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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.

References

Comments

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Comments 876 to 900 out of 1393:

  1. damorbel (RE: 869), "Now be so good as to answer my question:- What % of the heat tranferred to the atmosphere from the ground by radiation:- 14%?......40%?.......90%?" I would but I'm not quite sure exactly what you're asking. How much is transferred kinetically?
  2. damorbel (@872) "Fred, what I like about 'Back Radiation' is that it goes straight into the surface, nothing is reflected, even though most of the surface is water with a refractive index of 1,33; I'm sure Fresnel is weeping in his grave!" Care to explain what the refractive index of a substance has to do with its reflectivity? I think the key quantity you should be focussing on is absorptivity, which by Kirchoff's law is always equal to emissivity at a given wavelength. In the thermal infrared (the pertinent wavelength), the emissivity of water is about 0.95. Source: http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html
  3. Re #877 Stu you wrote:- "Care to explain what the refractive index of a substance has to do with its reflectivity?" All materials, transparent or not have a refractive index, it is just that some are so opaque that transmission is dominated by absorption and incoherent scattering. In practical terms the propagation of light in transparent materials is governed by the Fresnel equations you will see on the link that a beam of light arriving at a 'change in refractive index' - normally one thinks of air/glass with RI for air = 1.0 and RI for glass = 1.4 - 1.7. At the surface of the glass some of the light is reflected and some passes into the bulk of the glass. In this case the RI of air is very low so the amount reflected depends on 1/the RI of the glass and 2/the angle the beam is incident on the glass. So, if back radiation really exists, it would be reflected by any surface, liquid or solid, just like the sunlight in Trenberth's diagram. Further you wrote:- " think the key quantity you should be focussing on is absorptivity, which by Kirchoff's law is always equal to emissivity at a given wavelength. In the thermal infrared (the pertinent wavelength), the emissivity of water is about 0.95" The only problem I have is the figure 0.95 you cite for water. Yes, that is what it says in your link. I have no idea how this figure was obtained, it cannot possibly be valid for water, other transparent materials in the list have similar figures. Light incident on material can either be 1/reflected; 2/absorbed; or 3/transmitted. You clearly know that absorptivity(a) = 1 - reflectivity(r), and emissivity(e) = a What is missing here is transparency(t), light can also pass through material (glass etc.) without being absorbed (much) or reflected (much). So how the compilers of the table in your link can give a figure of 0.95 as the emmisivity of pyrex glass, I do not know, it is absurd, it is higher than the figure given for carbon, is someone joking?
  4. Re #876 RW1 you wrote:- "I would but I'm not quite sure exactly what you're asking. How much is transferred kinetically?" That is the question I asked you. But I will accept evaporation and convection as 'kinetic'.
  5. damorbel: "In practical terms the propagation of light in transparent materials..." I presume this might be where you're having trouble... in the relevant part of the spectrum (thermal infrared) water is opaque, not transparent or even translucent. It absorbs all thermal infrared incident upon in within a few tens of microns of the surface (apart from the small fraction that it reflects). Glass is also opaque to thermal infrared. No-one's joking, you just need to be aware that absorptivity is a function of wavelength. I haven't been following your discussion, only jumped in here, but I see you say 'if back radiation really exists...' Well, can you explain measurements of downwelling clear-sky longwave radiation without it being 'back radiation'? If it's not back radiation, what is it? Here's a paper that quantifies downwelling clear-sky longwave radiation as measured during Antarctica's polar night: http://journals.ametsoc.org/doi/pdf/10.1175/JCLI3525.1 I do hope you don't just dismiss these two points out of hand by simply saying 'it can't be valid' or 'it's absurd'.
  6. Re #876 Stu you wrote:- "Glass is also opaque to thermal infrared. No-one's joking, you just need to be aware that absorptivity is a function of wavelength." True, but the link doesn't say that or give any figures for different wavelengths, so I do not understand what you are driving at. In the case of water, evaporation is by far the dominant heat loss mechanism for incident radiation, even Trenberth's diagram show's this. You can easily discover the figures for yourself by checking world wide rainfall; the heat needed to evaporate water is deposited in the atmosphere when it condenses. Further "Well, can you explain measurements of downwelling clear-sky longwave radiation without it being 'back radiation'? If it's not back radiation, what is it" I have read the paper you linked plus the description of the radiometer used and it merely confirms what I wrote in #875. To summarise, the energy associated with heat is constantly being exchanged between (adjacent) molecules both mechanically by (elastic) collision and electromagnetically by absorption and emission of radiation; somewhere there is a link to a paper by Einstein in this connection. I know this sounds pedantic but it has to be said. What the radiometer used in your paper measures is the radiation from gases that emit electctromagnetic radiation because they have T>0K. What the radiometer doesn't do is measure the 'upwelling' radiation from the same gases. If you could measure the 'upwelling' radiation from the same gases then you would be able to determine how much energy was being transferred and in which direction; only then would you be able to work out what was happening to the temperatures at the various locations of interest. Measuring the 'downwelling radiation' tells you almost nothing because you are far from sure about what kind of material (pressure; density; temperature etc. - for a gas) is emitting this radiation (thus you don't know its emissivity) and where it is. The difference between 'back radiation' and the real world is that, in the real world radiation is emitted and absorbed locally, as explained by Einstein, and the world of back radiation where it comes from 'up there'!
  7. damorbel @881, I find your comment bizarre on two counts. First, the absorption strengths at different frequencies for all common, and most uncommon atmospheric components is well documented in the HITRANdatabase. Because of that, except in those areas of overlapping absorption, the actual molecules emitting the down welling IR radiation is easily determined, as indeed is the average altitude from which the gas emissions originated. So to a significant degree of accuracy, the downwelling IR radiation tells us a great deal about the kind of material - including its temperature (from its brightness temperature) and pressure (from pressure broadening)- that emitted the radiation. Second, you talk as though observations have not been made from above the locations of observed downwelling spectra. Indeed, you do so despite the fact that downwelling and upwelling spectra from the same location are shown in figure 1 of the intermediate version of the article above. These observations have been made not only by satellite, but also by aircraft from a variety of altitudes. Here, for example, are spectra obtained by an aircraft flying at 3.7 km altitude over the North Sea: And a comparison of those spectra with model results: The large spike at the point at which CO2 emissivity is highest is because the cabin temperature was slightly higher than the air at that altitude, as discussed in the paper. It illustrates the point, however, that except at the frequency of greatest emissivity, radiation emitted in the atmosphere will travel hundreds of meters, and even several kilometers before being absorbed (on average). So, "to summarize", energy is constantly being exchanged in the atmosphere between adjacent molecules by collision and radiation, and between very distant molecules at a much slower rate by radiation. In fact, the concept of "back radiation" is not premised on the radiation coming from high in the atmosphere (you are wrong about that); but your claimed refutation is based on an error. This may be a case of an ugly fact wrecking Einstein's beautiful theory, but I suspect the fact is just wrecking your misinterpretation of Einstein rather than the master himself.
  8. "In the case of water, evaporation is by far the dominant heat loss mechanism for incident radiation, even Trenberth's diagram show's this. You can easily discover the figures for yourself by checking world wide rainfall; the heat needed to evaporate water is deposited in the atmosphere when it condenses. Yes evaporation is a big consideration, but I'm not sure why it trumps radiation. Locally (or indeed regionally), the latent heat flux can be massive, especially where cold air flows over warm water, but Trenberth's schematic shows evapotranspiration having an average value of 80 W/m2, compared to 396 W/m2 radiation: Indeed if you take water to have an emissivity of 0.95 at 300 K, then it's emitting 436 W/m2 at that temperature. And regarding that value of emissivity for water: "True, but the link doesn't say that or give any figures for different wavelengths, so I do not understand what you are driving at." The problem then is just that you didn't read the part of the page that says "As a guideline the emmisivities below are based on temperature 300 K" A substance at 300 K will have its peak emissive flux in the thermal infrared. "I have read the paper you linked plus the description of the radiometer used and it merely confirms what I wrote in #875. To summarise, the energy associated with heat is constantly being exchanged between (adjacent) molecules both mechanically by (elastic) collision and electromagnetically by absorption and emission of radiation; somewhere there is a link to a paper by Einstein in this connection. I know this sounds pedantic but it has to be said. What the radiometer used in your paper measures is the radiation from gases that emit electctromagnetic radiation because they have T>0K. What the radiometer doesn't do is measure the 'upwelling' radiation from the same gases. If you could measure the 'upwelling' radiation from the same gases then you would be able to determine how much energy was being transferred and in which direction; only then would you be able to work out what was happening to the temperatures at the various locations of interest." When you say 'energy associated with heat is... exchanged... by absorption and emission of radiation', you need to qualify it with the fact that only greenhouse gases do any absorbing or emitting. Again, it's not just that gases emit because they have T>0K. N2 and O2 do next to no emitting and their energy has to be transferred by collision to a GHG in order to be radiated away. Also, radiation is emitted isotropically; if you measure a certain downward flux, you can be sure that the 'layer' of the atmosphere you're measuring is emitting the same flux upwards. I think Tom Curtis has answered your other points nicely.
  9. damorbel - I generally do not respond to your postings, as they seem to follow 'trolling' principles (lots of red herrings, for example), but I will on the following howler. I apologize for my tone, but this has gone on far too long. "Measuring the 'downwelling radiation' tells you almost nothing because you are far from sure about what kind of material (pressure; density; temperature etc. - for a gas) is emitting this radiation (thus you don't know its emissivity) and where it is." Measuring downward IR at the surface tells you exactly what you need to know - the amount of energy coming down from the atmosphere at the air-surface interface, and hence the appropriate values for the energy budget. That's 333 W/m^2 downward IR, repeatedly and accurately measured. Your quibbling does not change the data! Your objections are really just red herrings, damorbel, a pattern you have repeated over and over and over again on this thread for multiple months - every time you get corrected, you change the subject. We know the energies, we know (as per Tom Curtis's post) where those come from in the atmosphere, we know the upwelling radiation, and we really really do know what the IR absorptivity/emissivity of water, sand, dry and wet dirt, etc., are. And we know the physics of CO2, EM, and thermodynamics. The radiative greenhouse effect is fully supported by all of this - no other viable explanations have been put forth. ---- Back to the actual thread: The "2nd law" objection to the greenhouse effect is based upon a mistaken notion (As per Gerlich et al) that a cool object cannot add energy to a warmer object, since net energy transfer is in the other direction - a classic Fallacy of Division, as net transfer is a statistical effect, not a restriction on individual photons. Hence the "2nd law" objection is false. Are there any actual issues with this that are on topic? Moderators, might I suggest that topic adherence be strongly enforced, as we're at >850 postings on this topic, many of which are serious digressions?
    Response:

    [DB] "might I suggest that topic adherence be strongly enforced"

    Agreed. In the words of the King:

    "So let it be written, so let it be done!"

  10. Damorbel keeps digging while trying to impress by citing Einstein (in German, for added effect). "every time you get corrected, you change the subject." You reckon KR? I'm still waiting on how the energy of a photon is affected by the temperature of the source. I'm not counting on a quantum theory revolution any time soon. Damorbel knows just enough vocabulary to impress and confuse the gullible but he has repeatedly demonstrated the lack of understanding of his own words. I am unmoved with both him and LJR. So far, in a lot less than the 850 posts contained in this thread, I've seen the howler I cite here, the one you addressed above and LJR asking what the Earth emissivity is in the SW. But they're ready to give lessons to everyone and caution against the bold "hypothesis" of an atmospheric GH effect. What a farce. It's almost as if they are in disguise, on a campaign to completely discredit the ridiculous GH effect skepticism launched by G&T.
  11. Re 885 Philippe Chantreau you write:- "I'm still waiting on how the energy of a photon is affected by the temperature of the source." I think it is sumarised inWien's displacement law
  12. damorbel - Your reference clearly demonstrates that identifying the temperature of an emitting object requires a statistical analysis of a spectra of photons, meaning that an individual photon does not by itself identify the temperature of the source. Which therefore indicates that your assertion here is incorrect. See my earlier post on photon absorption in that regard. Photons, unlike Arizonan citizens, do not carry ID cards indicating their origin. Do you have any comments on the actual issue of this thread, the (observed) exchange of energies between cooler and warmer objects and the implications thereof toward the radiative greenhouse effect?
  13. Moderator/DB - I believe damorbel's last comment actually was on topic, if wildly incorrect in it's implications: multiple assertions have been made that the photons from cooler objects hitting warmer ones do not reach / don't get absorbed / are turned away by a restrictive bar-room bouncer after being carded, in violation of the 1st law of thermodynamics and the physics of absorptivity for individual photons. So while (as I said) quite incorrect, it is somewhat relevant - if only to highlight the errors made by "2nd law" objections.
    Response: [DB] Apologies then. My initial read found context for his linked reference lacking in his comment. Sometimes one because conditioned to expect certain things...
  14. To fill out my last post: Any argument that photons from a cool object will not add to the internal energy of a warmer object they impinge upon due to their origin are specious; there is no such information encoded in the energy of an individual photon, absorption is based solely upon that energy and the absorption spectra of the impinged object. Hence any such "2d law" energy transfer objections are clearly wrong, and based upon a misunderstanding of the physics involved.
  15. Re #887 (& 849) you wrote:- "identifying the temperature of an emitting object requires a statistical analysis of a spectra of photons" Temperature is the property of an individual particle since it is the measure of the energy in that individual particle. I have mentioned this before and nothing about statistics is necessary to establish this. What you are thinking of is the temperature of an ensemble of many particles that are interacting strongly with each other (in equilibrium) so that the particles have a distribution of energies, the distribution is called the Boltzmann distribution. Your #849 directly contradicts the basis of quantum theory. You write:- "An object, at 20C, has an 80% absorptivity for 6 micron photons" For a start - absorbing takes place at the particle level. Further, to be absorbed a photon has to encounter the absorbing particle in the correct phase etc. But, according to quantum laws, the photon cannot be absorbed at all unless it has sufficient energy E (E=hv, 'h' is Planck's constant and 'v' is the characteristic frequency) This was the basis of Einstein's 1905 paper on the Photo-electric effect It is an experimentally observed fact that photons must have sufficient energy i.e. a sufficiently high equivalent temperature (discovered from the Wien displacement law) before they can be absorbed and cause an electron to be emitted. This applies in all areas of physics and chemistry, ozone is formed when O2 is split into 2 x O atoms, with light; but this only happens with ultraviolet light (UV) shorter than 0.2microns
  16. damorbel - Quite a kettle of fish in that post, which is to say red herrings. I note that by comparing it to facts: Absorption does not require photoelectric emission of an electron; that's a completely different question - Herring. Individual photons have energies, not temperatures - False statement. Estimates of object thermal emission temperatures require a spectra (statistical knowledge) and some idea of the emission spectra. An individual photon is incapable of supplying sufficient information - Incorrect statement and herring. "...absorbing takes place at the particle level. Further, to be absorbed a photon has to encounter the absorbing particle in the correct phase etc." - Herring. My statement in this post regarding absorption points out that if the absorption spectra of an object has a value of 0.8 for 6 micron photons, it has an 80% chance of absorbing any 6 micron photon, no matter the origin. - Your post is flatly, completely, wrong.
  17. Re #889 KR you wrote:- "Any argument that photons from a cool object will not add to the internal energy of a warmer object they impinge upon due to their origin are specious; there is no such information encoded in the energy of an individual photon" But you are making much of the statistical distribution of energies in a given sample of particles with a given temperature. To start with, to have a defined temperature the particles must be in equilibrium, the average energy must be steady. In this condition the average energy is the total energy of all particles in the sample divided by the number of particles in the sample. But there is no need to have a certain number of particles to make a sample, so one particle with the same energy as the average energy of all the particles also has the same temperature as the whole sample.
  18. Wein's displacement Law: 1. A black body emits photons at range of frequencies. 2. The number of photons emitted (otherwise known as the intensity) at any particular frequency will vary. The plot of the intensity versus the frequency is the "black body spectrum" 3. The temperature of the black body can be calculated using the frequency of the most common (or more intense) photons. This is notated as υmax indicating the frequency at which there is maximum intensity. 4. The photons of this frequency do not have this, or any other, temperature.
  19. damorbel - "one particle with the same energy as the average energy of all the particles also has the same temperature as the whole sample. " (Emphasis added) Flatly wrong. That photon (the subject of the discussion) will have the average energy of the whole sample. Any individual photon can come from anywhere in the emission spectra of the object (high to low), and from any individual photon there is no fixed determination of temperature. damorbel - Are you asserting that an individual photon (with a particular energy) can be used to identify the temperature of the object that emitted it, thus affecting it's absorption? Or that the possibility of absorption is not a function of photon energy and absorption spectra? If so, you are sadly mistaken. Boltzmann thermal distributions are off-topic distractions. Energy transfer from cooler to warmer objects is the issue raised in this thread.
  20. Re #891 KR you wrote:- "Absorption does not require photoelectric emission of an electron; that's a completely different question - Herring." That was the point of departure for quantum physics in 1905, something I made quite clear. But it is now generally accepted that all electromagnetic interactions take place in the domain covered by quantum physics - it is called quantum electrodynamics (QED) http://en.wikipedia.org/wiki/Quantum_electrodynamics But you are making much of the statistical distribution of energies in a given sample of particles with a given temperature. To start with, to have a defined temperature the particles must be in equilibrium, the average energy must be steady. In this condition the average energy is the total energy of all particles in the sample divided by the number of particles in the sample. But there is no need to have a certain number of particles to make a sample, so one particle with the same energy as the average energy of all the particles also has the same temperature as the whole sample. you wrote:- "Individual photons have energies, not temperatures - False statement" We have been here before, the energy of a photon is available for electromagnetic particle interactions which are governed by quantum laws just the same, so it doesn't matter if the particle energy is defined in electronvolts or temperature, my use of the word 'temperature' to define the energy of a photon is common and justifiable. (In single particle interactions as in high energy physics individual particle energy is frequently just given in Joules. You wrote "Estimates of object thermal emission temperatures require a spectra (statistical knowledge) and some idea of the emission spectra. An individual photon is incapable of supplying sufficient information - Incorrect statement and herring" What do you mean by 'object'? (First line above.) Is a particle not an 'object'?
  21. damorbel - I am quite frankly appalled by your last post - I suspect I am not alone. The photoelectric effect (and in fact your entire post on it here) is a complete red herring, irrelevant to absorption of thermal energies. QED is the basis of absorption spectra determination, but that's a different question - also irrelevant. You are misusing "temperature" - particles have energies and velocities, photons have energies, an ensemble of particles have temperature, an ensemble of photons have spectra, the last of which can (with some idea of the emitting object spectra) be used to identify emitting object temperature. Your personal re-definition and misuse of a term is not in any way a compelling argument against thermodynamics. Do you have any answer to the two questions I posed at the end of my last post? The ones actually relevant to energy exchanges between cooler and warmer objects (atmosphere and surface)? Questions on the actual subject of this thread?
  22. Re #893 Phil you wrote:- " 3. The temperature of the black body can be calculated using the frequency of the most common (or more intense) photons. This is notated as υmax indicating the frequency at which there is maximum intensity. 4. The photons of this frequency do not have this, or any other, temperature." The diagram on this page Wien's displacement law (top right) has a temperature for each peak of the Planck curve - that is the temperature associated with the average energy of the photons.
  23. damorbel - "The diagram on this page Wien's displacement law (top right) has a temperature for each peak of the Planck curve - that is the temperature associated with the average energy of the photons. " Actually, the peak of the curve is the mode, not the average; the two are not identical unless the distribution is symmetric. That's a fairly common error. However, both mode and average are statistical values. You have now contradicted yourself; individual photons do not convey the temperature of the emitting object, as that requires a statistical ensemble.
  24. darmorbel @895 We have been here before, the energy of a photon is available for electromagnetic particle interactions which are governed by quantum laws just the same, so it doesn't matter if the particle energy is defined in electronvolts or temperature, my use of the word 'temperature' to define the energy of a photon is common and justifiable. (In single particle interactions as in high energy physics individual particle energy is frequently just given in Joules. So you agree that photons have an energy, which is related to their frequency, and you agree that the "back radiation" is emitted from molecules that have previously absorbed surface IR radiation (which must of the same frequency). So you agree that your previous notion of surface IR radiation of having "warmer photons than the back radiation" is wrong !
  25. Phil - "...the "back radiation" is emitted from molecules that have previously absorbed surface IR radiation (which must of the same frequency)." Actually, Phil, that's not correct. By my estimates each absorbing molecule at sea level collides with an absolute minimum of 1000 other molecules before it has a chance to emit, each modifying it's internal energy. (10^9 collisions per second, 10^-6 seconds minimum before emission) Emission spectra of the atmosphere depends on the temperature and makeup of the emitting gas - the varied energy level drops resulting in EM emission reflect the varied energies of the emitting molecules and their possible transitions through a radiating level. On average the photons from the surface will be of higher energy (shorter wavelength) than those from the air. But that's irrelevant to the fact that photons from the air are indeed absorbed by the surface, and that this absorption (by the 1st law of thermodynamics, conservation of energy) affects and slows the total, net energy transfer to the atmosphere and hence to space.

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