/m^2). Some of that energy is absorbed by the atmosphere, which radiates half of it upwards and half downwards. The end result (illustrative, if not numerically accurate due to model limitations) is that 240 W/m^2 come in, 240 go out, and the surface is emitting 267 W/m^2. A greenhouse gas atmosphere raises the temperature of the surface. The second is a more accurate radiative effect only zero dimensional model, which surprisingly (on my part) gets with 3% of real values. This uses the effective emissivity of the Earth, which drops as greenhouse gases rise (more re-emitted to the ground, also higher effective emission altitudes - both effects). Given an emissivity of .612 (as measured for Earth by satellites), 240 W/m^2 comes in, 240 W/m^2 goes out, and the surface is emitting at about 392 W/m^2 - just as expected. Radiative balances and emissivity decreases caused by GHG's drive surface temperatures to measured values. I'm not interested in convective greenhouses, with or without rock salt - those are red herrings in this discussion of radiative greenhouse effects. As to back radiation - the total energy received by the ground is the sum of solar and back radiation - both impinge on the ground. Arguing that the ground doesn't receive energy from back-radiation is the violation of conservation of energy, and hence the thermodynamic no-no.
  4. A number of these 'models', designed to contradict the standard physics - as well described by SoD - reminds me of the joke: A biologist, a physicist and a mathematician were sitting in a street cafe watching the crowd. Across the street they saw a man and a woman entering a building. Ten minutes they reappeared together with a third person. "They have multiplied", said the biologist. "Oh no, an error in measurement", the physicist sighed. "If exactly one person enters the building now, it will be empty again", the mathematician concluded. (although, personally, I'd expect better of the physicist) Even though the SoD series uses simplified models here and there to explore specific aspects of the physics, overall you can only understand what's happening by understanding the full system and physics. As said above, if a body is illuminated - with photons from any part of the spectrum - from another body, it'll reflect some, absorb others. Those absorbed (depending on the absorptance and spectrum) rise the temperature... the body will always radiate photons (not the ones absorbed, of course) depending on it temp, a la Boltzman + emissivity ... some or all of which (depending on geometry) will impinge on the original body doing the illumination, which will do the same physics. Then, you must account for the spectral nature of absorptance and emissivity, so that the respective conversion to heat and reflection won't be symmetric. Build the model properly, in your mind at least, otherwise you'll end up like our trio above.
  5. "R W Woods built two greenhouses – one rock salt, one glass" Call me curious, but a rock salt greenhouse? Watering time for the plants must be interesting. The experiment by RW Wood was done in 1909. WM Connelly aka Stoat pointed out the error in comparing this exercise to the greenhouse effect here.
  6. les - "A number of these 'models', designed to contradict the standard physics..." I hope you're not talking about the simple Excel models I posted earlier. They both support standard physics, and were intended to demonstrate to various people that greenhouse gases warm the surface.
  7. KR - not at all, I was alluding to 350 damorbel (where one moment all the radiation is returning to earth and the next it's radiating to space) and 351. Your excel and SoD's series are, IMHO, spot on. That people don't find they match their thought models is, IMHO again, because those thought models are wrong.
  8. Re 357 les You wrote:- "I was alluding to 350 damorbel (where one moment all the radiation is returning to earth and the next it's radiating to space)" No les, that's what you said, not me. I agree it isn't right but of course it's your interpretation, not what I said.
  9. Re 352 KR You wrote:- "First - Do you think thermal emission is monochromatic? " No. "Second - Absorptivity describes.... ...ground, for example, has about a 95% probability of absorbing a photon at 6 micron wavelength." Probably. Further you wrote:- "The thing is, photons do not carry ID cards" They certainly do. The energy of a photon is E = hv where 'h' is Planck's constant and 'v' is the source frequency. I suggest you check a book on thermal radiation before responding on this, you are clearly lost on this one. And further you wrote:- "So your statement "The temperatures are the same because the energies of the photons from both sources are the same" is incorrect. The sun provides... " Please check my 350; I was writng about an isothermal atmosphere, one with a uniform (vertical) temperature distribution. I didn't mention the Sun because I was on about atmospheric radiation: in an isothermal atmosphere there is no heat transfer of any sort because heat transport only happens with a temperature difference - standard 2nd law of thermodynamics - don't you think?
  10. 358 damorbel: Clearly that's what I said - and for good reason. Given (for the sake of argument): 1/ "Let us imagine for a moment that the surface and the upper atmosphere are at the same temperature. In this situation both surface and the UA are emitting photons with the same energy" and the assumption it requires... Then, this 2/ "The temperatures are the same because the energies of the photons from both sources are the same;" is a tautology - The spectrum of photons energies would be the same if the temperature, emissivity etc. are the same (said assumptions) - and we ignore it. So, two 'bodies', sufficiently identical to emit the same amount and spectrum of e-m radiation, and, we assume, the same absorptivity, so they identically absorb the energy will, by definition 3/ "there would be thermal equilibrium i.e. no energy transfer and no temperature change." Clearly, without interpretation, implies that no energy is escaping to space or any where else. You couldn't possibly mean that the surface is being headed by anything, like the sun, because you would have said so. yet in your "real" model: "bla bla... it is further radiated into deep space." QED. But really the problem is that you have given a qualitative description of how you feel things work and this leaves the door open to a range of errors - whether interpretation on one side, missing assumptions or just poor physics on the other. As SoD has shown, it is completely possible to build up mathematical models to describe how this works. If someone doesn't agree, the thing to do is present either alternative maths or, at least, show which assumptions or derivations are wrong in the original.
  11. Re 360 les You wrote:- "The temperatures are the same because the energies of the photons from both sources are the same;" is a tautology " Indeed it is a tautology because that s how you measure temperature remotely Then you wrote:- "So, two 'bodies', sufficiently identical to emit the same amount and spectrum of e-m radiation, and, we assume, the same absorptivity, so they identically absorb the energy will, by definition" And "Clearly, without interpretation, implies that no energy is escaping to space or any where else" "no energy is escaping"? That is not the case, the upper atmosphere (UA) is less dense than the surface and will exchange fewer photons with the lower atmospheric layers (and the surface) than pass through (the UA) on their way into space. Having a different density does not mean they cannot have the same temperature but 'the different densities' does translate directly into different amounts of energy. What you are missing is the fact that, with a uniform temperature, the lower atmosphere exchange photons but without any change in their relative energy.
  12. 361 : first, it's not "tautology because that s how you measure temperature remotely", it's a tautology because first you suggest, to paraphrase for clarity: that the em radiation is the same because the temp is the same, then you suggest that the temp is the same because the em radiation is the same. Nothing to do with measuring anything. Anyway, as you then go on to say that in fact they're exchanging "fewer photons", clearly the above is irrelevant, as I pointed out. Then, I'm sure I miss a lot in the real physics, but in this context I'm only looking at your description of how you see things. If something is missing, improve your description. e.g. in "Having a different density does not mean they cannot have the same temperature but 'the different densities' does translate directly into different amounts of energy." Clearly two bodies of the same temperature but different densities contain a different amount of energy. Equally clearly, if they are ideal Boltzmann black bodies they will radiate the same amount of em radiation per unit area. Then at two extremes there are: 1/ If both where in free space, the denser would take longer to cool than the less dense. But eventually both would cool to with a Planks whisker of absolute zero. 2/ If, somehow, all the radiation from each was absorbed by the other, they wouldn't cool. In between 3/ If the radiation from the denser was absorbed fully by the less dense, which in turn lost some energy to free space and some was radiated to the denser object, then, obviously, the amount of energy radiate into space wouldn't be going back to the denser object... it would cool - at some rate depending on it's energy density and the proportion lost in space - etc. till both where again withing a Planks whisker of absolute zero. Of course neither the earth (denser object) nor any part of the atmosphere are anything like that. They are not ideal Boltzmann black bodies. Nor are the em radiation flows so arranged... again, I recommend the SoD series to walk through the incremental complexities of reality.
  13. Re 362 les First you wrote:- "Clearly two bodies of the same temperature but different densities contain a different amount of energy." Then you wrote:- "Equally clearly, if they are ideal Boltzmann black bodies they will radiate the same amount of em radiation per unit area." Which is not the case at all. You are applying Kirchhoff's concept of a black body as a perfect absorber and emitter of radiation and it doesn't apply in this case, GHGs in the atmosphere get nowhere near this model because no gas, in any circumstances behaves like a black body. About your 'two extremes' 1/'free space' Really? Far too undefined; a gas in a vacuum? 2/'If, somehow, all the radiation from each was absorbed by the other, they wouldn't cool' Only if the density was the same and this is absolutely not the case for the vertical profile of the atmosphere. 3/'If the radiation from the denser was absorbed fully by the less dense' Simply impossible; see #2 From what you write 'Of course neither... ' you are clearly questioning the matter, like I am. Good luck! PS I have looked at SoD but he is far from having a good grip of the thermal characteristics of atmospheres - too much sloppy thinking.
    Response: [Daniel Bailey] If you believe SoD to be incorrect, please address that there, as SoD is well-established as an online reference tool known for accuracy in these matters. Until corrected, that status will remain.
  14. 363.... "Which is not the case at all. You are applying Kirchhoff's concept of a black body as a perfect absorber and emitter of radiation " Clearly. I was starting, as you observe, from an ideal model... always good to go back to basics - trying to clarify, at least for my self, your 'model'. which is why, in line with "and it doesn't apply in this case, GHGs in the atmosphere get nowhere near this model because no gas, in any circumstances behaves like a black body." I said that the earth and atmosphere are, indeed, not ideal black body objects. Did you actually read that?!?!? Just for clarity... 1/ no, I think it's clear the radiation is lost. end of. I said nothing of gases - I said 'object', it matters not what the object is made of (p.s. the atmosphere is a gas in a vacuum, but that's by-the-by) 2/... no. Read again they hypothetical - situation. In this highly simple model, not meant to represent the real world, but trying to clarify your 'explanation'... they are the same temp, the same surface area, same everything... except density... They would radiate and absorb the same amount of energy. simple. 3/ no it isn't. if I enclose a dense ball in a less dense shell, the shell will absorb all the radiation from the ball... where else is it going to go!?!? I am certainly not questioning the matter, nor - in general terms - the SoD model, with I rather like. But if you think his thermodynamics is amiss, I'd suggest you concentrate more on that then the e-m radiation bit... which, seems to me, you're struggling with.
  15. 364- in 2/ "same amount of energy" should read "same amount of energy / unit time".
  16. Re 364 les You wrote:- "Clearly. I was starting, as you observe, from an ideal model" If you start your model as a gas 'behaving like a black body' you really have nowhere to go because the odd CO2 (or H2O) molecule (at the density in the atmosphere - think of the thickness when brought to the surface and liquified) will never get anywhere near absorbing all the radiation from the surface. Even this observation is utterly irrelevant because any radiation absorbed by CO2 & H2O in the atmosphere is promptly re-emitted, if it wasn't the temperature of the intermediate layers would change. The atmosphere really is not, as far as radiation is concerned, different from a solid. Radiating solids also need internal heat transport to get the heat to the emitting suface. The emitting surface of solids is not an ideal, theoretical model; it is very complex and depends on the exact composition of the surface, which is seldom the same as the bulk material, it is an oxide or dirt or something.
  17. 366... clearly, and again, we all appreciate the complexities. And, again, I never said gas, I said 'body'. Boltzmann et al don't care what it's made of. Introducing both complexities and irrelevant factors (prompt photon emission? 'prompt' for e-m radiation, is a technical term regarding decay from excited states; does that matter here do you think?) prevent one from seeing the basic facts of the matter which bound such problems - which is the point of idealized models. I really did think going back to basics would help. But I can see that the road is to long and, really, rather than nowhere to go, I have better places to go!
  18. Re 363 You wrote:- "Response: [Daniel Bailey] If you believe SoD to be incorrect, please address that there, as SoD is well-established as an online reference tool known for accuracy in these matters. Until corrected, that status will remain." Sorry if I have upset anyone but I do not use links to put arguments I cannot support myself. Nor do I have any general position on what third party writers say; they may well be very good but giving a general approval seems to be fundamentally insecure since the question of interpretation arises. To put it another way, I require provable and reproduceable facts.
  19. damorbel The theory of evolution by natural selection is neither a provable nor reproducable fact; however that doesn't prevent the majority of biologists accepting that it is correct. Asking for definitive proof of any theory regarding the real world is unreasonable, as demonstrated by David Hume in the 18th century. We cannot observe causality, only correlation, and to move from an observation of correlation to assertion of causation we need to make assumptions. It is impossible to prove any theory about climate, they can only be disproved.
  20. Re 369 Dikran Marsupial Thank you for your contribution. I'm afraid I really don't see the connection between what you write and the application of the 2nd Law of thermodynamics to radiative heat transfer. I might add that the whole of quantum physics was started with the nummerous attempts to explain 'black body' radiation as identified by Gustav Kirchhoff in a way that matched experimental results.
  21. Damobel. Matching experimental results does not prove that the quantum explanation of black body radiation is correct (however it is repeatable). Quantum theory may be a very persuasive explanation for blackbody radiation, but it is not a provable fact. The point I was making is that if you require provable facts abdout the real world, you require the impossible. Science generally concentrates on the most plausible explanations, proof is generally reserved for mathematics. You wrote: "But all that you write makes it increasingly clear that the idea that the upper atmosphere (UA) can raise the surface temperature simply doesn't work." Nobody claims that the upper atmosphere does raise surface temperatures. The energy that causes the temperature of the surface to rise is from the sun, not the upper atmosphere, the upper atmosphere being warmer than outer space just means the surface looses energy to outer space less quickly and hence its equilibrium temperature is higher.
  22. Re 371 Dikran Marsupial, you wrote:- "Matching experimental results does not prove that the quantum explanation of black body radiation is correct (however it is repeatable)." I'm afraid I do not understand what you expect of quantum theory other than 'matches experimental results'. The essence of a good theory is that it enables new experiments to be devised which produce results that could not have been predicted with previous theories. Do you feel that quantum theory is deficient in this respect? Also you wrote:- "Nobody claims that the upper atmosphere does raise surface temperatures. The energy that causes the temperature of the surface to rise is from the sun" All significant climate energy comes from the Sun. The question is about how it is distributed. When you write:- "the upper atmosphere being warmer than outer space just means the surface looses energy to outer space less quickly and hence its equilibrium temperature is higher" Now there are various explanations for this and the presence of gases that radiate towards the surface in the infrared (GHGs) is the matter in hand. It is well established that matter cools when there is a net transfer of energy away from it. Correspondingly its temperature rises when energy is transferred into it. When you say "looses energy to outer space less quickly and hence its equilibrium temperature is higher", you are of course talking about a change in energy distribution. Now according to the 2nd law of thermodynamics a temperature difference is needed for a change of energy distribution and the also other way round, a change in energy distribution is always accompanied by a temperature difference. You seem to agree that it the CO2 in the atmosphere causes a surface temperature rise and many say that it is radiation from increases in CO2 that causes this. However this explanation would need the CO2 in the atmosphere to be warmer than the surface otherwise it will be the CO2 that is warmed by the surface (2nd Law). Personally I can see no other effect of CO2 that comes anyway near explaining temperature changes of any sort. Is there any other effect, in your opinion?
  23. damorbel - The point I was making is that you can reject any scientific theory regarding the nature of reality by demanding provable facts - but that doesn't make it rational or scientific behaviour. Consider a thought experiment, a blackbody object exists in a hard vaccum; initially the object is 273 degrees Kelvin (for the sake of argument), but it will cool by radiation to its environment which is at zero degrees kelvin. Now consider a second identical black body, also initially at 273 degrees kelvin, but now enveloped by a concentric hollow sphere of a blackbody material, leaving a gap all around of 1mm containing a hard vaccum. The shell is maintained at 272 degrees Kelvin. Which object will cool faster and why?
  24. Re 373 Dikran Marsupial, you wrote:- "Which object will cool faster and why?" The first because it radiates to a fixed 0K. Because of this it will eventually reach 0K The 2nd black body cools much less quickly because it radiates to a fixed 272K. Eventually the 2nd black body will reach the fixed 272K of the shell. Only a few people on the second body will notice much difference, 1K change is not very much. Those on the first body will not be caring much, they will have been frozen to death long ago! I am sorry but I can see no point in these examples. However there are interesting observations to be made. The first body has no heat source, internal or external and because it is 'black' it will cool at the maximum possible rate to 0K. If it is not black it will cool at a lesser rate, dependent on the surface emissivity. This could be very low indeed if it had a highly polished surface or even multiple surfaces; that is the principle behind using multilayer foil insulation (MFI) on spacecraft. The point is the MFI stops heat getting out of the craft by reflecting it back into it, so the craft does not cool down quickly. The MFI also reflects incoming radiation away from the craft, which is convenient because it stops the Sun heating it up too quickly. For these reasons MFI is sometimes called 'a thermal blanket'.
  25. damobel how does the first body know that it is radiating to 0K and know to emit fewer photons? BTW, this was the first stage in the thought experiment, lets not get ahead of ourselves, if we take it in small steps it will be easier to find out where our views diverge. The point of the examples is to find the point of divergence, which is the first step in determining which position is correct.

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