Heh. In response to a ridiculous claim making the rounds (I get comment bombed at WUWT daily with that nonsense) which I debunked here: A misinterpreted claim about a NASA press release, CO2, solar flares, and the thermosphere is making the rounds
Dr. Roy Spencer employs some power visual satire, that has truth in it. He writes:
How Can Home Insulation Keep Your House Warmer, When It Cools Your House?!
<sarc> There is an obvious conspiracy from the HVAC and home repair industry, who for years have been telling us to add more insulation to our homes to keep them warmer in winter.
But we all know, from basic thermodynamics, that since insulation conducts heat from the warm interior to the cold outside, it actually COOLS the house.
Go read his entire essay here. <Sarc> on, Roy!
UPDATE: Even Monckton thinks these ideas promoted by slayers/principia/O’Sullivan are ridiculous:
Reply to John O’Sullivan:
One John O’Sullivan has written me a confused and scientifically illiterate “open letter” in which he describes me as a “greenhouse gas promoter”. I do not promote greenhouse gases.
He says I have “carefully styled [my]self ‘science adviser’ to Margaret Thatcher. Others, not I, have used that term. For four years I advised the Prime Minister on various policy matters, including science.
He says I was wrong to say in 1986 that added CO2 in the air would cause some warming. Since 1986 there has been some warming. Some of it may have been caused by CO2.
He says a paper by me admits the “tell-tale greenhouse-effect ‘hot spot’ in the atmosphere isn’t there”. The “hot spot”, which I named, ought to be there whatever the cause of the warming. The IPCC was wrong to assert that it would only arise from greenhouse warming. Its absence indicates either that there has been no warming (confirming the past two decades’ temperature records) or that tropical surface temperatures are inadequately measured.
He misrepresents Professor Richard Lindzen and Dr. Roy Spencer by a series of crude over-simplifications. If he has concerns about their results, he should address his concerns to them, not to me.
He invites me to “throw out” my “shredded blanket effect” of greenhouse gases that “traps” heat. It is Al Gore, not I, who talks of a “blanket” that “traps” heat. Interaction of greenhouse gases with photons at certain absorption wavelengths induces a quantum resonance in the gas molecules, emitting heat directly. It is more like turning on a tiny radiator than trapping heat with a blanket. Therefore, he is wrong to describe CO2 as a “coolant” with respect to global temperature.
He invites me to explain why Al Gore faked a televised experiment. That is a question for Mr. Gore.
He says I am wrong to assert that blackbodies have albedo. Here, he confuses two distinct methods of radiative transfer at a surface: absorption/emission (in which the Earth is a near-blackbody, displacing incoming radiance to the near-infrared in accordance with Wien’s law), and reflection (by which clouds and ice reflect the Sun’s radiance without displacing its incoming wavelengths).
He implicitly attributes Margaret Thatcher’s 1988 speech to the Royal Society about global warming to me. I had ceased to work with her in 1986.
He says that if I checked my history I should discover that it was not until 1981 that scientists were seriously considering CO2’s impact on climate. However, Joseph Fourier had posited the greenhouse effect some 200 years previously; Tyndale had measured the greenhouse effect of various gases at the Royal Institution in London in 1859; Arrhenius had predicted in 1896 that a doubling of CO2 concentration would cause 4-8 K warming, and had revised this estimate to 1.6 K in 1906; Callender had sounded a strong note of alarm in 1938; and numerous scientists, including Manabe&Wetherald (1976) had attempted to determine climate sensitivity before Hansen’s 1981 paper.
He says, with characteristic snide offensiveness, that I “crassly” attribute the “heat-trapping properties of latent heat to a trace gas that is a perfect energy emitter”. On the contrary: in its absorption bands, CO2 absorbs the energy of a photon and emits heat by quantum resonance.
He says the American Meteorological Society found in 1951 that all the long-wave radiation that might otherwise have been absorbed by CO2 was “already absorbed by water vapor”. It is now known that, though that is largely true for the lower troposphere, it is often false for the upper.
The series of elementary errors he here perpetrates, delivered with an unbecoming, cranky arrogance, indicates the need for considerable elementary education on his part. I refer him to Dr. Spencer’s excellent plain-English account of how we know there is a greenhouse effect.
The Viscount Monckton of Brenchley (April 18, 2013)
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gbaikie says:
Which part of what the following sentence that I wrote did you not understand: “Neither of these differences should affect the average temperature (or, more precisely, the average of T^4), which is determined by radiative balance, but they do affect the range of temperatures”?
Let me try again: In your post that I was responding to, you talked about diurnal temperature ranges (e.g., the fact that it can during the day get much hotter on the Moon than on the Earth). That goes beyond considerations of radiative balance relevant to the greenhouse effect, which only constrains what the average temperature (average of T^4) must be over the planet. I explained to you that the moon can get hotter during the day (and colder at night) and still have an average surface temperature compatible with its emission temperature being what it has to be given its albedo and absence of a greenhouse effect.
Theory does not allow the average of T^4 to be larger…but it does allow the maximum diurnal temperature to be warmer. And greenhouse theory does not predict that diurnal temperature range, which will indeed depend on details like rotation rate and thermal inertia. So, it is you who apparently do not understand the basics of the theory.
The point is that you decided other people were brainwashed because, as it turns out, they understand science that you do not. Perhaps you need to entertain the idea that you are the one who is brainwashed.
I said “What you are missing is that emissivity is proportional to absorptivity”.
Actually, I should have said “is equal to” not simply “is proportional to”.
Question:
Make a large cube in vacuum of space and have the cube made from smaller cubes.
And have these small cube tightly stacked together.
So 1 km width, length, high cube comprised of a billion 1 meter cubes.
The amount energy this 1 km cube will radiate depends upon it’s temperature [say 300 K]
and it’s total surface area [6 square km].
Now, change it so all the small cubes are at certain distance from each other [say 10 meters
apart]. So have 1 billion 1 meters cubes with a 10 meter space between them and forming much larger cube. So get a cube shape which if looking at one side or any side of it, one mostly see space between them and mostly just see 1 meter cubes which are on outside of cube.
But depending angle looking at it, it can look like it’s solid- say looking at from angle the cube looks like pyramid. So depending angle looking at it, one see part cubes sides which in the interior. But for most part it’s just going to be the 1 meter cubes on the exterior radiating most of the energy.
So the cube which widely space cubes will radiate more energy than the tightly packed cube,
but I would guess it should not be a lot more energy.
Also with tightly stacked cubes one have heat conducted from interior to exterior, whereas if there is distance between them that only radiate energy can be transferred.
So question what would the differences be between tightly stacked blocks and spaced blocks be in terms how much and how they radiate energy?
For example it seems each of the blocks on the exterior of the widely spaced cube should radiate
more energy as compare to if tightly spaced and it also seems like the interior blocks would retain heat longer [as compared if tightly stacked].
Nullius in Verba and tjfolkerts,
Yes, sorry, a bit sloppy there. I did not include the concept of ‘work’ as a heating process simply because that wasn’t really part of the distinction I was trying to make. The ‘separate’ energy flows radiated between two BBs at different temperatures across a vacuum versus the net of the two (Q) was what I was getting at.
Bottom line, though: ONLY the net is a transfer of heat. The individual energy flows are NOT.
About ‘thermal energy’ vs. ‘heat’:
“Thermal energy is the part of the total potential energy and kinetic energy of an object or sample of matter that results in the system temperature. This quantity may be difficult to determine or even meaningless unless the system has attained its temperature only through warming, and not been subjected to work input or output, or any other energy-changing processes.”
“Microscopically, the thermal energy may include both the kinetic energy and potential energy of a system’s constituent particles, which may be atoms, molecules, electrons, or particles in plasmas. It originates from the individually random, or disordered, motion of particles in a large ensemble, as [a] consequence of absorbing heat.”
“Heat, in the strict use in physics, is characteristic only of a process, i.e. it is absorbed or produced as an energy exchange, always as a result of a temperature difference. Heat is thermal energy in the process of transfer or conversion across a boundary of one region of matter to another, as a result of a temperature difference.”
“When two thermodynamic systems with different temperatures are brought into diathermic contact, they spontaneously exchange energy as heat, which is a transfer of thermal energy from the system of higher temperature to the colder system.”
http://en.wikipedia.org/wiki/Thermal_energy
I find this definition/distinction very sound and easy to relate to. I don’t see how it confuses you, Nullius in Verba.
***
davidmhoffer says:
April 25, 2013 at 10:14 pm
***
David, your example is fine. The best IMO is the graph of IR leaving the earth. The integral under the curve is the total IR emission, and that must stay the same in equilibrium (since the sun’s input is constant). There’s a big “chunk” taken out by CO2. If the CO2 “chunk” is increased by increased CO2, then other parts of the curve MUST increase to maintain equilibrium (maintain the same total area under the curve). The surface emission is where it will increase — higher surface temp, while the TOA temp decreases (where CO2 emits). Same total IR leaving, just cooler at TOA & warmer at surface. Slayers need to understand this simple energy conservation law.
This does NOT address feedbacks/cloud & convection changes, etc — another story.
But the fact is, thermodynamics texts point out the refrigerator as a proof of energy flowing ONE way.
>>>>>>>>>>>>>>>>>>
This is a classic example of not understanding the difference between heat flow and energy flux.
As for the half dozen attack comments toward me overnight, I think it clear that a sincere attempt to help people understand the physics that is used to design and build everything from nuclear reactors to, yes, refrigerators, is unappreciated by those determined to wallow in their own ignorance, ignoring the math and data presented while accusing me of ignoring the math and data!
The ranks of the skeptics have become sickeningly infiltrated by those whose belief system trumps the facts, and who quote the laws of physics as if they say the opposite of that they do. These people are no better than the Phil Jones and Michael Mann’s of the world. They are blind because they choose to be blind.
Dropping thread. What a testament to the willful self imposed ignorance of the human race.
Gary Hladik says, April 26, 2013 at 2:49 am:
“I did read all of that, and it still doesn’t change the fact that, supposedly at equilbrium, power J is input to the system and power J1 = J/2 is radiating out. Either the temperature of the system rises (in which case it wasn’t at equilibrium), or energy is being destroyed. If the system is at equilibrium, you have to explain where the “extra” power J1 = J/2 is going. Note that if the input power is actually J1, inputs and outputs balance nicely and the system is indeed at equilibrium.
BTW, rather than telling me to read on to find “it”, why not just repeat “it”? That makes it a lot easier to figure out what you’re trying to say.”
Gary, according to your own words, you’ve read all the relevant passages multiple times and still you’re on repeat as if you hadn’t read it even a single time.
You’re completely stuck on the notion that the ‘system’ heat loss to space MUST equal the input to the sphere with or without the shell. As if the conditions in the two situations were equal. Even when it’s blatantly obvious (and stated several times) that with the shell in place, the steady input flux from the sphere’s internal heat source has more ‘work’ to do on its way to space than without the shell intercepting it. The entire flux from the surface of the sphere (J) cannot reach space, because it ’has to’ heat the shell on the way. Ergo J = J/2 + J/2, Q = Q’ + Q”, 400 W/m^2 = (400 – 200) + 200.
Q’ stays within the system. Q” escapes it. There is still perfect radiative balance.
Look at it this way: J is the entire and the only ‘real’ flux in the system. The two J1s are simply components of J, manifestations of the splitting of J upon interception by the shell (which has no other heat source than the sphere: J).
There are in effect TWO thermodynamic systems to consider here: the sphere and the shell. The sphere’s heat budget looks like this:
Hot reservoir (internal heat source): J –> || Sphere’s surface: J –> || Cold reservoir (vacuum between sphere and shell)
J = 400 W/m^2 (290K)
The shell’s heat budget looks like this:
Hot reservoir (sphere’s surface): (J – J1 =) J1 –> || Shell’s outer surface: J1 –> || Cold reservoir (space)
J1 = 200 W/m^2 (244K)
Let’s say the sphere’s surface is provided with a constant heat input from its source of 400 W/m^2. This warms the surface to an emission temperature of 290K. This temperature in turn spontaneously and naturally produces a flux emitted as outgoing heat loss of 400 W/m^2. This radiative flux is not something you can suppress or disallow the way you can suppress convective heat loss. It is dictated by the laws of physics.
Since the heat input is constant, the surface of the sphere will not cool in the meaning ‘temperature dropping’. As it would to cooler surroundings if the input were cut. This also means that there is no actual ‘cooling rate’ to reduce. The surface of the sphere would remain at 290K just as much with surroundings at 0 K as at 289K, it doesn’t matter. Neither of these two surrounding temperatures is capable of transferring heat to the sphere.
You are mixing Q (the heat transfer between two systems) into the concept of what sets the surface temperature of the warrmer object. If the warmer object is heated by a constant heat source, then the absorbed heat from that heat source is what sets the surface temperature of that object. And the heat loss flux will automatically correspond directly. Q doesn’t help. Q (or in the case of the shell, Q’ (J1)) simply reflects the speed by which the warm object can heat the cooler one, or at dynamic equilibrium how much net energy is being transferred per unit time from the warmer object in order to maintain the temperature of the cooler object.
Myrrh says:
April 25, 2013 at 4:52 pm
Myrrh, diffusion works for all molecules, no matter their specific weight. Diffusion is in fact Brownian motion at molecular scale. Wind and convection is what brings CO2 to long distances and height, up to the stratosphere. But the movements of real molecules of real gases in the real world is what keeps CO2 in the mixture.
In stagnant air, as is the case for e.g. air in open pores in snow, the heavier molecules and isotopes have a tendency to collect at the bottom of the column. For the Law Dome ice core, that resulted in about 1% increase of CO2 at closing depth, 72 meters below surface, after 40 years of settling time. For which is corrected in the measurements of CO2 in the ice cores.
In open air, the air movements even over a few years are more than sufficient to mix all CO2 everywhere and keep it mixed…
Kristian, I applaud your enthusiasm, but Gary’s explanations are much better than yours here.
You say:
Analogies are fraught with peril, but money can often be a very good analogy for energy.
Suppose “Mister Planet” has an income of Q = $400 one day. This is the “entire and the only ‘real’ payment into the system” (note: the income from the outside is NOT “J”). He in turn pays “Miss Shell” $400 that day = “J”. Miss Shell ‘splits J upon interception’ into J1 = $200 to Mrs. Space, and another J1 = $200 back to Mr. Planet.
Well, the next day, Mr Planet has an income of $400 (Q) AND another $200 (J1),. So now he pays Q + J1 = J = $600 to Miss Shell, who sends half to Space ($300) and half back to Planet ($300).
The next day: Q + J1 = $400 + $300 = $700 income and J = $700 passed on to Shell, who passes $350 back and $350 to Space. This is an infinite series approaching $400 + $400 = Q + J1 in to the planet and J = Q + J1 = $800 out from the planet (ie steady state with no temperature change). The shell has $800 in from the planet, $400 out to the planet, and $400 out to space (ie steady state with no temperature change).
Hmmm, I ran the calculation like this:
>>> ((3.84*10²⁶)/((5.6*10²³)*(5.6*10⁻⁸))⁰⋅²⁵
332.65
(Total Solar Output/(Total Surface Area*σ))⁰⋅²⁵
Doing that I got J/2 = J₁ for 1366 and 683, or thereabouts, as expected from problem 1023.
Hmmm…
Initially it would be:
1366⇒⇐0 | 0⇒, Net 1366 to shell, 0 to space, 0 emitted by shell
Then as it began heating the shell:
1366⇒⇐341 | 341⇒, Net 1025 to shell, 341 to space, 683 emitted by shell
1366⇒⇐683 | 683⇒, Net 683 to shell, 683 to space, 1366 emitted by shell
That fits problem 1023, and it seems like it should be stable.
If Net 683 to shell, 683 to space isn’t stable then perhaps:
1366⇒⇐1366 | 1366⇒, Net 0 to shell, 1366 to space, 2732 emitted by shell
If Net 0 to shell, 1366 to space is stable, I don’t see how you would get the star to heat up without:
*star heats up until the shell receives 1400*
1400⇒⇐1366 | 1366⇒, Net 34 to shell, 1366 to space
1400⇒⇐1400 | 1400⇒, Net 0 to shell, 1400 to space
*star heats up until the shell receives 1500*
1500⇒⇐1400 | 1400⇒, Net 100 to shell, 1400 to space
1500⇒⇐1500 | 1500⇒, Net 0 to shell, 1500 to space
*star heats up until the shell receives… etc, etc, etc*
Though it’s worth noting the scale of the shell radiation in comparison to the solar radiation:
>>> (5.6*10⁻⁸)*(5778⁴)
62,416,322.5146
>>> (5.6*10⁻⁸)*(5778⁴ – 395⁴)
62,414,959.2618
>>> (5.6*10⁻⁸)*(5778⁴ – 332⁴)
62,415,642.1521
*sings*
Around 62 million watts per square meter, around 62 million watts… you take one in and emit it right back, about 62 million watts left after all…
…
Oh, using 1026 I get that a shell at that distance reduces the rate of cooling by 0.999978222696
Why does the sun emit near blackbody radiation. I ask this question because the Sun is primarily made of hydrogen and a bit of helium which aren’t greenhouse gasses. Why does plasma radiate?
Now imagine a significant mass of nitrogen in outer space, held together by gravity which undergoes nuclear fusion. Does it emit any radiation? According to “ARGUMENT BY AUTHORITY” it’s not a GHG and doesn’t have any way to radiate heat to space does it? Does it then reach infinite temperature? What if we lower it’s temperature to 15c. Does it radiate then? What if we replace the iron core in Willis’s thought experiment with this Nitrogen ball surrounded by a metal shell. Any takers?
I thought that all matter with a temperature greater than 0K emitted radiation, the wavelength determined solely by it’s temperature, and NOT spectral emissions.
“I find this definition/distinction very sound and easy to relate to. I don’t see how it confuses you, Nullius in Verba.”
It doesn’t confuse me. I already have an intuitive understanding of what’s going on, whatever you choose to call the components of the system, and so it’s just a matter of relabelling. You can call it ‘Wakalixes’ if you like. But it’s a lot harder to simultaneously *construct* an intuitive understanding while struggling to figure out the usage of words being used to describe it, especially when they’re used in similar but not identical ways to their everyday meanings. It’s a poor teaching method.
In a textbook, where you can set out the definitions systematically, building up the complexity, it’s a valid way to describe/label a mathematical exposition to aid memory. What you’re actually teaching is the mathematics, you have to read the definitions from the maths, and the conventional names for them used in thermodynamics are a convenience because they’re easier to pronounce than mathematical symbology. But to explain a bit of physics casually without that carefully constructed framework, it’s confusing.
I should perhaps have asked what definitions were being assumed, or tried to write it in a way that avoided the confusing terms. To always say ‘energy’ rather than ‘heat’, for example. I don’t know if it helps, though.
It seems to me that the ambiguous terminology is sometimes used as a tactic and excuse to avoid answering the point, and people switch definitions freely to make whatever claim they want to at that moment. They’re not using it in a technical sense for a specialist concept – they’re just equivocating. I don’t have any patience for it. Whatever you call the things, the physics works the way it works, and even if people were using them in the technical sense they still couldn’t reach those conclusions by any valid logic. When somebody says heat flow is bidirectional, or that in a refrigerator heat flows from cold to hot, it’s perfectly obvious which convention they’re using and what they mean. To argue with that on the basis of language is avoiding the point; it’s not what the argument is about. To try to claim that this means a cold object can’t warm a hot object is invalid, whichever and whatever convention is used.
Anyway, when I said I thought it was even more likely to confuse, I meant it in the same sense as your “this seems very much to be where the confusion arises”. It wasn’t intended as a confession. 🙂
@paulinuk:
Here’s the basic deal: Isolated stationary atoms and molecules have emission spectra consisting of lines. So, that means that a tenuous gas of said objects still has the same except the lines start to broaden some due the molecular speeds (“Doppler broadening” http://en.wikipedia.org/wiki/Doppler_broadening ) and due to the interactions between molecules (“pressure broadening” or “collisional broadening”). As the molecules get closer together, i.e., the gas gets denser and denser, the lines become broader and broader. By the time that you get to densities of the liquid and solid state, you are really no longer talking about lines at all but rather a continuous spectrum.
For dense matter, like solids and liquids, this is essentially true. I say “essentially” because there is still a unitless factor between 0 and 1 called emissivity which is a function of wavelength. (In a sense, even a tenuous gas can be thought of in this way, except in such a case the emissivity will be an extremely strong function of wavelength and will be essentially zero except where there are the spectral emission lines.)
@ur momisugly Max April 26, 2013 at 10:25 am
What exact problem are you doing? It looks like you are putting a 1 AU shell around the sun, but then trying to compare that to “Problem 1023”, which only works for a shell near the inner sphere. Let’s assume you have replaced the sun with a 0.999 AU sphere @ur momisugly 394 K radiating 1366 W/m^2.
For Problem 1023 and the 0.999 AU sun, then your results do work:
“1366⇒⇐683 | 683⇒, Net 683 to shell, 683 to space, 1366 emitted by shell”
The one thing you seem to be overlooking is that the “net FROM the sphere” is now only 1366 – 683 = 683. So we can how power the 0.999 AU sun with only 683 W/m^2 rather than 1366 W/m^2 that would have been needed with no shell around it.
Exactly as Problem 1023 suggests. 🙂
[If you want a “tiny sun”, then your “⇐683” is wrong. Of the 683 emitted by the shell inward, ~682.99 will hit other parts of the shell, and only ~0.01 will hit the sun. The “net to shell” will be 1366 – 0.01 ~ 1366, while the “net to space” will still be 683, so the shell will keep warming up.]
Myrrh says (April 26, 2013 at 1:28 am): “We cannot feel Light as Heat.”
Myrrh is here! Now it’s really a party! 🙂
As for “feeling” light, stand in sunlight coming in through a glass window. Feel the heat? Most glass blocks infrared, yet you still feel the sun.
http://www.physicsforums.com/showthread.php?t=388609
As an experiment, you can buy a pane of glass guaranteed to transmit no infrared, put it between you and the sun, and see if you still feel the heat.
“…and your visible light from the Sun cannot raise the temperature of matter.”
Then a green laser shouldn’t be able to pop a balloon, right?
And a blue laser shouldn’t be able to burn paper, right?
Myrrh, doesn’t it bother you that your bizarro physics is so easily refuted?
I hadn’t noticed this til now…
Christopher Monckton says (regarding John O’Sullivan):
Perhaps if I wait long enough, Monckton will eventually admit he is wrong and I am right about everything that we disagreed on as he has now implicitly done about the “hot spot”. He had long argued that it was a fingerprint of warming due to greenhouse gases and had even badly misinterpreted contour maps in the IPCC report to bolster his claim, despite my patient attempts to try to set him straight. Now, he has reversed his position with no admission of his having been wrong and I right.
Of course, he still gets some things wrong:
(1) The IPCC has never asserted that “it would only arise from greenhouse warming”. He misinterpreted the IPCC report as saying this despite attempts by myself (and probably others) to explain to him why he was wrong.
(2) He misses the most obvious explanation for the hot spot being “missing”, which is remaining issues with artifacts in the multidecadal trends in some of the satellite and radiosonde analyses. In fact, we know this is the case because the analyses don’t even agree with each other, besides some of them disagreeing with the expected behavior.
(3) He ignores the fact that amplification of temperature fluctuations & trends as one goes up in the tropical atmosphere (the more technical description of the “hot spot”) has in fact been confirmed for temperature fluctuations on the monthly to yearly-timescales where artifacts in the data are not a problem. It is only for the long-term trends, where artifacts are a big issue, that there has been some difficulty in confirming such amplification.
Nullius in Verba says
” When somebody says heat flow is bidirectional, …… it’s perfectly obvious which convention they’re using and what they mean.”
Is it ?
A similar example would be a person who says ‘a whale is a fish’
Both show ignorance of the technical definition of the word.
It depends on whether they would understand the correct definition if it was pointed out to them.
This WUWT thread is attracting a reasonable level of scientific interest so it is worth sticking to the correct technical definition of the words.
richardscourtney says:
April 26, 2013 at 3:15 am
“This inhibits the passage of radiation from the Earth’s surface to space so the temperature of the Earth’s surface rises (for the same reason that an insulated house is warmer than an uninsulated house).,……………….”
Consider this:
Solar energy at the earth’s surface evaporates water [Heat of Vaporization]
The air, the water vapour and the CO2 with it heat up, expands and rises. [part of the water cycle].
The air, water vapor, and CO2 rises in the atmosphere and give off heat at higher altitudes.
The clouds form as the water vapour condenses. [Heat of Condensation], and possibly precipitation. The heat at the surface is of the earth is transferred from the surface to the upper atmosphere by conduction, convection and some radiation.
That does not look like “trapping” heat to me.
Alberta Slim:
re your post at April 26, 2013 at 1:06 pm .
You may be right, and to some degree you certainly are. But that changes the subject.
It is not relevant to what I was trying to explain.
Richard
Alberta Slim says:
So, what you are saying is that convection (and accompanying evaporation/condensation) is another way to transport heat away from the surface of the Earth. However, there are a couple of problems:
(1) This moves heat up in the atmosphere but not out into space. The only way to transfer energy out into space is via radiation.
(2) What convection does do is raise the temperature of the upper atmosphere relative to the surface. Since an important element of the greenhouse effect is that the temperature at the height at which the radiation escapes to space is lower than the temperature at the surface, why can’t convection “fix” this, i.e., why can’t it just transport heat up in the atmosphere until the temperature no longer decreases as you go up in altitude? The answer is that it turns out that the atmosphere is unstable to convection only as long as the lapse rate (decrease in temperature with height) is greater than what is called the “adiabatic lapse rate”. This means that convection can reduce the lapse rate to a certain degree but no further. If it could eliminate the lapse rate entirely, then the greenhouse effect would disappear…but it can’t, so it doesn’t.
richardscourtney says: April 26, 2013 at 3:15 am
Thank you for the dialog. I have some concerns regarding your latest comment. Specifically, your second point includes the statement
In reality, water vapour and some of the trace gases in the air are (GHGs). A little of the mostly short-wave radiation from the Sun is absorbed in the atmosphere by GHGs. But GHGs only absorb a little of this short-wave radiation. Importantly, the total heat input (from the Sun) to the Earth+atmosphere system is the same as in Case 1. Most of the radiation from the Earth’s surface is long-wave radiation which is efficiently absorbed in the atmosphere by GHGs. This inhibits the passage of radiation from the Earth’s surface to space so the temperature of the Earth’s surface rises (for the same reason that an insulated house is warmer than an uninsulated house).
[Radiation from Earth surface –> atmosphere (i.e. effectively insulation) –> space] (emphasis mine).
First, it is true that greenhouse gases including water vapor will absorb a small percentage of the incoming solar energy. [As an aside, I’m not sure exactly how much energy is absorbed, so I’m not sure if “small” is the correct adjective. In particular, for a blackbody at a temperature of 5780 Kelvin (the sun’s equivalent blackbody temperature), approximately 51 percent of the radiated energy is in the IR band (0.3 Terahertz to 430 Terahertz). The absorption bands of the various atmospheric greenhouse gases may correspond to a relatively small fraction of the total IR band, but at first glance it appears that greenhouse gases may act as an “insulator” (albeit a weak one) for incoming solar radiation.] However, water vapor forms clouds, and clouds reflect incoming solar energy. As I understand it, the average albedo of the Earth is 0.3; and most of this albedo is due to sunlight reflection from clouds. As such, although it is technically correct to say greenhouse gases absorb a small fraction of the incoming solar radiation, it is not correct to say greenhouse gases have minimal impact on the solar radiation absorbed by the Earth/Earth-atmosphere system. One way to see this is to replace the greenhouse gas water vapor with the greenhouse gas CO2. An atmosphere containing only the greenhouse gas CO2 will affect the total radiation absorbed by the Earth/Earth-atmosphere system differently than an “equivalent” amount of the greenhouse gas water vapor. This definitely makes the type of greenhouse gas an issue when discussing the total absorbed solar energy.
Second, if radiation from the Earth’s surface is the only way heat leaves the Earth/Earth atmosphere system, then ignoring cloud formation, I agree. Greenhouse gases would act “like an insulator” to outgoing radiation. However, in addition to absorbing IR radiation, greenhouse gases emit IR radiation. Furthermore, convection and conduction transfer the heat present in all gases near the surface of the earth higher altitudes where greenhouse gases can more efficiently radiate to space. It may very well be that the net effect of all these processes is an increase in the Earth surface temperature; but that is anything but obvious. As other commenters have pointed out, for a given Earth surface temperature fluid dynamics as well as radiation affect the rate energy leaves the Earth/Earth atmosphere system. These issues (and potentially other issues for which I am not even aware) are why I say “I don’t know the answer to any of the questions posed in my April 25, 2013 at 5:18 pm comment.
Reed
Kristian says (April 26, 2013 at 7:46 am): “Gary, according to your own words, you’ve read all the relevant passages multiple times and still you’re on repeat as if you hadn’t read it even a single time.”
Indeed, I have read it multiple times, and I understand exactly what you’re saying, probably better than you do. I just don’t buy it, because it contradicts known physics.
“You’re completely stuck on the notion that the ‘system’ heat loss to space MUST equal the input to the sphere with or without the shell.”
That’s true, I am completely stuck on the First Law of Thermodynamics. Funny thing, so is the rest of the universe. 🙂
“Q’ stays within the system. Q” escapes it. There is still perfect radiative balance.”
No. For one thing, you’ve left out the so-called “back radiation” J1 = J/2 from the shell toward the sphere (see the diagram for problem 1023). Since the shell is very close to the sphere (problem 1026 with R ~ r), J1 has nowhere to go but the sphere. What happens to it?
“There are in effect TWO thermodynamic systems to consider here: the sphere and the shell.”
With ONE input (e.g. a radioisotope inside the sphere), and ONE output (the shell). Either power in = power out, or the temperature changes. There are no other choices, unless you’re creating/destroying energy.
“The surface of the sphere would remain at 290K just as much with surroundings at 0 K as at 289K, it doesn’t matter. Neither of these two surrounding temperatures is capable of transferring heat to the sphere.”
Which is the entire problem. Bottom line is, you’re a disciple of the “imaginary second law of thermodynamics”, as scienceofdoom puts it (or “i2L” as I like to call it). From your comment April 26, 2013 at 3:30 am (“SoD’s confusion, for instance, about what the second law of thermodynamics is actually saying and not saying.”), I take it you’ve read his textbook examples of the REAL Second Law,
http://scienceofdoom.com/2010/10/07/amazing-things-we-find-in-textbooks-the-real-second-law-of-thermodynamics/
but still prefer to make up your own science. This is discouraging, but I’ll attempt to lead the horse to water one more time:
It turns out that a solar cooker, for example here:
http://solarcooking.org/plans/funnel.htm
can also act as a refrigerator at night, reaching temperatures below ambient atmospheric and even producing ice when air temp never appoaches zero degrees C. If you read the refrigeration section of the article, you’ll note the instruction to point the “cooker” at the open night sky rather than warmer objects like clouds, trees, walls, etc. This refutes your claim that radiation from the “surroundings” has no effect on an object’s temperature.
What Roy’s image of the thermal release from a house shows is that heat ALWAYS rises.
This means that the atmosphere must ALWAYS act as a surface cooling mechanism.
ie.. IT IS NOT A BLANKET AND IT DOESN’T HEAT DOWNWARDS
joeldshore writes “What you are missing is that emissivity is proportional to absorptivity (and both depend on the thickness of the gas layer). When it emits more, it also absorbs more.”
But that’s precisely NOT what I’m missing. So at the ERL in the case of more GHG it is both emitting more and absorbing more…for no net effect. So why does the ERL have to change with more CO2?
I’m not saying it doesn’t, I’m wondering why it “does”. And as per one of my earlier posts the only reason I can come up with is that opacity is not linearly proportional to distance/concentration which is entirely non-intuitive and furthermore I haven’t ever seen that result documented.
It worries me that the particular result has been derived from measurements of the atmosphere (and entered into modtran) when there is actually another reason for it.
One more thing joel, you said “What you are missing is that emissivity is proportional to absorptivity” but in fact emissivity is precisely equal to absorptivity for the same substance. So there is possibility of different emission/absorption from say H2O to CO2 but not from CO2 to CO2.