Guest Post by Ira Glickstein
This series began with a mechanical analogy for the Atmospheric “Greenhouse Effect” and progressed a bit more deeply into Atmospheric Windows and Emission Spectra. In this posting, we consider the interaction between air molecules, including Nitrogen (N2), Oxygen (O2), Water Vapor (H2O) and Carbon Dioxide (CO2), with Photons of various wavelengths. This may help us visualize how energy, in the form of Photons radiated by the Sun and the Surface of the Earth, is absorbed and re-emited by Atmospheric molecules. 
DESCRIPTION OF THE GRAPHIC
The animated graphic has eight frames, as indicated by the counter in the lower right corner. Molecules are symbolized by letter pairs or triplets and Photons by ovals and arrows. The view is of a small portion of the cloud-free Atmosphere. (Thanks to WUWT commenter davidmhoffer for some of the ideas incorporated in this graphic.)
- During the daytime, Solar energy enters the Atmosphere in the form of Photons at wavelengths from about 0.1μ (micron – millionth of a meter) to 4μ, which is called “shortwave” radiation and is represented as ~1/2μ and symbolized as orange ovals. Most of this energy gets a free pass through the cloud-free Atmosphere. It continues down to the Surface of the Earth where some is reflected back by light areas (not shown in the animation) and where most is absorbed and warms the Surface.
- Since Earth’s temperature is well above absolute zero, both day and night, the Surface radiates Photons in all directions with the energy distributed approximately according to a “blackbody” at a given temperature. This energy is in the form of Photons at wavelengths from about 4μ to 50μ, which is called “longwave” radiation and is represented as ~7μ, ~10μ, and ~15μ and symbolized as violet, light blue, and purple ovals, respectively. The primary “greenhouse” gases (GHG) are Water Vapor (H2O) and Carbon Dioxide (CO2). The ~7μ Photon is absorbed by an H2O molecule because Water Vapor has an absorption peak in that region, the ~10μ Photon gets a free pass because neither H2O nor CO2 absorb strongly in that region, and one of the 15μ Photons gets absorbed by an H2O molecule while the other gets absorbed by a CO2 molecule because these gases have absorption peaks in that region.
- The absorbed Photons raise the energy level of their respective molecules (symbolized by red outlines).
- The energized molecules re-emit the Photons in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
- This frame and the next two illustrate another way Photons are emitted, namely due to collisions between energized GHG molecules and other air molecules. As in frame (2) the Surface radiates Photons in all directions and various wavelengths.
- The Photons cause the GHG molecules to become energized and they speed up and collide with other gas molecules, energizing them. NOTE: In a gas, the molecules are in constant motion, moving in random directions at different speeds, colliding and bouncing off one another, etc. Indeed the “temperature” of a gas is something like the average speed of the molecules. In this animation, the gas molecules are fixed in position because it would be too confusing if they were all shown moving and because the speed of the Photons is so much greater than the speed of the molecules that they hardly move in the time indicated.
- The energized air molecules emit radiation at various wavelengths and in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
- Having emitted the energy, the molecules cool down.
DISCUSSION
As in the other postings in this series, only radiation effects are considered because they are the key to understanding the Atmospheric “Greenhouse Effect”. I recognize that other effects are as important, and perhaps more so, in the overall heat balance of the Earth. These include clouds which reflect much of the Sun’s radiation back out to Space, and which, due to negative feedback, counteract Global Warming. Other effects include convection (wind, thunderstorms, …), precipitation (rain, snow) and conduction that are responsible for transferring energy from the Surface to the Atmosphere. It is also important to note that the Atmospheric “Greenhouse Effect” and a physical greenhouse are similar in that they both limit the rate of thermal energy flowing out of the system, but the mechanisms by which heat is retained are different. A greenhouse works primarily by preventing absorbed heat from leaving the structure through convection, i.e. sensible heat transport. The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth.
That said, how does this visualization help us understand the issue of “CO2 sensitivity” which is the additional warming of the Earth Surface due to an increase in atmospheric CO2? Well, given a greater density of CO2 (and H2O) molecules in the air, there is a greater chance that a given photon will get absorbed. Stated differently, a given photon will travel a shorter distance, on average, before being absorbed by a GHG molecule and be re-emitted in a random direction, including downwards towards the Surface. That will result in more energy being recycled back to the Surface, increasing average temperatures a bit.
If backradiation was a phenomena that would heat up earth surface it would also heat up the sun up to the core.
By using Stefan-Boltzmann the non-GHG surface temperature is found by using E = sigma T^4 (SB) where E can only originate from the sun. If now T would go up further because of GHG, this would mean more energy input would be needed. So where does this come from, are GHG capable of generating energy? Many people seem to think so, but what about the following?
If we look at a CO2 layer as a plate with T2 above earth, you have something like the classic example of radiation between two plates here, where the sun would heat the lower one which is T1.
So yes radiation will go both ways, there is ‘backradiation’ and I go with that. But read that in the state (equation 19.2) where also plate number two would be T1, Qnet will be zero. Surface 1 will never get any warmer by the (back)radiation, despite the accumulated (heat)energy in plate 2 and radiation going everywhere. Read from this that radiation is energy but never heat.
So you can make calculations with multiple layers (like CO2 layers) and the result will be the same. T1 will not get any warmer with Qnet getting zero, and all the layers acting as insulation (like some people think CO2 does) and all plates actually permanently getting radiation from both sides. (Earth atmospheer would also be forced to become T1 from the surface up if it was’nt for gravity).
Now think about multiple plates where the sun’s surface is the first plate and earth the second etc. So the greenhouse effect says that the second plate is supposed to get warmer by backradiation from the plates next in line? Then also the first one, the sun’s surface will get warmer (no matter how tiny) by this backwards phenomena. The sun has many more layers, but this process will go on up to the core! The source of the energy!
Well, what is wrong here. The problem is that in our real universe the temperature of the sun in space is the driving force, telling all other layers down the line what temperature they must take on, in its path to the 3 Kelvin of space and not the other way around.
With new layers somewhere in the line, the temperature of the layers behind the new one will simply go down to establishing a larger dT maintaining the same energy flux.
Yes, it’s that pesky Second law, that in fact is constantly trying to lower the temperature of the sun.
So what are the tricks with radiation that many people don’t see? For starters I want to say that the photons of EM radiation can be regarde as cold, only interaction with matter gives any thermal energy (to get a temperature) which is taken away (partly) when a(nother) photon leaves.
Most greenhouse effect statements/sites claim an energy balance on earth surface because of the First law, the greenhouse theory makes fame by using energy in-out balances etc. convincing everyone who has no clue about thermodynamics. But this is dead wrong, the Second law rules here and this means simple energy conservation is not what determines radiative equilibrium temperature at the surface.
The earth could conserve it’s energy by accumulating it until it would be as hot as the sun, there is nothing against the First law with this.
So in GHG physics T of earth surface is supposed to go up with while receiving the same amount of energy from the sun. Now we know the Second law wants to establish a dynamic radiative equilibrium at the surface with a certain temperature but also that T = dE/dS, so this means that for T to be able to go up with backradiation Entropy must decrease during the absorption-emission process?? That will never happen.
Looking closer it works like this:
Why does the sun heat the surface to T? Because HQ radiation from the sun can leave earth as LQ IR radiation (to the cold space) in the process gaining entropy while raising kinetic energy as potential WASTE HEAT on the surface (and they don’t call this irreversible for nothing, the downgraded (photon)energy won’t be able to do the same trick at the surface again).
The radiation Quality is expressed as theRadiation Energy Density
Some LQ IR returns (backradiation), but it can never ever create more heat here. Because in that case it would have to leave with less energy so as even lower quality IR, and hey …… this can only happen at a LOWER surface temperature (think of the BB-spectra of earth next to the sun, all the energy flows from the high spectrum to the low one, anyone thinks some energy goes the other way here?).
See the contradiction, that’s the mighty Second law of thermodynamics in action that says: it’s all wrong with this heating by backradiation philosophy.
So backradiation will simply leave in a reversible proces (like reflection) without leaving waste heat, and with the same frequency as it had before and bounce whatever way it wants with no effect on the surface. It’s like in the Qnet = zero situation between the two plates (lotsa radiation, no more heating by the cold photons).
So what is ‘the heat’ that can’t go from low to high temperature, warming the earth surface and why the confusion?
Maxwell’s classic Theory of Heat states: heat is something which may be transferred from to another, according to the Second law of thermodynamics.
In any case, heat transfer needs matter which has a temperature (as kinetic energy) and another piece of matter at lower T. Radiation has no temperature and can never ever represent heat. And so the photons of radiation can be regarded as cold.
Heat is the waste from the irreversible radiation phenomena happening due to what the Second law dictates and is equivalent to the entropy gained and what is called dissipation of energy.
Heat can be released when a photon interacts with matter. Many think ‘the heat’ has to do with all the energy (E= h x v) related to the cold travelling photon. But we have seen that this is not the case, this is imaginary heat (which is not heat but energy) that is never released as long as matter is’nt involved. Energy is only released when the SWR photon disappears in the matter and after that heat from this matter may flow if other colder matter is present, but this heat is not equivalent with the photon energy but is a function of the temperature difference between surface temperature and colder matter above it (like Q = k*A*dT). If there is no colder matter, there is no heat (dT=0).
If earth had no atmospheer, the photons would warm earth surface to SB temperature but no heat would ever be exchanged. Then if there was a layer of CO2 around it at say 10 meters or 10 km, this would create backradiation, but the vacuum inbetween would be as cold as the 3K space, and the backradiation would not make the surface warmer, and there would be no heat despite all the radiation.
Now because heat is flowing from earth surface we know we have a second irreversible process (waste heat, making entropy rise) we also know for sure the new lower energy photon can never release heat at this temperature where it came from ever again, so to think backradiation can go beyond that and even create a higher temperature is impossible by the Second law of thermodynamics.
So the heat coming from the surface is the part of the solar radiation that did’nt get radiated away (directly) as IR (and the total energy of this IR is thus lower than the LWR that hits the surface). Heat is the part that has nothing to do with radiation, and the IR that left has nothing to do with heat.
And so we get the surface balance: IR = LWR – HEAT. So the IR that leaves the surface does’nt only have a lower frequency, but the total energy of it is also lower than the incoming LWR. This is the second blow for the imaginary effect of backradiation.
It is this HEAT only that warms the first layers of the atmospheer and gives the ground temperature’s measured (in air, so does not even represent the surface), and it can only take place by conduction from surface to air initially, it must be transported from the surface at T1 to matter at T2 and it finds its way up (conduction/convection – radiation) to space separated from the IR photons that left the surface.
Papers about the sun that are not infected with the greenhouse virus go along the lines of my view. Photons jiggle around for tenthousands of years from layer to layer, and they do not heat the core. This degradation of high quality X-ray photons to relatively low quality optical photons, is only to be expected from the Second law of thermodynamics.
Further: a photon can only travel a tiny distance before running into another hydrogen nucleus. It gets absorbed by that nucleus and the re-emitted in a random direction. If that direction is back towards the center of the Sun, the photon has LOST GROUND! It will get re-absorbed, and then re-emitted, over and over, trillions of times. The path it follows is called a “random walk”
Max gets a good part of it, but there are a couple of things missing, the higher you go in the troposphere, the colder it gets, and the slower the emission from collisionally excited CO2, water vapor or any other greenhouse gas. Second, the distance a photon emitted from CO2 can go is not very far, so effectively, the height at which emission in the CO2 bands can reach space is about 10 km, where it is very cold. You can see this in the modtran spectra (just hit submit the calculation) which show what would be observed from say the space station. The emission between 600 and 700 cm-1 is much lower than it would be without CO2, because it comes from a layer 10km up where the temperature is 220 K.
Joel Shore says:
“I am well-aware that gravity is a force and that gases are compressible.”
So, what happens to cooled air that descends?
“I, and other atmospheric and climate physicists are also aware of the constraints on lapse rates set by the decrease of pressure with height.”
If this is supposed to be some sort of appeal to authority, I’m not at all impressed. First, you have to say something that demonstrates you understand what you are talking about.
You say: “However, that does not allow you to violate the Laws of Thermodynamics at will.”
Nor does it allow you to ignore the system conditions with statements like this : “And, these Laws tell us that the surface temperature of an IR-transparent atmosphere will go toward a steady-state value determined by the condition that the radiative emission balance the radiation absorbed from the sun (in the absence of any other significant energy source).”; which totally ignores the influence of gravity on the initial state of the compressible gas that makes up your hypothetical IR-transparent atmosphere.
But, where have I violated any laws of thermodynamics?
When I got to the part where you demonstrated your complete lack of understanding the thermalization process I quit reading. The idea of an IR photon being absorbed and then quickly remitting with half going up and have down is absolute BS. Go back and study how the atmospheric thermal reservoir works.
I’m stunned that there is debate about the ability of one radiating object being able to heat another radiating object. It is so numbingly obvious that this happens that it begs the question: What the hell would keep it from happening?!?
As for which direction molecules radiate energy, try thinking of each of them as isotropic radiators (they’re not, but the principle applies). Except for geometry limitations of the molecule (asymmetric construction), they will behave like an isotropic radiator and radiate equally in all directions. In the atmosphere, where molecular orientation is assuredly random, even asymmetric molecules on average will not favor up vs down vs over there when it comes time to shed their packet of new energy.
Here’s another mental widget to help get your head around this. We’re talking about radiation here, remember. Picture a flash bulb sitting in free space. Imagine it is lit off. Imaging a sphere of light radiating away from the bulb. Imagine that light hits other things. Imagine some of those things are other lightbulbs. Some lit, some not. Now try to understand – there is no intelligence delivered with that sphere of light. It simply illuminates any surface it touches. That includes other lit surfaces. The bulbs don’t know this is going on and don’t care. They are just following the rules of thermodynamics.
Now the test – you put a very sensitive light sensor in the lab space and it has lenses so it sees but one single bulb. The entire field of vision of the sensor it filled with that bulb. Turn on the bulb and notice the sensor response. Now turn on another bulb that is conveniently located behind the sensor and so out of it’s sensing view. Note that the sensor detects that new light coming from the only thing in its view and it registers a higher light level than before. Some call it BS. In fact that just happened in this thread, but it is true that radiating objects light up other radiating objects in all parts of the spectrum including IR.
Ever wonder why three logs stacked in a pyramid will burn to ash but one log alone will quickly stop burning and turn cold? It is because the logs radiate heat onto each other while they burn and heat is one of the three essentials for combustion.
Nuther example because I sense a lot of ignorance here. You are sitting in your kitchen typing on your laptop, filling the WUWT blog with ignorance. It is dark out, and very cold – snow everywhere. You look at the outside thermometer and it is 5ºF – damn cold. You pull aside the drapes on your double paned sliding glass door so you can look out at the beautiful snowy wonderland. And after a few minutes you start to feel cold where before you were quite comfortable. How can that be? It is 72º in your kitchen – how can you feel cold? You granny comes over and pulls the drapes closed, calls you a dummy and that you’re going to freeze if you leave those drapes open. What does granny know? Did you notice too you were cold only on side of you facing the snowy winter wonderland?
Granny knows your body is radiating the maximum possible heat out of the house and into that winter wonderland. If you pull the drapes closed that heat loss stops. Why? Think about that then send your granny a card. A real card, not a lame iCard, and let her know how smart she is.
Ira Glickstein, PhD says:
March 29, 2011 at 4:12 pm
…there are at least three issues: 1) The small difference in surface area, 2) The large temperature increase due to the Atmospheric “Greenhouse Effect”, and 3) Convection/conduction/precipitation are not anywhere near 100% effective at transfering thermal energy from the Surface to the Atmosphere. Let us address each in turn:
1 – If the shell is at an altitude similar to the effective altitude of the Atmosphere, which is a teeny tiny fraction (less that 1/1000) of the radius of the Earth, the shell’s surface area will be almost the same as the Surface area of the Earth, so the supposed temperature difference will be miniscule.
2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1).
3- Even if (1) was as large as 30ºC, the combined sum of convection/conduction/precipitation would fail to transfer anywhere near 100% from the Surface to your shell.
Nice try though and THANKS for getting me thinking.
I wish to thank you Ira. I too enjoy and benefit from these discussions. They often make me consider something I had completely missed. In that vein, I would like to retract my statement that it’s a “waste of time” to discuss Earth surface temperature without including a discussion of conduction and convection. Discussions of incorrect analyses of physical phenomena often lead to better understanding and to viable formulations. BTW–the immediately above statement applies to my analysis. If my analysis is incorrect, I will likely learn from my errors; and in the process maybe give the person who corrects my errors insight into an issue he didn’t have before the discussion.
Now to address your three points.
First, “The small difference in surface area,”
It’s true a shell that has the diameter equal to the atmosphere diameter will have at most a slightly larger area than the Earth. However, we’re discussing the simultaneous existence of backradiation and a cooler active object surface temperature. As such, there is no restriction on the radius of the shell. It’s also true that the larger the radius of the shell, the smaller will be the backradiation from the shell to the active sphere. The reason a larger shell radius results in less backradiation is as follows. For a thin shell, (a) the area of the inside shell surface is close to the area of the outside shell surface, and (b) for a highly conductive shell, the temperature of the shell inner surface will be close to the temperature of the shell outer surface. In a steady-state condition, the area of the shell outer surface must radiate energy away from the shell at a rate equal to that produced by the active-object internal thermal source. Since the inner shell surface area and temperature are close to the outer shell surface area and temperature, the rate energy is radiated from the inner shell surface will be close to the rate energy is radiated from the outer shell surface, and hence close to the rate of internal energy supplied by the active object. However, as the radius of the shell increases, more and more of the energy radiated from the inner surface of the shell will be directed toward another part of the shell inner surface, not toward the active sphere. As such, a corresponding drop in the backradiation to the active object will take place. Backradiation will still exist, but the rate of backradiation energy flow decreases as the sphere radius increases.
Second, The effect of “Greenhouse” warming of the Earth is over 30ºC
The approximate 30ºC number is obtained by assigning to the Earth a reception albedo of approximately 0.3 and an emissivity of approximately 1 [i.e., an albedo of approximately 0). If you treat the surface of the Earth as a graybody (i.e., you use a modified Planck blackbody radiation law with an emissivity (1 minus albedo) somewhere between 0 and 1], then you must use the same albedo for both reception and emission. For example, if you completely enclose an inert graybody object (non-zero albedo) within a surface that is maintained at a uniform fixed temperature, the steady-state temperature of the enclosed graybody object will equal that temperature. However, if to that scenario you apply Planck’s graybody radiation law with an emission albedo different from its reception albedo, the result will be that in radiation balance (rate of energy absorbed equals rate of energy emitted) the inert enclosed object’s steady-state temperature will be either higher or lower than the temperature of the surrounding material. The steady-state temperature of the inert object cannot both be “equal to” and “not equal to” the temperature of the surrounding (enclosing) material. I believe of these two choices (assuming one of them is correct), the correct answer is “equal temperature”. If true, then the use of different reception/emission albedos and Planck’s graybody radiation law are incompatible. As others have pointed out, I question the use of Planck’s graybody radiation law for a gas/surface interface. It may be valid for minute amounts of gas, but as the gas density increases, I have a feeling the law starts to break down. [As an aside, for a gas there is no clearly defined “surface”. Since one term in Planck’s blackbody radiation law is a differential area at a known temperature, any use of Planck’s blackbody radiation law or a graybody version of that law for gases is inherently impossible.] If the same albedo is used for both reception and emission, the difference between the Earth’s actual temperature and it’s graybody temperature without an atmosphere is more like 7ºC, not 30ºC.
Third, “Even if (1) was[n’t] as large as 30ºC, the combined sum of convection/conduction/precipitation would fail to transfer anywhere near 100% from the Surface to your shell.” The rate of energy transfer via mechanisms other than radiation doesn’t have to be anywhere near 100%. The active object temperature will be decreased if the rate of energy transfer via non-radiative mechanisms is greater than the rate of backradiation. Since the rate of backradiation decreases with increasing shell radius, for a sufficiently large shell, it won’t take much conduction to produce a lower active object temperature. I believe experiments using eight or nine silver struts connecting the active object to the shell can be constructed that will likely transfer a sufficient rate of conductive thermal energy. If you don’t believe so, turn on your stove, place the blade of a looooong sterling silver knife on the element, and see how long you can hold onto the knife handle.
Bottom line, I don’t claim greenhouse gases won’t increase the surface temperature of the Earth. However, I do argue that the backradiation argument is insufficient to make such a claim. This gets us back to my statement to the effect that any meaningful discussion of the Earth’s surface temperature must include a discussion of all thermal transfer phenomena, not just radiation.
Well ok, I guess, although many here don’t seem to think it is ok. But why is the sensitivity of atmosphere to an additional molecule of CO2 logarithmic? Your model implies that it wouldn’t be.
In this essay you have qualitatively described a small part of a complex nonlinear system which has huge amounts positive and negative feedbacks. Without numbers coming from theory and/or measurements and without describing connections of your radiation physics to the other parts of the system it is not possible to assess its validity and relevance. It is good that you did not have any conclusions or advice to the policymakers.
Nice graphics, anyhow 🙂
@DP
It’s a shame none of your examples support your argument.
The first experiment is simply completely wrong.
The three logs burn better because a pyramid allows much more air into the fire and increases the surface area exposed to the fire promoting combustion. Boy Scouts 101. Nothing to do with radiation.
The reason it is cold near the window is because:
a) The window is a much poorer insulator than the walls. This means the air near the window is cooled by conduction and radiation to the outside and convection on the inside surface of the window.
b) The window is also further from the fire and receives much less thermal energy according to the inverse square rule of radiation.
c) The window has a very low albedo so it doesn’t reflect heat back into the room.
If you repeated the experiment with triple glazed windows it wouldn’t work.
How is it possible for a pyramid of logs to allow more air to the burning part of the log that a log standing alone without competition for that air from other logs? It really is the mutual heating that sustains the burn.
FAIL
Multiple panes of glass work against conduction. They are useless against radiated heat unless they are designed (coated) to radiate IR back into the room in which case it is exactly the same as drawing the drapes closed but much more expensive. Pick up a good science book and don’t put it down until you understand how radiated energy works. Then teach your kids.
It’s more than 20 years since I did any spectroscopy so I’m pretty rusty. However we certainly weren’t told about any (alleged) heat trapping abilities of molecules by our PhD qualified chemistry lecturers.
The author seems to have confused spectrophotometry which only involves the outer electron orbitals and IR spectroscopy which involves changes in the vibrational modes of molecular bonds.
IR spectroscopy is primarily used for non-destructive testing of carbon rich solids such as coal, plastics and pharmaceuticals.
It should be remembered that spectroscopy is an empirical tool for measuring structures, quantities or concentrations of a substance. It isn’t designed for determining the existence of a greenhouse effect.
DP;
How is it possible for a pyramid of logs to allow more air to the burning part of the log that a log standing alone without competition for that air from other logs? It really is the mutual heating that sustains the burn.>>>>
Thank you thank you thank you thank you! What a wonderful, obvious, easy to understand example! Mind if I borrow it from time to time? I’ve been relying on my igloo example of cold things radiating heat to warm things and have become exasperated with the nonsensical objections.
As for bananabender….
If you’d actually taken Scouts 101 pretty much the first thing you learn is that any logs of a decent size simply will not burn well unless the burning faces are used to heat each other up. In your own example, if the pyramid of logs promoted surface area exposed to air, why wouldn’t the logs burn from the outside in instead of the inside out? If you took Scouts 201 you’d also know that arranging the logs in layers, each layer a set of parallel logs at right angles to the layer below, makes a fire that puts the pyramid to shame. It creates a whole series of surfaces radiating at each other to achieve much higher temps.
George E. Smith says:
March 29, 2011 at 5:38 pm
I did not say that all the earth’s energy was absorbed by the atmosphere and re-radiated. It is only true for a small range of wavelengths associated with the absorbtion bands of the greenhouse gases. I have said this many times in different ways (including other comments of mine on this thread) but Eli Rabbet covers it pretty succinctly a few comments back. This is not a theory. It has been measured to be the way the world is.
DavidM
I’m just reading this article. This section below seems at variance to your very dubious Igloo example.
tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
“Thermodynamics says that no object in the universe can heat itself by its own radiation, nor can heat flow from cold to hot”.
BigWaveDave says:
I’m not ignoring gravity. I am just accepting the fact that gravity doesn’t magically create energy out of nothing. You write P/R=rho T and say that “density decreases at a lesser rate than pressure” in the troposphere and conclude that temperature must rise. However you fail to note that:
(1) This is just an observed feature of the troposphere. It is not a necessary feature and, in fact, in other levels of the atmosphere (like the stratosphere) it is not the case that the temperature decreases with height.
(2) The fact that temperature decreases with height does not imply that gravity somehow sets a constraint on the surface temperature. And, indeed, it does not…In the absence of an IR-absorbing atmosphere, the constraint on surface temperature is set by simple radiative balance considerations. (In the presence of an IR-absorbing atmosphere…i.e., the greenhouse effect, it is more complex but is still set by energy balance considerations, although one has to consider energy balance within the atmosphere too and convection is also involved.)
You are proposing that the earth’s surface can be at a temperature where the earth would continuously emit much more energy than it absorbs from the sun, which is not possible for it to do without rapidly cooling down unless there is a significant other energy source that comes into play.
RJ says:
March 30, 2011 at 3:13 am
DavidM
I’m just reading this article. This section below seems at variance to your very dubious Igloo example.
tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
“Thermodynamics says that no object in the universe can heat itself by its own radiation, nor can heat flow from cold to hot”.>>>
NET heat cannot flow from cold to hot.
If it didn’t then you couldn’t stay warm inside of an igloo anymore than you could standing outside the igloo. Igloos work. Period.
Pour boiling hot coffee into a thermos at room temp. Which is hotter? The thermos of the coffee? The coffee obviously. Put a bowl out on the counter and fill it with an equal amount of the same coffee, cover it to prevent heat loss via convection. Wait six hours. Coffee in the thermos is still hot. Which is hotter? The coffee or the thermos? The coffee. What about the coffee in the bowl? Room temp. Huh.
So what heat source was it that kept the coffee hot? Given that it was in a thermos that was at a lower temperature than the coffee and there were no other heat sources being put into the thermos, why did it stay hot?
The answer is that it got some of its own heat back. It radiated heat to the thermos, which, as a consequence of warming up, radiated some back into the coffee.
According to davidmhoffer an igloo is an example of ‘a cold thing radiating to a warm thing’.
Snow is an excellent insulator as it contains pockets of air and, like a greenhouse, prevents convection. This means that a warm object (a human body, for example) will warm the interior of the igloo by conduction and radiation. The snow cannot add energy to a warmer body. The molecules of the warmer body are vibrating and this is the signal of its temperature. The radiation from a cooler body cannot vibrate the molecules more than they are already vibrating as it has less energy. The snow is not an extra radiator adding to the heat created by the warm body. If it were otherwise a an infra-red reflector (a mirror) would be a heater.
davidmhoffer says:
March 30, 2011 at 2:11 am
As for bananabender….
If you’d actually taken Scouts 101 pretty much the first thing you learn is that any logs of a decent size simply will not burn well unless the burning faces are used to heat each other up. In your own example, if the pyramid of logs promoted surface area exposed to air, why wouldn’t the logs burn from the outside in instead of the inside out? If you took Scouts 201 you’d also know that arranging the logs in layers, each layer a set of parallel logs at right angles to the layer below, makes a fire that puts the pyramid to shame. It creates a whole series of surfaces radiating at each other to achieve much higher temps.
I was never a boy scout. That means I didn’t learn any of their incorrect explanations for how fires work.
A pyramid or a “log cabin” fire forms a venturi drawing fresh air in from outside the base like a chimney. Hot air rises via convection up the the inside of the structure. This causes the evaporation of volatiles which then ignite. The outside doesn’t burn well because wood is a very poor conductor of heat and there is very little direct heating from the flames.
In the case of a horizontal log the heat is carried upwards by convection so the wood below is not well heated. Charcoal is an exceptionally effective insulator so the wood below the flame doesn’t reach combustion temperature and the fire exhausts the fuel supply.
Radiation is a negligible contributor to flame propagation in small fires. Heat flow is always from hot to cold.
A number of burning logs in contact with one another will certainly aid combustion by aggregating their heat.
However the jumble of logs will also obstruct and redirect and locally accelerate the flow of air between them so it is also true that accelerating the flow of air between the logs will fan the flames further.
dp says:
March 29, 2011 at 11:28 pm
Multiple panes of glass work against conduction. They are useless against radiated heat unless they are designed (coated) to radiate IR back into the room in which case it is exactly the same as drawing the drapes closed but much more expensive.
Curtains reduce heat loss primarily by preventing convection currents near the window surface. This in turn reduces the effectiveness of the glass as a conductive heat exchanger. Curtains have very little effect on radiative heat loss. Triple glazing is far more effective at preventing heat loss than curtains in a cold climate
Triple glazed windows are effective primarily because they prevent conductive heat loss. Radiative heat loss across triple-glazed windows is low because glass is only partly transparent to IR radiation.
Pick up a good science book and don’t put it down until you understand how radiated energy works. Then teach your kids.
I’ve got three science degrees and have studied physics and chemistry at university. Unlike you I actually understand how radiation works.
Igloos are temporary structures that work primarily by preventing wind chill and heat loss by convection. Snow is also a fairly good insulator reducing conductive heat loss to the outside air. The cold igloo doesn’t have any warming effect at all on the inhabitants.
Ira says “2 – The effect of “Greenhouse” warming of the Earth is over 30ºC, which is many, many times the maximum possible temperature difference due to (1). ”
This is only true if you disregard that STP is 0 C. If taken into account then GHG is 15 C max.
A few things that stick out in this discussion:
1 – The concept that the Earth surface warms by back radiation is a mis-understanding. The surface warms because of a rather large object 93 million miles away. The issue is that the OLR radiation leaving the Earth’s surface is slowed in its escape through the atmosphere because of absorption of certain wavelengths including the one about 14 microns that corresponds to an absorption band in CO2. This has the effect of increasing the energy and temperature of the near surface atmosphere, so slows the heat loss from the surface. If this is a violation of the 2nd law of thermodynamics, so are all forms of insulation…
2 – There is a confusion between attempting to explain things at an atomic or quantum scale and in the macro world of classical physics.
3 – the logarithmic effect of absorption (or at least that it is a reasonable approximation at the current CO2 concentrations) is easily demonstratable in absorption spectroscopy (i.e. can be verified by experimental observation). Absorption of a photon is probablistic, and dependent on how close the photon’s energy is to the absorption band of the CO2 molecule – this is why the argument that the CO2 wavelength is ‘saturated’ with regard to IR is a bit over-simplified, those CO2 molecules that have not absorbed a photon ideally suited to its absorption band can still absorb a photon at a markedly lower or higher energy, although the likelihood of absorbing this photon reduces the further from the ‘ideal’ wavelength the photon is.
4 – Despite the above, all of this discussion is rather theoretical until adequate consideration is given to the dominance of convection as a bulk heat transfer mechanism in the lower atmosphere, and to the moderating effects of the oceans as a heat sink and source, remembering that the issue from a human experience perspective really is how much is the temperature increasing in the biosphere.
from Wikipedia
While I certainly do not trust Wikipedia completely, since anyone may post almost anything to obscure topics, I know from personal experience that important scientific topics are monitored by domain experts who protect them from spurious information. Reed Coray, please cite a respected source for your “7ºC” estimate, and same for mkelly regarding your “15 C max”.