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Climate Hustle

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

Last updated on 5 July 2015 by skeptickev. View Archives

Printable Version  |  Offline PDF Version  |  Link to this page

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 601 to 650 out of 1494:

  1. each time a collission occurs, there is a net loss of energy to gravity What?? gravity gains energy? I dont understand this comment at all.
  2. Camburn @598, the atmosphere does not cause the Earth to accelerate in space, so there is no loss of energy to gravity. In fact, the paths of all molecules are accelerated to towards the surface, but because the atmosphere is evenly distributed over the Earth (to a reasonable approximation), the Earth is accelerated in all directions equally by this interaction, and hence not accelerated at all. The energy gained by gravitational acceleration of particles in the atmosphere is returned by elastic collisions at the surface.

    Further, collisions do not expend heat energy. You are assuming the statistical properties of a group of molecules must also be the properties of individual particles. The energy involved in the collision of particles is the kinetic energy of the particles, plus their individual rotational and vibrational energies. Collisions can result in the redistribution of energy amongst the particles, and the rotational/vibrational states, and can result in the emission of IR radiation. But there is no "heat energy" as a distinct property from the kinetic energy, and rotational/vibrational energy of the particles; and sum of the energies of the particles (including photons) involved is always conserved.
  3. A loss of energy to gravity???!!??? What in the world is that?
  4. Tom C@602:
    1. Conservation of energy.
    Gravity is a force. Everything in our atmosphere/earth sphere is affected by gravity. It consumes energy. Without it, we would all float etc.

    Yes, kinetic energy is kinetic energy. But even kinetic is the result of heat and can be converted to heat. Friction is an example.

    When two molecules, even tho of minisucle weight, collide, energy is not only transferred, it is expended because they do not bond, rather they collide and go a different direction and they will go in a different direction at a slightly slower speed. That energy is absorbed by gravity.

    Even our atmosphere, while of atomic weights, requires an expenditure of energy to stay aloft. This energy comes from the sun as I know of no other source.

    A physics prof was trying to get this into my mind in college. Not sure I ever really totally grasped what he was trying to show, but rattling my old cobwebs I am trying to understand it still.

    It gets back to including all energy in any equation talking about energy.

    Maybe he was all wet, but the older I get the more I think he understood something quit clearly that I didn't.
  5. Tom Curtis (RE:597),

    "1) The energy absorbed at the surface is the Incoming Solar Radiation absorbed at the Surface (approx 161 w/m^2) plus the Back Radiation absorbed at the surface (approx 333 w/m^2)."

    OK, here is my question then to you: If you agree that the atmosphere cannot create any energy of its own, and 78 W/m^2 of the 239 W/m^2 (239 - 161 = 78) entering the system never reaches the surface, then where is the 333 W/m^2 of back radiation coming from if 239 W/m^2 is also leaving the system? 40 + 78 + 121 = 239 W/m^2 leaving. 396 W/m^2 - 239 W/m^2 = 157 W/m^2 emitted down to the surface. 161 + 157 = 318 W/m^2 at the surface (396 W/m^2 required). Also, 157 W/m^2 + 97 W/m^2 = 254 W/m^2 (333 W/m^2 required).
  6. RickG @ 591

    "Specifically, what is it that you guys don't understand about this diagram? You are trying to make it into something that it is not."

    I'm not trying to make this diagram into anything. I question the basis by which the energy is stored in the atmosphere...if not temperature. So I ask again how do you know is there...can it be measured? And by what means is the atmospheric energy stored?
  7. Sorry Camburn, it still makes no sense.
    You say "Gravity is a force."
    Then you say "That energy is absorbed by gravity."

    So energy is absorbed by a force. Please elaborate on the physical process there, I'm at a loss.
  8. Tom Curtis (RE: 597),

    "Saying that 239 w/m^2 becomes 396 w/m^2 is to directly assert the non-conservation of energy."

    How do you figure? I said the 239 W/m^2 entering the system becomes 396 W/m^2 at the surface. The amount leaving is still 239 W/m^2. Power in = power out = Conservation of Energy.
    Response: [DB] Fixed unclosed html tag.
  9. Tom Curtis (RE: 597),

    "5) The surface radiation is a function of temperature and emissivity, which is not 1 at any location, though very close to 1 at most."

    OK yes, temperature and emissivity, which for all practical purposes is 1 because the surface is a near perfect black body radiator.
  10. Philippe@607:
    Now you are understanding why I didn't understand. I was trying to explain his reasoning, which I did a poor job at doing.

    His lectures etc on that still tickle my brain tho. I know there is something there that I didn't understand.
  11. Tom Curtis (RE: 599),

    "RW1 @595, all energy "emitted" from the surface is radiative only because we do not talk about "emitting" convection, or evapo/transpiration."

    OK, this is the crux. Is 396 W/m^2 radiated from the surface or not?

    "Not all energy flux from the surface, however, is radiative. In fact, only 80% of it is."

    Agreed.

    "And some of the energy flux carried by convection and evapo/transpiration makes its way to space. Do you deny that?"

    No. As I said, trade offs do occur but an equal and opposite amount less is then returned to the surface. Less energy returned to the surface than what initially left in kinetic form will cool the surface, which reduces the amount of emitted radiation by an equal and opposite amount.
  12. Camburn @604:

    1) First and most importantly, the action of a force on an object never consumes energy. It only ever changes its form, and its location. This concept is actually very hard to grasp for most people because it is counter intuitive based on our everyday experience. In our everyday experience, a body in motion will come to rest unless an external force acts on it. This is the experience fairly well captured by Aristotle's laws of motion, but it is an illusion based on not taking into account the effects of friction and air resistance. In space (or by careful experiment) we can see that Newton's laws reign supreme, and that:

    a) A body at rest or in a state of steady motion will remain at rest or in a state of steady motion unless an external force acts on it;

    b) Force equals mass by acceleration; and

    c) For every action, there is an equal and opposite reaction.

    So, if we consider a gas molecule heading towards space. Gravity indeed acts on it, and it decelerates, losing energy in the process. But gravity equally, and oppositely acts on the Earth at the same time, accelerating it so that it has more energy. In fact, it gains exactly as much energy as the molecule loses.

    If we follow the path of the molecule under gravity, and ignore all the other collisions (which cancel out in effect), eventually the molecule will collide with Earth, resulting in another exchange of energy that cancels out the exchanges that took place under gravity. (To tell this story completely accurately, I would need to include gravitational potential energy, which shows up in General Relativity as very small changes in mass; but this is just a comment on a blog so you'll have to settle with the short and dirty version).

    Anyway, gravity does play a crucial role in the atmosphere. It is because of gravity that the atmosphere is dense near the surface, and thin away from the surface. A secondary consequence of this is that molecules move rapidly (the gas is warm) near the surface and slowly (the gas is cool) away from it; and it is possible to predict those temperature relationships using Newtons theory of gravity. But it does not result in a loss of energy, because all forces only result in the exchange of energy, never its loss.

    2) Heat that is generated by friction is just kinetic the energy of the molecules that make up the substance being warmed as they vibrate in position; or in the case of a gas, move around in their container. But because that heat is just the motion of those small particles, the small particles themselves do not lose energy to friction.

    3) The molecules in a gas do need to have significant kinetic energy to stay aloft. That is the energy of motion that they have because of the temperature of the gas. If the gas cools, they slow down, and will eventually not be able to escape the surface (which will mean the gas has condensed as a liquid or solid). However, they have this energy, the do not need to expend energy because any forces just rearrange the energy, not dissipate it. Further, they do not have frictional energy losses. The do lose energy by the radiation of IR gases, which needs to be replaced directly or indirectly from the sun - but for practical purposes they do not lose energy over and above that.
  13. Tom Curtis (RE: 599),

    "And some of the energy flux carried by convection and evapo/transpiration makes its way to space. Do you deny that?"

    Let me try to explain this a little better. Yes, some of the kinetic energy flux into the atmosphere from the surface by thermals and latent heat can end up being radiated out to space. However, if this occurs, it also must result in an equal amount of energy less being returned to the surface. If less energy is returned to the surface than the amount of energy that left the surface, the surface will cool. As a result, the cooler surface will radiate an equal amount less than the amount radiated out to space from latent heat and thermals.

    Thus, the net effect of latent heat and thermals on the radiative budget is zero, as Conservation of Energy dictates, because all the energy leaving at the top of the atmosphere is radiative.
  14. Tom@612. I'm saving that one.

    Your physics explanations are always good, but this one has bells, bows, balloons and buzzers.

    Outstanding.
  15. RW1, would you consider the following simple model:




    The model is a simple box with mirrored back and sides. We will assume the mirrors are 100% efficient, and reflect all light. The box is covered with a material that is completely transparent to all light coming from outside the box. However, it is half mirrored on the inside, reflecting exactly 50% of all light from the inside of the box, and transmitting without loss the remainder.

    The box is not a model of the greenhouse effect; but it does have the virtue that any thermodynamic issues raised by the greenhouse effect are also raised by this box, but in a simplified form.

    In this box, we have the following equalities:

    1) Incoming light (A) = Outgoing light (C) (by virtue of conservation of energy).

    2) Light reflected from the lid (D) = light transmitted by the lid (C) = Outgoing light (by virtue of the defined half mirrored property of the lid).

    3) Light reflected from the back = light striking the underside of the lid (B) = light transmitted by the lid (C) plus light reflected by the lid (D) (by virtue of conservation of energy).

    Therefore
    4) Light reflected from the back of the box (B) = light reflected from the back of the lid (D) plus Incoming light (A) = 2 x A (again, by conservation of energy.

    By simplifying the situation, ie, by getting rid of any concerns about convection and light absorbed by the atmosphere etc, we should be able to raise any issues you have with the consistency of the GHE with the laws of thermodynamics without getting hung up on trivia. Do you agree? Do you also agree with me that this simple model does not violate any laws of thermodynamics?
  16. 615 Tom Curtis

    I like, understand and agree with your model, but..
    The box is not a model of the greenhouse effect; but it does have the virtue that any thermodynamic issues raised by the greenhouse effect are also raised by this box, but in a simplified form.

    it doesn't do the thermodynamics justice.
    (For fun) To do that the 'mirrored' walls should be perfectly black, perfectly insulated and with infinite heat capacity; the radiation hitting them would be perfectly converted into heat and the walls would black-body radiate. Then if the half-mirrored front was, instead, transparent at wavelength λ (give or take), and opaque else where: the box would heat up until radiance from the walls reaching the window at λ equaled the radiance coming in...
    With a little more messing around (2nd window, transparent at a different wavelength, etc.) you'd have a green house 'box'.
  17. les @616, with some minor amendments, and one major one, your variant box would much better model the greenhouse effect. However, it is not clear that it raises any issues of thermodynamics not raised with my simpler box. So, unless it is clear that we cannot get agreement that no additional thermodynamic arise in the simple model, than I would rather stick with that. If it becomes clear that we cannot get that agreement, we should introduce one variation at a time until we can isolate the actual issue in dispute.

    The one major disagreement, by the way is the infinite heat capacity. An infinite heat capacity would imply the walls of the box would neither heat nor cool, no matter who much radiation they absorbed or emitted, which would itself violate the laws of thermodynamics (I think) and certainly not accurately model any real physical process.
  18. Tom@617 ... absolutely, hence the "for fun" remark - it was only that, sometimes, one good model deserves another.

    as for the heat capacity - I hesitated long and... then well, didn't have time to think that one out. Indeed the heat capacity would determine how long it would take to "heat up until" - and an infinite heat capacity would result in it taking an infinite time to heat up.

    Boundary conditions, hu? always a good way of doing a sanity check on a model - which is the point of your model; so back to that.
  19. Tom Curtis@615

    1. A = C ok

    2. C can not equal D without violating the 2nd law. Otherwise you have doubled your light/energy with a mirror and a filter.

    Light can not brighten due to it's own reflection. If your box was fully enclosed such all surfaces are reflective save two small aperture. One aperture to receive light the second to radiate light, do you really think the light/energy will increase beyond it's input?

    Now change your perfectly reflective interior to one with an emissivity of 1 (black body). Do you believe C will be grater then A. Again no. Black body emission represent the maximum conferred energy for light input. Therefore, a surface with emissivity less then one will NEVER radiate more then it's black body equivalent...regardless of it's own reflection or it' own re-radiation (back radiation).
  20. 619 Ryan - hint: the walls are reflective.
    All the photons which do D (i.e. don't escape when they hit the front window) will bounce around (lets say B') till they hit the window again and are either C' or D', then B''/C''/D'', B'''/C'''/D''' etc. till they do.
  21. "till they do." - oops, sorry should be "till the sum of Ds = A"...

    Or do you think it's possible that a significant number of photons will bounce around the box for ever without leaving?
  22. les@621

    Like I said C can equal A, but C cannot equal D.
  23. 622 Ryan.. Yes, and I'm pointing out that, really - they are (all) integrals.

    That may be a problem of notation rather then violation of a fundamental physical property.
  24. L.J.Ryan

    C will equal D. This scenario is very similar to the spreadsheet I posted here. You can model Tom's box using a 5 column spreadsheet thus;

    In row 1 type A, blank, B, C, D (to represent the quantities on Tom's diagram)

    In row 3 type 100, 0, =A3+B3, =C3*0.5, =C3*0.5
    In row 4 type 100, =D3, =A4+B4, =C4*0.5, =C4*0.5
    Copy row 4 into the next 30 lines of the table. You will find equilibrium reached after about 17 iterations and that Tom's calculations match in every detail.

    To check the conservation of energy you must let the accumulated energy in the box dissipate. To do this copy row 30 into 31 and set cell A31 to 0. Copy row 31 to the next 15 or so cells. If you sum column A and D (don't forget, column D represents Tom's arrow C) you will find they are equal.
  25. Tom, I really like this. Same idea as Science of Doom example but much simpler.
  26. L.J.Ryan, I forgot in my post 615 to mention the equalities only apply in the equilibrium condition.

    To explore the non-equilibrium condition, let us assume time steps equal to the average time it takes light to cross the depth of the box once. We use the average time because the photons may be at different angles, and hence have different path lengths. Let us also assume that 100 photons enter the box in each time step. We set the initial time step,0, to the time photons first start entering the box, but before they strike the back wall of the box. In that case, the number of photons in each of A, B, C, and D for progressive time steps are:

    Following Phil's suggestion, I have modeled this on a spreadsheet, using the following formulars:

    Column B2 and subsequent: 100, column C2: 0, Column C3 and subsequent is the sum of columns B and D for the preceding row. Column's D and E2: 0; and for columns D and E3 and subsequent, 0.5 times the value of column C in the preceding row.

    The first twelve steps show as follows:


    STEP A B C D
    0 100 0 0 0
    1 100 100 0 0
    2 100 100 50 50
    3 100 150 50 50
    4 100 150 75 75
    5 100 175 75 75
    6 100 175 87.5 87.5
    7 100 187.5 87.5 87.5
    8 100 187.5 93.75 93.75
    9 100 193.75 93.75 93.75
    10 100 193.75 96.875 96.875
    11 100 196.875 96.875 96.875
    12 100 196.875 98.4375 98.4375

    Clearly there is a problem for the spreadsheet that it allows fractional photons. What would happen in the real case is that occasionally 99 photons would leave the box, and occasionally 101, but typically 100 would leave the box. Furthermore, the mean value of photons leaving the box once the equilibrium state is reached would be 100. So, ignoring the quirk of fractional values, it is plain the system quickly approaches the state described in my 615.

    Do you have any problems with that?
  27. C Truth @627, the ideal gas law is included in atmospheric physics in calculating the lapse rate, as shown in this university lecture, and as explained by me very briefly in 563 above. It is also included in analysis of convection, but as convection in the atmosphere is what establishes (on average) the lapse rate, that is saying the same thing. It follows that any explanation of the green house effect that incorporates the environmental lapse rate already incorporates the gas law.

    As previously discussed in this thread, the standard theory of the greenhouse effect incorporates the lapse rate as an essential element of the theory. So, yes, understanding the gas law can provide insights into the greenhouse effect, and those insights were discovered decades ago, and are the basis of the modern understanding of the greenhouse effect.
  28. Tom Curtis @626

    "Do you have any problems with that?"

    As I said A=C, so with that we agree. However, energy can NOT accumulate within the box, B can not equal 2A, no way can't happen. Your scenario is a light/energy doubler.

    If you change your filter to reflect 75% what happens?

    Take it further, enclose a flashlight within a completely and perfectly reflective box, at what point is there infinite energy therein?
  29. Tom Curtis (RE: 615),

    "By simplifying the situation, ie, by getting rid of any concerns about convection and light absorbed by the atmosphere etc, we should be able to raise any issues you have with the consistency of the GHE with the laws of thermodynamics without getting hung up on trivia. Do you agree?"

    No, I don't agree that it is trivia. Understanding the energy flows relative to the radiative balance is absolutely fundamental to the entire GHE and ultimately surface temperatures (i.e. how much surface emitted radiation is coming back from the atmosphere and how much is passing through).

    All I'm saying is latent heat and thermals are just redistributing energy around the thermal mass of the system - mostly from the tropics to the higher latitudes. The bulk of this energy condenses to form clouds, weather systems and returns as precipitation. Any amount of it that ends up radiated out to space is equally offset at the surface by a lesser amount returning, which cools the surface. All the energy flows are constant, thus this effect is already accounted for in the 396 W/m^2 emitted at the surface.
  30. LJ@629>However, energy can NOT accumulate within the box,

    This is not correct. Don't forget that the flashlight is continuously outputting radiative energy (converted from energy stored in batteries). If the energy cannot escape, then yes of course the radiative energy accumulates in the box. Otherwise you would violate conservation of energy, because if the flashlight is outputting energy and it does not escape and it does not accumulate, then it must have been destroyed.

    It would of course stop accumulating after the flashlight runs out of battery or shuts off. Fortunately in the earth system analogy our "flashlight" will not run out of juice for a very long time.
  31. e @631

    Let's assume the flashlight radiates with a 1W bulb for 100 hrs. How much energy is contained within the box at the end of a 100 hrs? How long to accumulate a gigawatt?
  32. 1J/s for 100 hours, thats 360kJ accumulated. Of course your torch also absorbs energy so guess that is going melt at some point.

    RW1 - this lightbox example is simple demo of how not to make inappropriate inference about energy from energy flux through different surfaces. Do you agree with light box as TC has set it up?
  33. Keep in mind, everyone, that the gain of the box is 0.5, less than one, and hence a run-away feedback is not possible. Please see Does positive feedback necessarily mean runaway warming for details.

    Tom Curtis, I like your example. I did much the same thing on this thread earlier, except adding a value (which could be a column) of emissivity (0.612 for Earth, as measured), where your "C" was (1.0 - emissivity) * B, and "D" was emissivity * B.

    If you do that with 240 as input, the results are quite interesting, as per Trenberth 2009.
  34. RW1 @630, as it is difficult to carry on two discussions at once on the same thread, do you mind holding of on the discussion of the relevance of the light box until we have settled that it does not violate any law of thermodynamics?

    And to that end, do you agree that the light box does not violate any law of thermodynamics?
  35. scaddenp @633

    360kJ is at minimum expended by the batteries. Surely you would accumulate more then 360kJ within the box. After all, the claim is reflected light (B from Tom Curtis's diagram) is twice the input.

    Why the discrepancy?

    So lets step it back, if your box was fully enclosed such all surfaces are reflective save two small aperture. One aperture to receive light the second to radiate light. Close the output while receiving 1W at the input.

    The energy accumulated within the box after 100 hrs is what, 360kj?

    If the first aperture is then closed does the box now contain 360kj of light?
  36. L.J.Ryan @629, let us consider this step by step:

    1) Consider the box as described, but without any lid. In this case all the light will reflect of the wall of the box and exit through the aperture where the lid was.
    Is that correct?

    2) Now consider the case in which we place the lid on the box, but at an angle so that all light reflected of the lid will leave the box through some other aperture. In this case, the amount of light leaving the box through the lid will be half of that which enters, while the amount that is reflected by the lid and leaves through the other aperture will also be half of that which enters the box.
    Is that correct?

    Do either of these scenarios violate any law of thermodynamics?
  37. 636 Ryan "One aperture to receive light the second to radiate light" Ah, some more thermodynamics...
    ... have you met Maxwells Demon?
  38. CTruth @627 said
    I agree with RW1 and a few others that observational data seem to indicate the availability of a significant kinetic energy content in the lower troposphere than cannot be accounted for by the solar input alone.

    CTruth, my position (and if I may be so bold to suggest, I think the position of Tom Curtis, KR, scaddenp et al) is this;

    ... that observational data indicates the availability of a significant energy content in the lower troposphere than cannot be accounted for by the solar input alone but is accounted for by an exchange of heat between the surface and the atmosphere. This heat has accumulated in the planetary system in the distant past in the process of reaching (approximate) thermal equilibrium

    If you tried the spreadsheet model I suggested earlier (and Tom Curtis improved @626), one thing you can discover from it is that at the time P(in) = P(out) then ΣTP(in) - ΣTP(out) is at its maximum and is equal to the capacity of the system to hold P (so as to conserve energy or matter) These statements hold true for Tom's half-mirrored box (where P is photons), KR's rivers running in and out of a reservoir (where P is water) and the earths energy budget (where P is heat). T is time, of course, and - heads up to les @623 - yes of course the summations should really be integrals !
  39. Phil, interesting point @639. I have included two new columns in my spreadsheet. One, titled Gain, is Bx-Dx for each row, x. The second, titled "Stored" is the sum of the values in the Gain column from row 2 to row x, for each row, x. The exact formula is "=SUM(G$2:G2)" where G is the column for Gain. As expected, with each progressive iteration, Gain falls to 0 while Stored increases to 400 when the number of photons exiting the box matches the number entering the box. When the incoming photons are reduced to zero, the Gain immediately becomes -100 before slowly increasing to zero, which it reaches approximately as Stored reaches zero and photons leaving the box reaches zero.
  40. Tom @640
    Cool - thats shows it nicely - but hey I don't need convincing :-(

    Kudos has to go to KR, who first alerted me to the idea of using spreadsheets to do these sort of simple iterative demonstrations. On a more general note I wonder if the SkpSci team have considered trying to let contributers share "live" spreadsheets via "cloud" providers like GoogleDocs. It would need to protect the documents from abuse and hide users email addresses, but it might be a useful additional resource if it could be made to work.
  41. LJ >After all, the claim is reflected light (B from Tom Curtis's diagram) is twice the input.

    Why the discrepancy?


    Amount reflected off a surface is not the same as accumulated energy in the box. If a single photon "bounces" back and forth from one wall to another, then you are going to count multiple reflections even though the energy content is still a single photon.

    The amount reflected tells us how many times photons have bounced off the walls, while the accumulated energy tells us how many photons are in the box. Do you see the distinction?
  42. Re #580 RickG You wrote
    "The diagram is not about temperature. Its about incoming solar radiation expressed in W/m^2 and how it is distributed throughout the Earth's climate system, which is the proper unit of measure for that particular type of energy (Incoming Solar Radiation)."

    The thread is about 2nd Law of Thermodynamics which states the direction energy is transferred WRT temperature.

    Trenberth's diagram is all about energy transfer (W/m^2) without any reference to temperature anywhere, thus it says absolutely nothing about atmospheric thermodynamics or the possibility of CO2 having any influence on climate in any way.

    You write further:-
    "Why would Ternberth or anyone for that matter want to use 12 year old data when more up to date data is available? And again, the diagram is about the distribution of energy, not temperature."

    The age of the data has no relevance, Trenberth's diagram does not present any useful information for any discussion on climate change (anthropogenic global warming - AGW) because it is completely deficient in temperature information, the driving parameter in the 2nd law of thermodynamics.
  43. 643 damorbel - "Trenberth's ... is completely deficient in temperature information"

    Someone might correct me; but, seems to me, the diagram includes "surface radiation", "back radiation" from GHGs, "Emitted by Atmosphere" ... all of which are temperature dependent.
  44. Tom Curtis@637

    You are avoiding questions posed @636.

    If your box was fully enclosed such that all surfaces are reflective save two small aperture. One aperture to receive light the second to radiate light. Close the output while receiving 1W at the input. The light source occludes the reflected light from "leaking" out the input.

    The energy "accumulated" within the box after 100 hrs is what, 360kj?

    If the first aperture is then closed, does the box now contain 360kj of light? Asked otherwise, can the "accumulated" energy in the box be captured?
  45. e@642

    You asked:

    "The amount reflected tells us how many times photons have bounced off the walls, while the accumulated energy tells us how many photons are in the box. Do you see the distinction? "

    Using Tom Curtis 615, what is the value of each at equilibrium i.e.

    1. accumulated energy

    2. reflected energy..if you can quantify it
  46. LJRyan @645, in an ideal fully mirrored box, the accumulated energy would be 360 kilojoules. Of course, in practise you would not have perfectly reflecting walls, and given the high speed of light, and consequent very large number of reflections in a short period, the light would decay to zero very quickly. Likewise, again because of the high speed, the light would escape the aperture before you could close it. But practical difficulties do not prevent us from exploring theoretical possibilities in ideal cases.

    @646, let the time interval be the time it takes a photon to travel from the lid to the back wall. Then

    1) the accumulated photons at equilibrium is 4 times the number of photons that enter the light box at each time interval (see 640); and

    2) at equilibrium the number of photons reflected in each time interval is 3 times the number that enter the light box in each time interval. Of those, 2 times that number are reflected of the back wall, and a number of photons equal to the number that enter are reflected of the lid. This ignores reflections of the side walls which are irrelevant to the overall issue.

    The answer has to be in terms of photon numbers, not energy because the wavelength of the photons has not been specified. If we specify that all photons have the same wavelength, then the multipliers for photons in the answers above can be used for energy.

    Now, can you answer my questions @637
  47. Re #644 les you write:-

    "... all of which are temperature dependent."

    Too true, I couldn't agree with you more. So do you not think, to make a useful contribution to a discussion on temperature change, the temperatures should be mentioned?

    Also, the emitting materials do not all have the same emissivity; Trenberth should have inserted the emissivity that applies.
  48. damorbel 648 - fine. We agree on so much including, it would seem, that the diagram is not "completely deficient in temperature information".
    Why don't people say what is apparent without exaggeration? The diagram doesn't explicitly mention temperatures and piles of other information. Indeed it's a cartoon. It clearly does "present ... useful information for any discussion on climate change (anthropogenic global warming - AGW) ".

    Why the hysteria, then? Same with LJRyan, giles et al. If you guys where half the scientist scientists you'd have be to participate in tgis discussing, you'd be far more easy with the shorthand notations, use of approximations, anstract models,partial perspectives and all the other tool we use on a daily basis to understand things.

    Take a chill pill.
  49. damorbel @ 643,

    I'm well aware of the title of the thread. However, my original comment was about Trenberth's diagram and what you said about it. For what Trenberth is demonstrating temperature is neither necessary or relevant in that diagram. I have no problem understanding the diagram myself.

    Trenberth has a presentation in which that diagram is described here on pages 13 & 14. Other diagrams with "temperature" are described in the presentation as well, in their appropriate place.

    If you still have a problem with the diagram, then perhaps you should contact him personally and take the matter up with him at the National Center for Atmospheric Research (NCAR).. I'm sure he is open to new ideas and wants to be sure his diagrams convey the proper information and documentation.
    Response: [Muoncounter] damorbel has been given this suggestion a number of times to little if any effect. Instead, we have more pointless repetition of the same tedious argument, which we may surmise is damorbel's actual intent.
  50. Tom Curtis@647

    "accumulated energy would be 360 kilojoules"
    So the accumulated "boxed" light would radiate 1W for a 100 hrs, once the second aperture is opened?


    "the light would decay to zero very quickly"
    Sounds like a violation of the 1st law...remember perfectly reflective walls. Sounds as if you don't believe your own answer.


    "The answer has to be in terms of photon numbers, not energy because the wavelength of the photons has not been specified. If we specify that all photons have the same wavelength, then the multipliers for photons in the answers above can be used for energy."

    ok lets assign three wavelengths and redo @646

    1. 11364 nm
    2. 10062 nm
    3. 2898 nm


    @637 questions
    1) Consider the box as described, but without any lid. In this case all the light will reflect of the wall of the box and exit through the aperture where the lid was.
    Is that correct?
    1a)I not sure I understand your question...if the lid included the aperture wouldn't it also be removed with the lid...?

    2) Now consider the case in which we place the lid on the box, but at an angle so that all light reflected of the lid will leave the box through some other aperture. In this case, the amount of light leaving the box through the lid will be half of that which enters, while the amount that is reflected by the lid and leaves through the other aperture will also be half of that which enters the box.
    Is that correct?
    2a)Again, I'm not sure I understand your question. Is the box partially open? How open?

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