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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.

Last updated on 22 October 2010 by TonyWildish.

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Related Arguments

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 251 to 300 out of 1406:

  1. damorbel - Your physics are so far off it is difficult to know where to start. But I would strongly suggest that you read the following, by Dr. Roy Spencer, climate skeptic, which directly addresses this topic:

    Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still

    A quote from this, involving heated plates as a thought experiment: "Since the temperature of an object is a function of both energy gain AND energy loss, the temperature of the plate (or anything else) can be raised in 2 basic ways: (1) increase the rate of energy gain, or (2) decrease the rate of energy loss. The temperature of everything is determined by energy flows in and out, and one needs to know both to determine whether the temperature will go up or down. This is a consequence of the 1st Law of Thermodynamics involving conservation of energy."
  2. Re #251 you wrote:-

    " I would strongly suggest that you read the following, by Dr. Roy Spencer, climate skeptic, which directly addresses this topic:"

    I've read Roy Spencer on this matter, he's wrong.

    He uses the 'insulation' argument for CO2 etc. keeping heat 'in' (the atmosphere).

    This insulation/blanket argument is invalid because insulators only contain heat when it is in the container already, either because it was put there from outside, like putting hot soup in a flask, or there is a source of heat like combustion or radioactivity 'contained' by the insulation.

    Heat that arrives from outside the container, like the Sun/Earth arrangement is just as effecively kept out of a container (flask etc.) as it is kept inside.

    Eventually the contents of an insulated container revert to ambient temperature, your soup or coffee gets cold and your ice cream melts. What a shame!
  3. damorbel
    as I said and you apparently did not read, part of the spectrum. My impression is that you're not much confident with these simple physical concepts but you presume you know better.
  4. damorbel, If you're contesting Dr Roy's very basic physics, you need to go back there, read his post again, skip all the comments but read all Dr Roy's responses to those comments.

    Your argument about insulating heat being "in" the container already is irrelevant. What matters is that the earth has constant input of heat / radiation. It just happens to be from the sun. KRs reference @251 really is a fantastic one.

    Even if it takes you all day or longer, read it, reread it, copy it by hand, rewrite it - leave it for a while and then read it again. Whatever. If study techniques need to be applied, apply them. Dr Roy is a good teacher.
  5. damorbel - You should really read that article again. Carefully.

    Sunlight (like the electric heater in Spencers blog) passes through the atmosphere essentially unaffected by greenhouse gases, due to its spectra (primarily visible light). Thermal IR from the Earth, on the other hand, is strongly affected by greenhouse gas presence, which act as the 'blanket'. GHG's change the rate of loss, while input energy remains almost unchanged by them. Hence the atmosphere acts as a true "one-way mirror".


    Note that most sunlight passes through the atmosphere (affected mostly by Rayleigh scatter), while most IR does not (from Barrett Bellamy Climate)

    You've been pointed to this information multiple times, by multiple posters, yet you insist that your grasp of the physics is superior to the other contributors. I suggest you go read some of the very informative links you've been pointed to.
  6. BP "If OHC is supposed to be the true indicator of global warming and we have only seven years of reliable OHC data, then it is not cherry-picking to use what we have, is it?"

    Well actually I think true total OHC is long way from being tied down. However, can we assume that say 7 years down the track from here, and with good OHC data, if that OHC shows the warming trend, you will finally accept that we have a warming planet and its not just some measurement error?
  7. damorbel, you wrote:
    Even higher up in the atmosphere the gas density becomes so low and the chance of a photon being reabsorbed becomes correspondingly low.

    For thin atmospheres many photons emitted by H2O & CO2 do not get reabsorbed by adjacent H2O & CO2 molecules, some are reabsorbed by the surface but others are absorbed by deep space.
    I'm not sure you are thinking three dimensionally. Even way high up in thin atmosphere, H2O and CO2 molecules are emitting radiation in all directions, including down. So "adjacent" molecules include the ones below. Energy transferred by radiation down, warms the atmosphere below. That layer of atmosphere also radiates in all directions, including down, thereby warming the atmosphere below it. That cascade continues down to the surface.
  8. damorbel, you wrote "Heat transfer due to radiation goes only in one direction only, out into deep space."

    Yes, but the energy (via radiation) transfer in the direction down to the surface reduces the heat (net transfer of energy) from the surface up toward space. Which means that activity by greenhouse gases slows the cooling of the surface. Without greenhouse gases providing that offsetting (dare I say "back"?) radiation toward the surface, the heat (net transfer of energy) from the surface up would be larger, so the surface would cool faster, outstripping the replenishment of energy from the Sun, and consequently the surface would end up colder.
  9. FWIW, posted some molecular visualizations of the various states of the CO2 molecule here.

    The Yooper
  10. damorbel, you wrote "Since emitting (GH) gases absorb also there is no chance that any imbalance in thermal energy transfer will arise as described by 'back radiation.'"

    You are wrong. Greenhouse gases do not absorb 100% of the radiation they emit; they do not create a closed cycle that traps the energy within the greenhouse gases. They emit radiation in all directions, including down. If other greenhouse gas molecules in that downward direction absorb that radiation, those molecules in turn radiate in all directions including down. That downward cascade of radiation continues all the way to the bottom of the atmosphere, where due to the closeness of the surface, a substantial amount of the downward radiation avoids reabsorption by other greenhouse gases and so makes it to the surface.

    Also, greenhouse gas molecules transfer the energy they acquire not just by radiation but also by conduction. As KR explained on another thread, a CO2 molecule on average has 1000 collisions with other molecules in which it transfers energy, before it emits a photon. Those collisions are with not just other greenhouse gas molecules, but with any molecules--non-greenhouse gas molecules, liquid molecules, solid molecules. The recipients of those energy transfer in turn collide with other molecules of all kinds, thereby transferring the energy again. Thus collision (conduction) is an additional path for greenhouse gases causing warming.
  11. Re #253 Riccardo you wrote:-

    " part of the spectrum. "

    Your argument is that something can be 'black' in part of the spectrum. No it can't. On that basis a bright yellow surface can be called 'black' because it has nothing in the blue part of the spectrum, green has nothing in the red or blue part of the spectrum.

    When doing thermal and energy calculations black must mean 100% of the spectrum or you will get a false answer.

    A black body absorbs 100% of the spectrum by definition.
    A black body (above 0K) emits (with different intensity) in 100% of the spectrum, by definition.

    'Real' black bodies fall short of this 100% property for each of two possible reasons, they reflect like Earth or a mirror thus never quite 100%, or they transmit like glass or CO2 but never quite 100%.

    Glass is particularly interesting because it is obvious the reflection is where the refractive index changes. All materials, even gases, have a refractive index >1, consequently no material substance can behave according to the definition of a black body.
  12. Re #254 adelady you wrote:-

    " Your argument about insulating heat being "in" the container already is irrelevant."

    Not in reality. Read what Dr. Spencer puts in his OP:-
    "Imagine a heated plate in a cooled vacuum chamber, as in the first illustration, below. These chambers are used to test instruments and satellites that will be flown in space. Let’s heat the plate continuously with electricity"

    Dr. Spencer's model of a vacuum chamber makes nothing clear. I am familiar with the working of the type of chamber he shows. He shows an electric heat source, said to be at 160°F. How so? Lets say it is regulated but is its temperature uniform? Parts are much closer to the 0°F walls. The walls must be regulated to be a uniform 0°F over the inside surface (this is frequently done with liquid N2 but of course at −321 °F)

    Even if the heater is a perfectly uniform 160°F the temperature of the additional bar will only be uniform if it is a perfect (thermal) conductor, otherwise there will be a thermal gradient in it, a thermal gradient that depends greatly on the geometry of the entire installation.

    So is the 160°F source regulated? In which case the temperature is completely unaffected by the presence of the 2nd bar.

    In the case where the heater is a real heater i.e. it also has a geometrical thermal gradient, its temperature is not a uniform 160°F.

    But the nub of the matter is, what will change with the introduction of the brown bar? There are a number of scenarios:-

    1/ Suppose the brown bar is a very good but not perfect coductor of heat and it is so big it touches the container wall and the heater so that heat has a fairly easy passage. The temperature between the heater and the bar now depends entirely where you measure it because it depends on the thermal conductivity of the materials of the heater and the bar.

    2/ Let us now consider the case where the heater is a point and the bar also, so now conductivity has no role.
    This is a false proposition because a point has no surface area so it cannot, at 160°F or any other temperature, emit any energy.
    Lets ignore that and say there is a temperature gradient from a point at 160°F in the centre of Dr Spencer's container to the edge at 0°F. The temperature of the brown point in such an arrangement would depend on the position of the point and, since the point also has no surface area, it isn't affected by any emmission or absorption, just by the local photon intensity (the energy of the photons coming from the heater is not changed by the distance from the heater but the intensity i.e. photons per cm^2 is) falls according to the inverse square law.

    You will have noticed that there are many ifs and buts associated with my explanation but at least it tries to make something out of Dr. Spencer's quite unrealistic proposition.
  13. Re #257 Tom Dayton you wrote:-

    "So "adjacent" molecules include the ones below. Energy transferred by radiation down, warms the atmosphere below."

    Only if it exceeds the photon energy coming up. This should be obvious from the formation of the stratosphere where the energy absorbed by O2 & O3 warms the atmosphere with characteristic results, there is a consequent temperature inversion (the temperature rises in the stratosphere from about -50C to about 0C, depending on where you look) and suppression of convection.

    This stratospheric warming phenomenon when compared with the tropospheric lapse rate should make it very clear that the role played by absorption/emission via GHGs in the tropospheric temperature profile is non existent.
  14. Tom Dayton #257: "Energy transferred by radiation down, warms the atmosphere below."

    damorbel #263: "Only if it exceeds the photon energy coming up."

    Nonsense.

    There is just no logical way to arrive at that conclusion. You aren't just spouting ridiculous violations of basic physics, but also basic math.

    Let's say the "photon energy coming up" is 5 units per time X and the energy going down is only 1 unit per time X. The down photons are less than the up photons so you claim they cannot result in a warmer surface. Put the starting energy at the surface at 100 units.

    Ignoring incoming energy for simplicity, after 1 X has elapsed this would yield;
    100 - 5 + 1 = 96

    You claim this is no warmer than;
    100 - 5 = 95

    Which is just wrong.
    96 > 95

    Adding an extra step for incoming energy would obviously yield the same result. The surface is warmer with the down photons than without them. Early grade school level mathematics. Inescapably true. Yet you deny it.

    That's just pathetic.
  15. Re #260 Tom Dayton you wrote:-

    "That downward cascade of radiation continues all the way to the bottom of the atmosphere, where due to the closeness of the surface, a substantial amount of the downward radiation avoids reabsorption by other greenhouse gases and so makes it to the surface."

    So you maintain that the few cold photons coming down from any altitude can counteract the effect of the larger number of warmer photons radiated upwards to the extent that they raise the temperature of the surface 33C? Forgive me if I don't agree!
  16. damorbel - A few small notes:

    - Photons don't have temperatures (cold, warm), they have energies.

    - The warm plate in Spencer's example receives a certain amount of power from the electric heater; the temperature in that example is not 'regulated' (no thermostat), but is rather the temperature where the amount of thermal radiation from the plate (determined by the object shape, emissivity, and it's temperature) match incoming power.

    - An object (even if rather cool) warmer than the absolute zero of space will radiate some thermal energy; when that hits the 'warm' object, the incoming power to the warm object changes.

    - Power in matches power out, or the temperature of an object (the accumulated energy) will change.

    - A cool nearby object (or atmosphere) is still much warmer than absolute zero, changes the incoming power to your 'warm' object, and the temperature of the object will change until, once again, output power matches input power.

    It's really that simple.
  17. No, damorbel, neither I nor anyone else has ever said that. Instead, the photons that hit the surface add energy to the surface sufficient to compensate for (replace) some of the energy lost from the surface by the photons leaving the surface. The net result is that the surface cools less than it would have without the incoming photons. Therefore the surface ends up being warmer when it has incoming photons than it would have been without incoming photons.
  18. Its almost as though damorbel views IR radiation coming up from the ground as exerting a pressure such that any back radiation downwelling from CO2 cannot overcome. A good analogy would be the Deepwater Horizon oil spill in the Gulf of Mexico earlier this year. Pressure from the oil upwelling from below made it initially difficult for engineers to force drilling mud back down the pipe to stem the spill. If up-pressure is 100 GPS, how could forcing anything back down the stack work unless it first overcame the up-pressure from below? Unfortunately for damorbel, photons can and do travel simultaneously through the same "stack"...
  19. A few more comments on "hot" and "cold" photons.

    - Heat is the internal energy of molecular matter due to vibrations of the atomic nuclei with respect to each other. Photons are much more basic wave-particles and so cannot hold heat.

    - As KR says, photons do have energy, this is equivalent to frequency (see here for example. The energy of photons emitted by any molecule are equal to the separation of their quantum states and therefore fixed under all conditions.

    - Temperature affects the amount of photons emitted, since in a colder substance fewer molecules are in excited quantum states. This is, of course, my first point restated slightly.
  20. Re #266 KR you wrote:

    "Photons don't have temperatures (cold, warm), they have energies."

    The temperature associated with photons comes from the Planck energy distribution, the peak of the curve tracks the Wien displacement law

    In the same way as a gas has a distribution of energies following the Maxwell-Boltzmann distribution according to its temperature; photon energy follows Planck's law.

    But, since the bulk gas has the average temperature of all the molecules, this also means that the individual (isolated) molecule has its own temperature. In the same way a photon also has its own temperature, related of course, to its energy.

    This goes further; a planet orbiting a star is immersed in photons emitted by the Sun, the number of photons intercepted by a planet is reduced by the inverse square law but this is the only reduction, making the equilibrium temperature of a planet a function only on the Sun's (photon) temperature and the planet's distance from the Sun.

    The idea that planetary temperature is affected by its albedo is quite mistaken.
  21. damorbel #270: "The idea that planetary temperature is affected by its albedo is quite mistaken."

    Your belief that you have any idea what you are talking about is quite mistaken.

    If albedo is irrelevant to temperature... why exactly is it that black asphalt gets hotter than white cement on sunny days? Or what magical property would be at work such that a planet with no atmosphere covered in white cement would be just as hot as an otherwise identical planet covered in black asphalt?

    That's the thing which gets me about nearly everything you say here... it isn't just that it is wrong, it is that any person capable of observing the world around them should know it is wrong.

    First you argue that a non-zero energy flow produces zero heating... now that reflection and absorption yield the same result. It's gibberish.
  22. Re #270 CBDunkerson you wrote:-

    "If albedo is irrelevant to temperature... why exactly is it that black asphalt gets hotter than white cement on sunny days? "

    You must not confuse 'rate of heating' with 'final temperature'.
    Highly absorptive materials heat up quickly because they absorb a large % of the incoming radiation. Switch the radiation off and they cool correspondingly quickly.

    Highly reflective materials (high albedo) heat up slowly and cool down slowly in the absence of input; an example of this is a thermos flask with its highly polished surfaces.

    In either case, black or shiny, the final temperature, after stabilisation from whatever the initial temperature was, will be the average of the fluctuating sunshine or whatever thermal input there is.

    You must see from this that, with a fluctuating radiation input, the temperature of the asphalt will fluctuate about the average temperature far more than the contents of the flask but both will have the same average temperature.
  23. damorbel, I understand what you believe. I also know it is false. How you can not know it is false is a great mystery to me.

    If you fire identical lasers at a black iron plate and a mirror the black iron plate is going to get hotter than the mirror... no matter how long you wait. This is basic and obvious, because the mirror reflects more of the laser light and thus absorbs less energy than the black iron plate. Less incoming absorbed energy means a lower temperature at which the emitted radiation equals the incoming radiation (i.e. equilibrium point).

    The same is true of sunlight or any other radiation and any other matter it is striking. The lower the energy absorption rate the lower the final temperature of the object will be.

    Again, let's look at it mathematically;

    Incoming radiation: 100 units
    Object A reflectivity: 90%
    Object B reflectivity: 25%

    Object A reflects 90 units and absorbs 10. That 10 absorption heats up the object until it is emitting 10 units. At that point the 90 units reflected + 10 units emitted equals the 100 units incoming and the object is at equilibrium.

    Object B reflects 25 units and absorbs 75. That 75 absorption heats up the object until it is emitting 75 units. At that point the 25 units reflected + 75 units emitted equals the 100 units incoming and the object is at equilibrium.

    At equilibrium object A is emitting 10 units of energy and object B is emitting 75 units. Object B is thus much hotter than object A. Albedo has a direct and obvious impact on temperature.
  24. damorbel, the law about equilibrium temperature uses only the energy that the object actually absorbs. Energy contained in photons that reflect off the object is excluded by that law.

    Think about it: Energy that the object did not absorb does not exist inside the object, and so cannot be emitted by the object. The object does not "need" to emit energy it never absorbed.
  25. damorbel writes: This goes further; a planet orbiting a star is immersed in photons emitted by the Sun, the number of photons intercepted by a planet is reduced by the inverse square law but this is the only reduction, making the equilibrium temperature of a planet a function only on the Sun's (photon) temperature and the planet's distance from the Sun.

    The idea that planetary temperature is affected by its albedo is quite mistaken.


    damorbel then explains this idea further: Highly reflective materials (high albedo) heat up slowly and cool down slowly in the absence of input; an example of this is a thermos flask with its highly polished surfaces.

    Look, this is just wrong. It really is. If you're (understandably) reluctant to accept that from a bunch of anonymous strangers on the Internet, please just stop by whatever university is nearest to where you live and talk to someone in the physics, atmospheric science, astronomy, or earth science departments, and see if they can explain it to you.

    The earth receives short-wavelength radiation from the sun, and radiates away long-wavelength radiation. If the albedo of the earth increased, it will receive less short-wavelength radiation (visible, near-infrared). But this doesn't imply an immediate, corresponding reduction in outgoing long-wavelength radiation. Instead, the planet will gradually cool. As it cools, the flux of outgoing long-wavelength radiation will gradually decrease, in accordance with Stefan-Bolzmann, until incoming and outcoming radiation fluxes are once again in equilibrium, with the planet at a lower temperature.
  26. I suppose this discussion about the role of albedo on equilibrium temperature should be moved to the thread about albedo. Coincidentally, Rovinpiper just posted a question about exactly that, exactly there. I replied there.
  27. Re #273 CBDunkerson you write:-

    "Object A reflects 90 units and absorbs 10. That 10 absorption heats up the object until it is emitting 10 units. At that point the 90 units reflected + 10 units emitted equals the 100 units incoming and the object is at equilibrium.

    Object B reflects 25 units and absorbs 75. That 75 absorption heats up the object until it is emitting 75 units. At that point the 25 units reflected + 75 units emitted equals the 100 units incoming and the object is at equilibrium."

    Fair enough. But then you write:-

    "At equilibrium object A is emitting 10 units of energy and object B is emitting 75 units. Object B is thus much hotter than object A. Albedo has a direct and obvious impact on temperature."

    How so? A has only 10% absorption capacity, B has 75%. Now the absorption capacity is always equal to the emission capacity, after all the reflection part cannot emit as well as reflect, can it?

    So both objects have the same temperature, any difference would clearly break the 2nd Law of thermodynamics.
  28. Re #275 Ned you cite my post:-
    "Highly reflective materials (high albedo) heat up slowly and cool down slowly in the absence of input; an example of this is a thermos flask with its highly polished surfaces."

    Then you write:-
    "Look, this is just wrong. It really is."

    So thermos (vacuum) flasks don't work this way? Care to explain how they do?
  29. damorbel, at face value your statement is correct that "the absorption capacity is always equal to the emission capacity." But I think you meant something else--something incorrect.

    Here is a correct rephrasing: The reflected photons are irrelevant to the absorption and emission of the object, and therefore are irrelevant to the temperature of the object. The only contributors to the temperature of an object are the photons absorbed and the photons emitted.

    You will get exactly the same temperatures, absorptions, and emissions of objects A and B that you get in the scenario that CBDunkerson described, in this different scenario:

    (1) Make both objects A and B perfectly absorptive--no reflection, in other words albedos of zero.

    (2) Isolate A from B.

    (3) Give object A its own radiation source--a source that sends only 10% of the radiation that CB's original source did. Object A is absorbing 100% of that, so Object A is absorbing the same radiation (and therefore the same energy) that it was getting in CB's original scenario.

    (4) Give object B its own radiation source--a source that sends only 75% of the radiation that CB's original source did. Object B is absorbing 100% of that, so Object B is absorbing the same radiation (and therefore the same energy) that it was getting in CB's original scenario.

    (5) The temperature of Object A will be lower than the temperature of Object B.
  30. damorbel, a vacuum flask has an inner chamber inside an outer chamber. If the inner chamber is filled with a hot liquid, emission from that chamber can be reduced by making the inner surface of that inner chamber reflective. But emission from that inner chamber is not reduced by making the outer surface of that inner chamber reflective. Once that radiation has escaped from the inner chamber, it must get through the walls of the outer chamber, which can be reduced by giving the inner-facing walls of that outer chamber a reflective coating; that bounces the radiation back from the outer wall into the gap between the inner and outer chambers.
  31. Re #279 Tom Dayton you wrote:-

    "1) Make both objects A and B perfectly absorptive--no reflection, in other words albedos of zero."

    This destroys the whole matter. If you consider the case when albedo is zero there can be no effect due to albedo and there can be no confusion arising from the influence of albedo and 2nd law of thermodynamics.

    "(2) Isolate A from B."

    Why?

    "3) Give object A its own radiation source--a source that sends only 10% of the radiation that CB's original source did"

    I think you should be more precise and define the source better. I really do not understand why you need two sources to explain these concepts.
  32. damorbel, I posed my alternate scenario so you can see that the temperatures of the objects in that scenario are identical to the temperatures of those objects in CBDunkerson's original scenario. That should help you understand that photons reflected don't contribute to temperatures of the objects. Only the absorbed photons matter.
  33. Consider Tom's example using one source, but objects A & B are spaced so that they receive the amount of energy described and are isolated from each other (i.e. directly opposite each other & fully obscured by the source).
  34. Re #280 Tom Dayton you wrote:-

    "But emission from that inner chamber is not reduced by making the outer surface of that inner chamber reflective."

    A polished metal outer surface is an excellent insulator, the old fashioned silver coffee pot is a a good example, the modern chrome model is just as good because it doesn't need polishing so much.

    Another application of this principle is multilayer insulation Multilayer insulation stacks up reflective surfaces and is extremely effective.
  35. damorbel writes: So thermos (vacuum) flasks don't work this way? Care to explain how they do?

    The problem is that planets don't work this way. Read the rest of my comment.

    The point is that the incoming and outgoing radiation fluxes have different spectral distributions. A change in the visible/NIR albedo doesn't imply a corresponding change in thermal infrared emissivity.

    A lot of your comments in this thread seem to involve trying to analogize the earth-sun radiation balance to some object, like a thermos or asphalt or a coffee pot. With all due respect, that's not necessarily the best approach.
  36. damorbel, the reflective outer surface of an inner layer of multilayer insulation does not help by reducing that layer's emission. It helps instead by reducing that layer's absorption of the radiation emitted by the next-most-outer layer--radiation that this inner layer emitted, that was returned by the outer layer.
  37. Tom Dayton writes: damorbel, at face value your statement is correct that "the absorption capacity is always equal to the emission capacity."

    ... at a particular wavelength. Part of the problem with damorbel's argument here is that the incoming solar radiation has a very different spectral distribution from the outgoing longwave radiation. Absorptance in the visible/near-IR is not necessarily equal to thermal infrared emissivity.
  38. damorbel - The last 15-20 postings you have presented have made it increasingly clear that you do not have a firm grasp of the physics involved. That's not an insult - we all start somewhere.

    I, like Ned, strongly suggest you go check with your local university or other institute of learning, and find out some more of the basics.
  39. damorbel writes: Another application of this principle is multilayer insulation Multilayer insulation stacks up reflective surfaces and is extremely effective.

    Good grief! Did you even read that wikipedia page you linked to? How could you not have noticed that their explanation of how multilayer insulation works is exactly the process whereby backradiation from CO2 in the atmosphere raises the temperature of the earth above what it would be in the absence of that CO2!

    The process that you yourself cite as "extremely effective" is the exact same process that you claim violates the second law of thermodynamics!
  40. KR writes: damorbel - The last 15-20 postings you have presented have made it increasingly clear that you do not have a firm grasp of the physics involved. That's not an insult - we all start somewhere.

    Yes. And I would note that, as we saw with the Evil Waste Heat Thread, the usual SkS "skeptics" are once again standing by on the sidelines. Apparently they're willing to quibble endlessly over things like UHI, who wrote what in a snippet of somebody's email, etc. But they're not willing to speak up and help address the problems with even the most appallingly confused argument coming from the "skeptic" side.

    As always, it turns out that "climate skepticism" is rather asymmetric around here. The unwritten rule seems to be that "No SkS skeptic shall ever publicly disagree with another SkS skeptic."

    IMHO that's pretty depressing.
  41. Ned @ 290 - there's always the exception to the rule, albeit very minor, BP waded into Ken Lambert a few weeks back over a "theoretical observed" comment.
  42. And took hits for not observing the Skeptics' "Code Duello"

    The Yooper
  43. Re #287 Ned you write:-
    "... at a particular wavelength. Part of the problem with damorbel's argument here is that the incoming solar radiation has a very different spectral distribution from the outgoing longwave radiation."

    The wavelenth difference is indeed great but what that count for? Sure it indicates that the Sun/Earth system is in considerable disequilibrium. But the only significance of this is the nature of the disequilibrium, which is precisely what we are talking about, the contradiction of AGW/GHE 'science' and the 2nd Law of thermodynamics, exactly the OP topic of this thread.

    Further you write:-
    "Absorptance in the visible/near-IR is not necessarily equal to thermal infrared emissivity.

    A statement like that just confirms what I am arguing. If they weren't equal there wouldn't be an equilibrium temperature of any sort. If emissivity always was different from 1-a (a is albedo) then the temperature would never be stable, rising or falling according to the sign of the difference.
  44. damorbel, I can't get past even your second paragraph, which seems to be gibberish.
  45. Tom, it is gibberish. No "seems" about it.

    The fourth paragraph I kind of get -- damorbel is confused about the difference between emissivity and absorptance, on the one hand, and emitted energy and absorbed energy, on the other. The first two are unitless fractions, and the latter two are radiant fluxes. That confusion probably explains the seemingly erroneous conclusion about rising or falling temperatures.

    But the second paragraph? Yeah, it's nonsense. Let's try a few substitutions:

    "The salmon difference is indeed great but what that count for? Sure it indicates that the Estonia/marshmallow system is in considerable hypothermia. But the only significance of this is the nature of the hypothermia, which is precisely what we are talking about, the contradiction of platypus/unicorn 'badminton' and the 2nd Earl of Ambergris, exactly the OP topic of this thread."

    Does that make any more or less sense than the original? Hard to say!
  46. damorbel #277: "How so? A has only 10% absorption capacity, B has 75%. Now the absorption capacity is always equal to the emission capacity, after all the reflection part cannot emit as well as reflect, can it?

    So both objects have the same temperature, any difference would clearly break the 2nd Law of thermodynamics."

    I'm sorry, but what part of 75 units of energy is greater than 10 units of energy don't you understand?

    Yes, an object cannot emit more energy than it absorbs. Ergo, if the more reflective object is only absorbing 10 units of energy it can only emit 10 units of energy. Those 10 plus the 90 reflected equal the 100 total incoming and thus incoming and outgoing energy are in balance. Ditto the less reflective object except that it is absorbing and emitting 75 units of energy. 75 > 10. It has absorbed and is emitting more energy. Higher energy absorption and emissions equals higher temperature. For the 90% reflective object to be the same temperature it would have to be emitting the same 75 units of energy... which added to the 90 units reflected would be 165 units total... which runs afoul of the law of conservation of energy... an extra 65 units of energy can't just spontaneously appear from nothing.

    You seem to be arguing that absorption and emission are the same for all objects... rather than that they are the same for each object. That clearly isn't the case because an object can't absorb energy it has reflected away.

    Taking Tom's example of a theoretically 100% reflective object it is clear that it would absorb no energy... and thus would be at absolute zero. The more energy an object reflects the less it absorbs and the colder it is.
  47. Re #294 Tom Dayton you wrote:-
    " I can't get past even your second paragraph, which seems to be gibberish."

    My 2nd para. goes like this:-

    'Sun/Earth system is in considerable disequilibrium.'

    You may not be familiar with the thermodynamic meaning of the term 'equilibrium'; a thermal system is out of equilibrium when there is a temperature difference inside the system. This means that the entropy is below the maximum and there will be energy transport within the system according to the 2nd law of thermodynamics.

    The reason why the wavelength of incoming radiation is of no great importance is fairly simple; incoming radiation is either scattered (the albedo or reflected, if you like) or absorbed; the third possibility, transmitted, is not generally considered in planetary physics for reasons that should be self-evident.

    By definition the absorption does not affect the scattering, it is the scattering that affects the absorption. However it remains true that the scattering that gives the albedo its characteristic wavelength function i.e. its spectral characteristic.

    From this you will realise that the total scattering depends only on the amount of scatteing material present and the magnitude of the scattering is independent of the direction of arrival of the scattered wave; meaning the material that causes the albedo (scattered solar radiation) will have the same total effect on the emitted radiation, even though the response is in a different part of the spectrum. It is this that makes the emissivity and absoptivity the same in terms of power, even if not at the same frequency.
  48. No, damorbel, you are incorrect that "the total scattering depends only on the amount of scattering material." Scattering does depend on frequency of the radiation and the size of the reflecting matter.

    But your obsession with scattering is not relevant to absorption, which is the problematic behavior of greenhouse gases. Just to get you off of your reflection obsession, let's assume that you are correct that the same amount of radiation emitted by the atmosphere, water, and land toward space are reflected back, as the amount of radiation coming from the Sun that is reflected by all those. As that emitted radiation is on its way toward space, before it is reflected back down, some of it is absorbed by greenhouse gases. The absorbed radiation's energy can't be reflected, because it's not in the form of radiation any more. Only some of that energy immediately is turned back into radiation. So right there you've got a greenhouse gas trap of radiation and therefore a trap of energy, completely in addition to any reflection. Even if you were correct about reflection (you're not), the greenhouse gas absorption effect would exist, so increasing greenhouse gases would trap more energy.
  49. OK. So, damorbel recently wrote this gem:

    The wavelenth difference is indeed great but what that count for? Sure it indicates that the Sun/Earth system is in considerable disequilibrium. But the only significance of this is the nature of the disequilibrium, which is precisely what we are talking about, the contradiction of AGW/GHE 'science' and the 2nd Law of thermodynamics, exactly the OP topic of this thread.

    Now, he/she tries to explain it, but the only explanation is:

    (1) The difference in the wavelengths of radiation emitted by the sun vs. by the earth means that the sun and the earth are not at the same temperature.

    (2) This temperature difference means that heat will flow from one to the other.

    It should be obvious that this contributes nothing whatsoever of value.

    None of this justifies damorbel's nonsensical claim that planetary albedo is irrelevant to temperature ... and none of it has anything to do with AGW, let alone proving a "contradiction" between AGW and the 2nd law of thermodynamics.

    Damorbel, did you ever read the last paragraph of this comment? Did you understand it?

    I'd also note that damorbel has still not explained why he/she approvingly cites an explanation at wikipedia that explicitly relies on the exact same mechanism that he/she thinks violates the 2nd law of thermodynamics.
  50. #299: "nonsensical claim that planetary albedo is irrelevant to temperature"

    We went through a week or so of back-and-forth on the Chaos theory and global warming thread over 'climate calculators' that show specifically how albedo influences temperature. Seemed like a no-brainer at the time.

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