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)
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
joeldshore try this explanation. Maybe it’ll help you see where I’m coming from with this and perhaps you can spot the flaw (if there is one) with this reasoning.
Taken from the point of view of a CO2 molecule at the ERL ready to emit. The photon emitted is heading up and above it is a haze of CO2. But at the ERL that photon has a 50/50 chance of escaping (by definition) and so maybe it makes it, maybe it doesn’t.
Now double the CO2 levels. Again taken from the point of view of the CO2 molecule at the (original) ERL , the photon released upwards sees a doubly thick haze of CO2 and so now it only has a 25% chance of escaping. But with double the CO2, right alongside the first CO2 molecule is a second CO2 molecule also ready to emit. And the photon produced also has a 25% chance of escaping. So as independent events, together there is still a 50% chance of one of them escaping.
This represents no net effect and the ERL doesn’t need to change for the same amount of radiation to escape. Where is the flaw? Please dont parrot back the standard speil, Use my example to illustrate where it goes wrong.
One might argue for example that the second CO2 molecule might actually radiate downwards instead and the answer to that is to simply choose another one that is radiating upwards. When the CO2 levels are doubled, the second half equally radiates both up and down and you can (on average) pair up ups and downs from each half.
“But with double the CO2, right alongside the first CO2 molecule is a second CO2 molecule also ready to emit. And the photon produced also has a 25% chance of escaping. So as independent events, together there is still a 50% chance of one of them escaping.”
And if you quadruple the CO2 level, there would be a 100% chance of one escaping? And what happens after that?
It helps to think of a more intuitively familiar situation like a layer of fog, or a glass full of diluted milk. To infra-red eyes, that’s what the Earth’s atmosphere would look like. It is somewhat fuzzy, but it is as if the ‘visible’ surface radiating to space were the top of the fuzzy layer. If you add some milk to the water, you can see into it about an inch. If you add more milk, the original surface is obscured.
For optically thin materials (i.e. that you can see right through) it does work somewhat as you say. But as the material becomes optically denser, the surface becomes more “solid” and the light emitted approaches a maximum, but the light escaping diminishes in geometric progression. Eventually the substance may be considered ‘opaque’, and the absorbing surface layer is only a few tens of thousands of molecules thick. All the light that escapes comes only from the surface layer. All the lines of sight from deeper layers to the outside are blocked.
If the material was isothermal, then this would make no difference. Every line of sight ends in some absorbing molecule, so all that changes is the thickness of the translucent surface layer. The light emitted (once thick enough to be entirely opaque) is the same. But when the temperature varies, the thickness of the translucent layer comes to matter. A thin layer only emits at the upper surface temperature. A thicker layer emits at a range of temperatures.
“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.”
This graph:
http://en.wikipedia.org/wiki/File:Solar_Spectrum.png
Shows how much of the solar spectrum is blocked from incoming solar energy.
Graph is linked form
http://en.wikipedia.org/wiki/Sunlight
In the graph the yellow is how solar energy reaches the top of Atmosphere and the red
is how much reaches the surface at sea level. Assuming sun is at zenith and it’s a clear day.
The graph indicate 5 chunks taken out by the H20. And a little bit by oxygen, CO2, and ozone [O3].
The vertical part of graph relates to amount watts square meter per nanometer of spectrum.
Also note a large amount spectrum is missing in visible light spectrum- which due no gas absorbing this spectrum of light but instead the sunlight scatter/reflected/diffused going thru Earth’s thick atmosphere.
So sunlight is 1360 watts per square meter at top of atmosphere and by time reaches earth surface it’s around 1000 watts per square meter- losing about 360 watts per square meter.
Or more correctly prevent 360 watts per square meter from directly reaching the sensor which measuring how sunlight is reaching the surface- some that sunlight could be scattered so reaches other parts of earth [or arriving at different angle]. So all it’s indicating is the energy is
not directly reaching the sensor at the ground [pointed at the sun].
Anyhow one can make a guess of how much watts of sunlight is blocked by H20.
So looks like about 50 nm and at about 1/4 watt per square meter- so about 12.5 watts for first two chuncks taken out. So total of about 25 watts. Third chunk remove seem like most amount
and it’s less than 1/4 watt per square meter removed but about 100 nm. So it about 25 watts.
And the two chunks look less than 25 watts in total. So somewhere around 100 watts per square
meter is blocked by H2O. This energy is not reflected but is absorbed and would be re-radiated
in some direction.
Nullius writes “And if you quadruple the CO2 level, there would be a 100% chance of one escaping? And what happens after that?”
No. Still 50%. There will be 4 tries to escape each with a 12.5% chance of succeeding because the amount of CO2 above has again doubled and so the chance for each escape has halved. Each are independent events.
Hmm. Yes. I see what you mean.
Independent probabilities don’t quite add like that. n molecules each with escape probability p result in a probability (1-p)^n that none will escape, and hence 1 – (1-p)^n that at least one will. With two molecules it’s 44%, not 50%. With 4 molecules 41%. However, since there is a chance of more than one photon escaping it works out the same on average.
However, that’s only the case for optically thin materials. The total cross section of all the molecules cannot exceed 1.
TimTheToolMan: Notice that your argument only works if we were right at the level where there was a 50-50 chance for the radiation to escape to space.
Let’s consider emission from other levels: For example, emission from a level lower in the atmosphere where there was originally only a 25% chance of escape. Now, doubling CO2 reduces that to only a 6.25% chance of escape per photon, so even with double the photons emitted, you are still down in total emission from this level by a factor of two.
On the other hand, consider a level higher in the atmosphere where the chance of escape was originally 90%. Doubling the CO2 reduces that to 81%, and since you have double the photons emitted, you are now actually up in emission from this level.
So, what is the net effect? The net effect is that you have more emission from levels higher in the atmosphere and less from those lower in the atmosphere. This would be a “wash” if the atmosphere were isothermal, but it’s not: The levels lower in the atmosphere are warmer and emit more while those higher in the atmosphere are colder and emit less.
Hence, the result is more “weight” from levels of the atmosphere that emit less and less “weight” from levels of the atmosphere that emit more.
Joel Shore said:
“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”)”
I say:
Thanks for the reply but this only complicates things for me. It doesn’t mention gas density in Thermal Doppler Broadening: the Doppler shift is related to the spectral frequency, velocity (or temperature) and mass of the particle.
If you accelerate electrons in a radio antenna they emit radio waves and the same is true if you accelerate protons. If a neutrally charged atom is accelerated, due to it having a temperature, then won’t it too emit radio waves because it contains internal charge? This radiation has got nothing to do with spectral line broadening because electrons and protons don’t have a spectrum as far as I know!
Nullius in Verba says:
Of course, optical thickness depends on wavelength and there are wavelengths in the atmosphere for which it is optically-thin. However, you still don’t end up with more emission at these wavelengths as the density of optical absorbers/emitters increases. The reason is because you have the Earth below (sort of the ultimate optically-thick backstop) and so you always end up in net subtracting from the Earth’s emission. [I believe if you did have an optically-thin gas with no as-warm or warmer solid emitter behind it then you could increase the emission by adding more gas.]
Here’s an experiment: Inject Nitrogen gas at 1atm and at 400K into a large chamber full of N2 at 1atm and at 288K. Observe the hot plume with an infra-red camera as it enters the chamber. It should be invisible as “N2 can’t radiate heat”???. If the plume is visible then there’s something terribly wrong with atmospheric radiation theory.
Nullius writes “However, that’s only the case for optically thin materials. The total cross section of all the molecules cannot exceed 1.”
The definition of ERL from Mosher’s link (which seems good enough to be a working definition) is
“ERL: The lowest level in the atmosphere from which infra red radiation is able, on average, to escape upwards to outer space without being reabsorbed”
“On average escaping” means 50% escape and so that defines the opacity quite clearly. If instead of using the ERL, we used ground level then the chance of a photon escaping would be very low. And doubling the amount of CO2 above it would still only halve the chance but when the chance is minuscule, it hardly matters conceptually does it.
“Here’s an experiment: Inject Nitrogen gas at 1atm and at 400K into a large chamber full of N2 at 1atm and at 288K. Observe the hot plume with an infra-red camera as it enters the chamber. It should be invisible as “N2 can’t radiate heat”???. If the plume is visible then there’s something terribly wrong with atmospheric radiation theory.”
The hot air could heat the walls- that should be visible.
Not sure if replace hot N2 air with hot CO2, you would see the gas.
No.
You don’t need any physical experiment or “Einsteinium thought games” to find a fundamental problem with these infinitely-continuing CHG discussions:
From above:
For simplicity, let’s assume those values are correct.
Now. If every second of every day, 360 of Trenberth’s 1360 watts DON’T penetrate a supposedly “completely transparent” atmosphere, but since their energy does arrive at top-of-atmosphere but doesn’t arrive at bottom-of-atmosphere, then how do those 360 watts get “lost” from the earth’s overall system?
Clearly, they ARE absorbed by the oxygen, nitrogen and argon and dust and aerosols, and ARE re-radiated by those same gasses and dusts and solids. However, just as clearly, they ARE also ignored by the CAGW community because their theory does not allow such inconsistent, inconvenient maths.
“Now. If every second of every day, 360 of Trenberth’s 1360 watts DON’T penetrate a supposedly “completely transparent” atmosphere…”
The term isn’t “completely transparent”, it’s just “cloudless”.
And in what sense are they being “lost” or “ignored”? They appear in the Kiehl-Trenberth diagram, in which the numbers add up. It’s simply re-radiated, ultimately to outer space. It doesn’t actually matter that much whether it was stopped on the way in or the way out, all that matters is what level it is radiated out from. And the CAGW community go on about aerosols quite a lot.
joeldshore writes ” emission from a level lower in the atmosphere where there was originally only a 25% chance of escape. Now, doubling CO2 reduces that to only a 6.25% chance of escape per photon”
If you’re at a level where there is 25% chance for a photon to escape and you double the CO2 then you halve the chance, not quarter it. Hence it becomes 12.5% chance per photon and there are two of them still totalling 25% chance to escape for independent events.
Are you sure you have your probabilities right ? Or am I the one making a mistake in this?
“If you’re at a level where there is 25% chance for a photon to escape and you double the CO2 then you halve the chance, not quarter it.”
You square it. Probabilities multiply. Think of it as two 25% chances, one after the other.
Nullius writes “You square it.”
Of course. Silly me, clearly a blonde moment. So its molecules radiating NOT at the ERL that make the differences. sheesh.
If it’s any comfort, I had the same moment. I knew the probabilities didn’t add, but when I saw that it worked out fairly close for p = 0.5, I got myself mixed up for a bit.
You can see a certain depth into any translucent material. The upper part of that layer isn’t so strongly obscured, but isn’t dense enough to be very visible, either. As it gets thicker, it both gets more visible and more obscured by the material above it. The obscuration increases exponentially faster so it soon disappears from sight. But in the shallow surface layer, you can initially get some partial cancellation from the two effects. I’ve learnt something from having to think about it. Thanks.
RACookPE1978 says:
Yes, there was actually a meeting held in a super-secret location organized by Maurice Strong in which all the groups that do atmospheric modeling pledged to not include in their models things that are inconvenient like this. It’s amazing that you have found this out!
Boy, and that guy that you all hate because he claimed a correlation between belief in AGW being a hoax and belief in conspiracy theories was clearly completely out-to-lunch!
How long do you think it takes the energy represented by one surface photon to escape the Earth’s atmosphere? Unimpeded by collision, it would be nearly instantaneous, but slowed by what could be many millions of collisions and absorption/emission cycles? Sure, there is a delay which is enormous from a photon POV, but it will be small from perception POV. There would be a distribution, but what’s the median value of the distribution? Five milliseconds? Ten? You cannot build a practical heater from this delay. And, as should be obvious, anything capable of increasing the Earth’s surface temperature by 33C would be a very practical and useful heater.
Put a gallon jug of water in your front yard and assign yourself the trivial challenge (trivial compared to Mother Nature’s hard work of a 33C increase) of increasing its average temperature by 10C or even 5C using “back radiation”.. You academic climatologists can’t do it. Not even close. Oh, I forgot, the GHE heating machine only works when the the air is rarefied and cold. Right.
Don’t take it from me, take it from Maria Bykova, a PhD chemist from Novosbirtk, RU. “There is no mechanism,” she said to me. You AGW academics and true believers are running a flea circus.
tjfolkerts says, April 26, 2013 at 9:14 am:
“Kristian, I applaud your enthusiasm, but Gary’s explanations are much better than yours here.”
Your post is written in such a ridiculously condescending tone that I really shouldn’t be bothered
answering it. But never mind.
We (hopefully) all agree that the flux from the surface of the sphere cannot be suppressed or disallowed in any way by a countering flux. It could not physically obstruct or block it, reducing it that way. No, we are simply subtracting it budgetwise (OUT-IN) at the surface of the sphere.
So in this case we have:
400 –> 400
Planet alone in space.
400 –> || 400 –> <– 200 || 200 –>
Just after the shell is emplaced.
The heat loss from the sphere (and from the shell) is now too small (400-200=) 200 W/m^2. So it has to increase.
The surface temperature of the sphere rises to 321K, resulting in an outward flux of 600 W/m^2 and a net flux (heat loss) toward the shell of (600-200=) 400 W/m^2:
400 –> || 600 –> <– 200 || 200 –>
But remember, the incoming to the shell is still physically the 600 W/m^2 leaving the sphere. It’s not been reduced. So it will split into:
400 –> || 600 –> <–300 || 300 –>
For a net heat gain for the shell of (600-300=) 300 W/m^2 and a corresponding heat loss (from the outer surface) of 300 W/m^2.
Still no radiative balance, though. So the sphere needs to heat up some more:
400 –> || 700 –> <– 300 || 300 >
Well, we have the same thing going, the shell will warm up further and then the sphere will have to warm up even more and so on …
Let’s go straight to the equilibrated situation, then:
400 –> || 800 –> <– 400 || 400 –>
Now compare this to mine:
Sphere – 400 –> 400
Shell – (400-200=) 200 –> 200
What is the difference? Not much really. There is just one thing to notice. I am not adding any heat to the sphere from the shell. I am only adding heat to the shell from the sphere.
How so?
Let’s deconstruct the process.
For the shell it works out fine for Tim. The heat received is (800-400=) 400 W/m^2 to the inner surface, from the sphere. The corresponding heat ejected from the outer surface to space is 400 W/m^2. The same. We have energy AND heat balance. This is exactly equivalent to my shell’s steady state.
What about the sphere? I have heat balance. Why? Because the incoming flux from the internal source (the heat gain) is exactly matched by the resulting outgoing flux (the heat loss). I also have energy balance. Why? Because the outgoing flux of 400 W/m^2 is transferred to the other thermodynamic system, the shell, in whose budget it is fully accounted for (in a reduced heat gain and heat loss). We cannot BOTH subtract the outgoing flux from the shell’s inner surface from the incoming flux from the sphere (reducing the shell’s heat gain) AND from the outgoing flux from the sphere’s surface (without taking into account the still existing opposite flux from the outer surface of the shell). This would constitute double counting. Or rather, you would turn Q = Q’ + Q” into Q = 2Q’, skipping Q” and rather doubling Q’. This is what Tim does.
What you actually have for the flux leaving the surface of the sphere with the shell in place, i.e. between sphere and space, is:
400 out from sphere, minus 200 in from the shell, plus 200 out from the shell. You cannot include the minus without the plus: Q = (Q – Q’’) + Q’’, J = (J – J1) + J1.
You can see this quite clearly if you compare with the Carnot cycle:
http://i1172.photobucket.com/albums/r565/Keyell/Carnot_zps4049e783.jpg
If you’re standing in place of the shell (or the Carnot engine), facing inwards toward the surface of the sphere (the hot reservoir) and with space (the cold reservoir) at your back, what do you see? You see the incoming flux from the sphere/hot reservoir (Q = J = Qh). This is absorbed in its entirety and split upon re-release. When the shell is heated to its steady state temperature, it sends out J1 (J/2) from its inner surface and a similar flux J1 (J/2) from its outer. The outwardflowing J1 flux (Q’’ = Qc) is the heat loss of the shell. But what about the inwardflowing J1 flux? Where does that fit into it all? It reduces the heat gain of the shell: J – J1 = J1, Q – Q’’ = Q’ = Q/2. In other words, it reduces the net energy coming in to the shell (J – J1), not the originally incoming flux from the sphere itself (J). So here’s the take-home message: The inward J1 reduces the heat gain to the shell. The outward J1 increases its heat loss. J, though, is not touched by any of the J1s. If it were, then there would be no J to come in to the shell in the first place. It would’ve been reduced (halved) even before arrival.
Tim agrees. He’s quite intent on letting the full outward flux from the sphere (800 W/m^2) hit the inner surface of the shell even after he’s cut the heat loss flux from the surface of the sphere in half ((800-400=) 400 W/m^2).
It is actually easier to follow this line of thought with the Carnot cycle (above):
There you would have:
Qh
V
Engine –> W (Qh/2)
V
Qc (Qh/2)
Half of Qh from the hot reservoir is used for work in/by the engine (W), half is shed to the cold reservoir (Qc). Nothing at all at any time is ’flowing back’ to the hot reservoir to reduce its heat loss (Qh = J) or, worse, to somehow add to its heat gain. W is the equivalent to Q’ (J – J1, Qh – Qc). Energy is being continuously utilized by the intercepting ’layer’ and this cannot ever reach the cold reservoir. Hence the reduced Qc/Q’’.
I have a hard time seeing how this simple circumstance could be such an impossible thing to comprehend.
After the incoming net flux of energy to the shell is reduced by half by the inward J1, there is still no net flux flowing from the shell to the surface of the sphere, meaning going IN. The resulting net flux is still positive for the shell. It is still moving outwards … ultimately from the outer surface of the shell to space: Q = Q’ + Q’’ = (Q – Q’’) + Q’’. Note from this, the heat transfer rate to the shell from the sphere is reduced as the heat transfer rate to space from the shell is increased.
Tim does have an apparent heat balance for the sphere’s surface:
400 –> || 800 –> <– 400
(400 IN; (800-400=) 400 OUT).
But upon closer inspection his problem readily reveals itself. How? Well, how on earth did he get to the 800 outgoing W/m^2 when the heat input is only 400 W/m^2? The 800 W/m^2 after all has to come from a rise in surface temperature from 290K to 345K (a 19% increase). How is this accomplished?
We know already that the internal heat source itself only supplies 400 W/m^2 to the surface. This corresponds to an emission temperature of 290K. We also know that the incoming 400 W/m^2 from the shell can not in any way disallow or suppress/block the resulting BB emission flux from the surface of the sphere toward the shell of 400 W/m^2.
So what are we left with?
The incoming 400 W/m^2 from the shell. Extra HEAT. Only these could have provided the heat to raise the surface temperature of the sphere past the Stefan-Boltzmann dictated emission temperature from the internal heat source input.
So in the end, there is no subtraction, no reduced heat loss from the sphere. There is only addition of extra heat from the shell to the sphere’s surface. This is the only way to raise its temperature beyond 290K.
And this is of course a complete and blatant violation of the second law of thermodynamics. And it goes directly against what Spencer (and all the rest) is claiming, that the reason why the sphere is heated is because its heat loss is reduced by the shell, not because its heat gain is increased by it. A cooler object simply cannot increase the heat gain of a warmer one.
Gary Hladik says, April 26, 2013 at 3:08 pm:
“Indeed, I have read it multiple times, and I understand exactly what you’re saying, probably better than you do.”
Haha, you’re quite a piece of work, aren’t you Gary? No, you’ve understood diddly-squat. You think (and claim) you understand what I’m saying, but quite clearly you’re not. You’re simply wallowing in your own self-satisfied ‘I’ve already got the answer, the Truth, so you don’t even have to try to prove me wrong’ kind of mind-set. Completely blinkered. Well, good for you. I guess it keeps you happy.
“That’s true, I am completely stuck on the First Law of Thermodynamics. Funny thing, so is the rest of the universe. :-)”
Only, this is not about the first law of thermodynamics. 🙂
The first law of thermodynamics does not say in any way or fashion that the heat loss of one thermodynamic system needs to balance the heat gain of another. How come you’re so willing to misrepresent it?
Do you know the concept of thermodynamic systems, Gary? And how it relates to the first law of thermodynamics?
Do you know anything about the Carnot cycle? About ‘work’ and ‘heat’? And how they relate?
One starts to wonder.
““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?”
For crying out loud! Q’ is J – J1! There’s your J1. If I ‘left out’ the back radiation J1 = J/2, then Q’ would be the same size as Q and Q” would be 0. We would have || J –> || 0 –>. J = J + 0. How’s that for a radiative balance?
““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.”
No, no, no! The sphere has one heat input and one heat output (400 –> 400), the shell has one heat input and one heat output ((400-200=) 200 –> 200). How hard is this to get?
““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” (…)”
Nope. You’re a disciple of the ones who believe such an ‘imaginary law’ exists, having ultimately created it for themselves, as a strawman to focus on and cling to instead of dealing with what the real objections to what they’re doing are actually 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- f-thermodynamics/ but still prefer to make up your own science.”
Yes, I’ve read them. SoD is the one imagining things. But I can see the source of his confusion.
Regarding that, I also take it you’ve read my ‘modern’ physics references of how ‘heat’ is defined as opposed to ‘thermal energy’? How is the shell tranferring heat to the sphere, Gary, increasing its content of thermal energy, elevating its kinetic level, raising its temperature beyond what the input from the internal heat source can muster?
Do tell me.
“Put a gallon jug of water in your front yard and assign yourself the trivial challenge (trivial compared to Mother Nature’s hard work of a 33C increase) of increasing its average temperature by 10C or even 5C using “back radiation””
Sigh. The greenhouse effect as implemented in the climate models doesn’t work by back radiation. It’s not the mechanism the science proposes. So you’re completely wasting your time attacking it. Even if you disproved the existence of back radiation (which you can’t, because it does exist) it would have absolutely no impact on AGW, because it’s totally irrelevant. All the argument does is discredit scepticism to no effect.
Increasing the temperature of a jug of water with backradiation is easy – it’s even easier to do it with a whole swimming pool of water. Just put it at the bottom of a solar pond.
Gary and Tim,
The idea that you people seemingly subscribe to, is that the half of the outgoing flux from the surface of the sphere that does not end up as heat loss to space from the outer surface of the shell, has the power to do work twice. First it is spent warming the shell. Then it sort of reawakens to do a second warming job. It goes back to the sphere whence it originally came, and warms that some more too. (Of course leaving the shell bereft of thermal energy in the process – i.e. it cannot maintain its temperature AND warm the shell at the same time.)
This is loony accounting.
Inserting an opaque layer between a heated surface and space will make the layer warm, not the surface warmer than before. The surface is the layer’s source of heat. The layer simply ’takes’ some of the heat from the surface that used to go straight out to space for itself, spending it on heating up. There is still perfect radiative balance between the surface and its surroundings (heated layer + space). There is also radiative balance between the heated layer and its surroundings (surface + space). The whole situation is simply different from before the layer was inserted.
How come this fundamental knowledge of the inner workings of thermodynamic systems completely bypass seemingly grown and, one must assume, well-educated people when it comes to the magic of the radiative GHE?
Ferdinand Engelbeen says:
April 26, 2013 at 8:50 am
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.
Ferdinand – AGWScienceFiction doesn’t have real gas molecules. So Brownian Motion is not applicable.
They use Brownian motion “explanations” because they don’t have convection, if they explained convection they wouldn’t have the AGW Greenhouse Effect, so they have to avoid it. They avoid it by saying their gases are ideal, and then the rest is just adding confusion by giving explanations which ‘appear’ to explain well mixed, but when you break it down you can see it’s not applicable.*
What I am trying to explain here is that AGWSF has substituted “ideal” gas for real gas, ideal gas has no properties – it does not have volume, it does not have attraction, it does not have weight, it is not subject to gravity – because it is not real. Ideal gases are imaginary hard dots of nothing with no mass so nothing for gravity to act on. It remains a useful beginning point in calculations but only makes sense when all the missing properties and processes are added back in. This is a great source of confusion in these discussions, when those from traditional physics and applied science don’t know this is what they have done in AGW fisics…
AGWSF have empty space instead of our real world atmosphere which is composed of real gases, not ideal.
Here, real world gases with volume is how we can hear sound: http://www.mediacollege.com/audio/01/sound-waves.html
“Note that air molecules do not actually travel from the loudspeaker to the ear (that would be wind). Each individual molecule only moves a small distance as it vibrates, but it causes the adjacent molecules to vibrate in a rippling effect all the way to the ear.”
This is not the “empty space with molecules miles apart” of the AGW SF ideal gas narrative, but a medium created by the juxtapositions of molecules with volume squashed together by gravity.
They have no sound in their impossible world, so they can’t hear anyone of this..
This is our real atmosphere, real gases with volume under gravity. These real gases expand when heated and become lighter than air and rise, they condense when they get cold, become heavier than air and sink. This is basic bog standard meteorology, this is how we get our winds, by differential heating of volumes (packets) of air.
This is how we get our rain, by heated water evaporating and rising because less dense lighter than air carrying that heat away from the surface, and when it reaches the colder heights the real gas water vapour releases its heat and condenses into liquid water or ice, and precipitates out, it rains.
That is the Water Cycle, the “greenhouse gas” water vapour takes heat away from the surface into the colder heights of the troposphere, this is how we get our weather. Rain is carbonic acid, because real gases have attraction and water and carbon dioxide are greatly attracted to each other – water has a residence time of 8-10 days in the atmosphere, and carbon dioxide in that is fully part of it.
I am also trying to point out that there is no internal joined up logic in their explanations, they use “ideal gas diffusion by their own molecular momentum travelling at great speeds through the empty space atmosphere of ideal gas bouncing off each other in elastic collisions and so thoroughly mixing”, and, they use Brownian Motion which is applicable to particles in volumes of fluid, gas or liquid, and ideal gases have no volume, and, they use “turbulent mixing by winds” – none of these are applicable in the real world of real gases.
I’ve just come indoors, there is not a breath of wind outside – as I look out of the window the tops of the trees are still, as I look into the sky the clouds are not moving. Brownian motion hasn’t moved the dust on my desk.. The major wind systems do not cross hemispheres, there is no “global mixing by turbulent winds”. They don’t have winds anyway! Because winds are volumes of atmosphere moving, and they have only empty space. Because to get wind you have to have real gases which can get lighter and heavier than air which separate out.., which means heat transfer by convection, etc. etc.
These are AGWSF “memes” to explain “well mixed” which their fictional fisics based on ideal gas is incapable of producing.
They have no rain in their carbon cycle because it spoils their AGW narrative, not that any of them have noticed the whole of the real Water Cycle is missing.., but they couldn’t have it anyway with their gases which go straight from the surface to empty space..
How do they get clouds with all the ideal gas molecules miles apart from each other bouncing off each other at great speeds?
Their Earth and its atmosphere is a complete and utter fiction, it is created by sleights of hand mixing and removing and renaming and lots of other techniques, manipulations using terms from physics.
*AGWSF introduced this into the general education system some time ago, they began by teaching new infant/junior level teachers who weren’t specialised, and it’s these which began the classroom “explanations of well mixed” by opening bottles of scent and saying it was Brownian motion which spread the smell, and by pouring coloured ink into water to show that this was “how carbon dioxide thoroughly mixes by diffusion so it can’t be unmixed”. These, in the real world of real gases, are by spread convection currents.
“Separation diffusion from convection in gases
“While Brownian motion of multi-molecular mesoscopic particles (like pollen grains studied by Brown) is observable under an optical microscope, molecular diffusion can only be probed in carefully controlled experimental conditions. Since Graham experiments, it is well known that avoiding of convection is necessary and this may be a non-trivial task.
“Under normal conditions, molecular diffusion dominates only on length scales between nanometer and millimeter. On larger length scales, transport in liquids and gases is normally due to another transport phenomenon, convection, and to study diffusion on the larger scale, special efforts are needed.
“Therefore, some often cited examples of diffusion are wrong: If cologne is sprayed in one place, it will soon be smelled in the entire room, but a simple calculation shows that this can’t be due to diffusion. Convective motion persists in the room because the temperature inhomogeneity. If ink is dropped in water, one usually observes an inhomogeneous evolution of the spatial distribution, which clearly indicates convection (caused, in particular, by this dropping).”
http://en.wikipedia.org/wiki/Diffusion
The confusion is deliberate, only some/one who knew real world physics very well indeed could have come up with this complex, subtle, scam.
Kristian says:
April 27, 2013 at 8:43 am
And it goes directly against what Spencer (and all the rest) is claiming, that the reason why the sphere is heated is because its heat loss is reduced by the shell, not because its heat gain is increased by it. A cooler object simply cannot increase the heat gain of a warmer one.
As I said before, not my preferred subject of discussion (too long ago…), but what happens if you simply wrap the sphere in some kind of insulation? With a constant heat source, the sphere temperature simply will go up much higher than without insulation, until a new equilibrium is reached where energy source = energy loss.
Something similar happens with the shell: that hinders the radiation to the outside world, no matter if that is by direct transfer of heat (via a real insulator) or indirect via radiation. In all cases, all the energy of the source must be radiated to the outside world, or one of the parts of the system will heat up, either the sphere or the shell or both. As the system is in equilibrium when energy out = energy in, the shell need to emit the same amount of energy as the source within the sphere, nothing less.
No, no, no! The sphere has one heat input and one heat output (400 –> 400), the shell has one heat input and one heat output ((400-200=) 200 –> 200). How hard is this to get?
The 200 radiated from the shell to the sphere is additional to the 400 from the inner source. The sphere thus has two heat inputs. The sphere receives 400+200=600, but only emits 400. Thus the sphere heats up. That is what happens.
The sphere doesn’t know that the source of the outside 200 is cooler than itself, energy is energy, no matter if that is from vibrating molecules or from radiation. Still no violation of any physical law, as the radiated energy flows from warmer objects to colder objects are larger than reverse. All what happens is that the total energy loss of the warmer object is reduced, like as if it were insulated.
Some molecules in a material have higher vibration (thus a higher “temperature”) than others, even in a cooler object compared to a warmer. That doesn’t violate any law, as the average energy flow is from a warmer object to a cooler object.