Guest Post by Willis Eschenbach. [See Update at end]
The CERES dataset continues to yield new insights. My joy is to graph different relationships and then see what I can learn and find out from the graph.
Inter a lot of alia, the CERES dataset allows us to calculate the amount of the very poorly named “greenhouse radiation”. This is the downwelling longwave (thermal) radiation from the atmosphere. It’s commonly called “DLR”, which stands for downwelling longwave radiation.
(NOTE: This is not a thread for disputing the existence of downwelling longwave radiation from the atmosphere. DLR exists, it’s measured both manually and automatically all over the planet every day, and it doesn’t violate the 2nd Law. Get over it.
If you wish to dispute that, fine, I encourage you to do so … but this thread is not the place to do it. There are dozens of threads on the subject, the web is a large place, pick one and go. Please, don’t test me on this. I don’t want to have to snip comments claiming no DLR or that the “greenhouse effect” violates the 2nd Law, but I sure will … END NOTE.]
As first proposed by Ramanathan, the downwelling longwave greenhouse radiation (DLR) can be measured from satellite and ground measurements. It’s the amount of upwelling longwave surface radiation absorbed by the atmosphere. It’s calculated as the upwelling longwave radiation from the surface minus the upwelling longwave radiation measured at the “top of atmosphere” or TOA, meaning in this case at the satellite.
The CERES dataset has values for all-sky and clear-sky radiation. The difference between those is the effect of the clouds.
Now, the amount of downwelling longwave radiation changes over time. And in theory, part of this change is from an increase in CO2. We can calculate the theoretical change in DLR resulting from the change in CO2.
Putting all of that together, we get the following plot of the changes in downwelling “greenhouse” radiation since the turn of the 21st century.
Figure 1. Changes in downwelling “greenhouse” radiation by source.
Now, there are several things of note in this graph.
First, the total change in downwelling radiation from the atmosphere (“greenhouse radiation”, or DLR) is more than three times the change due to CO2. Presumably, this must be from a combination of changes in water vapor, latent/sensible heat loss, atmospheric solar absorption, and increased surface temperature.
Next, CO2 is the minor player in the greenhouse DLR game. Non-CO2 clear-sky DLR increase over the 2000-2022 period was ~ 3.2 W/m2. Cloud DLR decrease was ~ 1 W/m2. But the CO2 change was only 0.7 W/m2, the smallest of the three.
Next, I theorized a couple of decades ago that thunderstorms, clouds, and other emergent phenomena act to oppose temperature changes and thereby stabilize the temperature. Since then, I’ve provided a variety of evidence to back up that claim. Here, in Fig. 1 you can see that while greenhouse DLR from CO2 and from water vapor are increasing, to the contrary, the greenhouse DLR from the clouds is decreasing. This is evidence in favor of my theory.
In fact, the change in the clouds over the period has more than offset the change due to CO2 … who knew?
Finally, we can compare the change in greenhouse DLR to the change in temperature. This gives us the “TCR”, the transient climate response to a change in DLR.
In this case, the TCR is 0.2°C per W/m2, which would be equivalent to 0.7°C per doubling of CO2. This is markedly smaller than the usual value for the TCR, which is on the order of 1.5W/m2 per 2xCO2 or so. Looking at Figure 1 we can see why that is so. The total change in observed greenhouse DLR is ~ three times the theoretical change from CO2. Since the change is larger, the sensitivity to the change perforce must be smaller.
And since TCR values are typically about 55% of the equilibrium climate sensitivity (ECS), that would make this estimate of the ECS about 1.4°C per 2xCO2.
Finally, there’s been little change in total greenhouse DLR since ~ 2016 … another unsolved mystery of the sea.
[UPDATE:
I note that despite my polite request, some impolite folks want to hijack this thread to proclaim that the “greenhouse effect” isn’t real and downwelling radiation doesn’t exist.
I said I’d snip such comments, but I realized that if I do, the authors will likely style themselves as martyrs rather than just common thread hijackers. So I’m gonna pass.
In hopes of expanding their concepts of the greenhouse effect, however, let me recommend the following posts of mine:
The Steel Greenhouse 2009-11-17
There is a lot of misinformation floating around the web about the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, or a “mirror” that reflects part of the heat back to Earth, or “a pane of glass” that somehow keeps energy from escaping. It is none of these things.
People Living in Glass Planets 2010-11-27
Dr. Judith Curry notes in a posting at her excellent blog Climate Etc. that there are folks out there that claim the poorly named planetary “greenhouse effect” doesn’t exist. And she is right, some folks do think that. I took a shot at explaining that the “greenhouse effect” is a…
Let me also repeat my request. Please, guys. There are lots of places to debate these questions. But this is not one of them.
Sadly,
w.
Best to all,
w.
To Avoid Misunderstandings: When you comment please quote the exact words that you are discussing.
“ I don’t want to have to snip comments claiming no DLR or that the “greenhouse effect” violates the 2nd Law . . . .”
I don’t understand Mr. Eschenbach. The Second Law only applies to isolated systems. Clearly the Earth and the Earth’s atmosphere are not isolated systems. There is no requirement for the Earth’s atmosphere to obey the Second Law–although it may. You should be able to shut down any argument on that fact alone.
I find this an odd comment. My thermo book uses the atmosphere as and example of a large reservoir for use in development of the second law.
As the second law also deals with entropy is Jim also saying the atmosphere doesn’t have follow the idea of entropy?
I watched a video where Neil deGrasse Tyson was explaining the Second Law. At the beginning of the video he firmly and correctly states that the Second Law only applies to isolated systems. Then all the examples in the show demonstrating the Second Law were not of isolated systems. What can I say?
Entropy is defined as dS = squiggly dQ divided by T. The equal sign only applies to reversible systems. If you can find an actual reversible system, then let me know.
DeGrasse differs from my thermo book so I think he is wrong.
You didn’t answer my question. You said the second law only applies to isolated systems and the atmosphere wasn’t isolated so according to your claim the atmosphere doesn’t follow the idea of entropy.
i don’t think you can have it both ways.
“i don’t think you can have it both ways.”
I have no idea what you are talking about. The 1st Law only applies to closed and isolated systems; the 2nd Law only applies to isolated systems; and the 3rd Law only applies to closed systems.
Technically that is true, Jim. However, nothing in our universe satisfies any of those conditions. Yet we still manage to use the 3 Laws to describe how the universe works. It as all about “close enough approximations”, like everything else in physics (there are no frictionless surfaces or ideal springs either, but we can still figure things out close enough for practical purposes).
“Technically that is true, Jim.”
Thanks.
“However, nothing in our universe satisfies any of those conditions.”
So, considering that the Universe is an isolated system doesn’t qualify the Universe as an isolated system? The Earth and the Sun aren’t open systems? TYS’s hermetically sealed water glass isn’t a closed system? I think you need to rethink that statement.
“Yet we still manage to use the 3 Laws to describe how the universe works.”
It’s four laws, but isn’t it amazing how we can do that with our stupid physics?
“close enough approximations”
Sounds like government work.
Four laws, yes, my bad, the 0th tends to escape my memory. Never mind, carry on!
I said “nothing in our universe satisfies any of those conditions”, and the universe isn’t really “in” the universe. It “is” the universe. But yes, the universe (as far as we know today) satisfies the condition of being an isolated system subject to the 2nd law.
Technically, we can’t use the 2nd law to refer to anything smaller than the universe. But the same reasoning that gives rise to the whole-universe law (statistical probability) also applies to more-or-less isolated subsets of it. Right?
Your original point was that the Earth’s atmosphere need not obey the 2nd law. Neither does Earth itself. That is true. But what do you think is the exact probability of a colder object the size of Earth’s atmosphere making an object the size of Earth (which is warmer to begin with) even warmer still? It’s not 0, for sure, but how close to 0 is it?
Do you think that Willis is trying to stake his case on “The laws of probability are against me, not just astronomically, but universally – but come on, it could happen” ???? He could say that, but who would fall for it if he did? Besides you, I mean?
“Sounds like government work” Yes, the whole climate scare is indeed based on government work. And most of it is false.
“Four laws, yes, my bad, the 0th tends to escape my memory. Never mind, carry on!”
Thanks. The 0th law allows us to measure thermodynamic systems. Unfortunately, climate science misuses the concept.
“Technically, we can’t use the 2nd law to refer to anything smaller than the universe. . . . Right?”
Right.
“Your original point was that the Earth’s atmosphere need not obey the 2nd law.”
Thanks for remembering my original point. I get annoyed when people say that radiation from a colder object can’t be absorbed by a warmer object. If that was true then how could we see things colder than our eye retina? I doubt that the light from a firefly–even on a warm summer evening–is coming from an object warmer than my retina.
There’s a difference from saying that radiation from a colder object can be absorbed by a warmer object and that a colder object can warm a warmer object. The “net” flow of radiation will be from warmer to colder.
The warmer object will lose entropy as it cools and the cooler object will gain entropy as it warms. And the total entropy will increase in an isolated system per the 2nd Law.
“And most of it is false.”
True. I agree.
Quantum radiation absorption is not the same as classically measured power, which I think might be the point you are making. Fireflies produce short-wavelength light through some quantum trickery that avoids having to heat their body up to 6000 K, fortunately for them. But if you try to measure thermal power from them via a sensor that is at human body temperature (around 38 C), you will get a zero or negative answer (the firefly isn’t warmer than you are, at least I don’t think so). The Earth-surface pyrgeometers say the same thing (negative power) when pointed at the colder atmosphere. That is what I keep trying to teach Willis, but he refuses to listen.
“The warmer object will lose entropy as it cools” Don’t you mean gain entropy? The cooler state is the more statistically likely one (given an environment to lose heat to), hence higher entropy. The entire universe gains entropy as it cools to a uniform temperature. That’s the 2nd Law.
““The warmer object will lose entropy as it cools” Don’t you mean gain entropy?”
Seriously? Heat added to a system is positive heat transfer.. Heat removed from a system is negative heat transfer. That’s the usual standard. Entropy is delta S = delta Q/T (for a reversible system). Negative delta Q = negative entropy as T is always positive–absolute temperature. You’re not trying to tell me that a crystalline solid has more entropy than a gas or liquid? That would violate Boltzmann’s formula.
Let’s say that there’s a net transfer of heat, delta Qnet, from object B to object A. Object B is warmer than object A, so Tb > Ta. Therefore delta Qnet/Ta > delta Qnet/Tb. The gain of entropy by the cooler object is greater than the loss of entropy by the warmer object. And that would satisfy the 2nd law for an isolated system.
“The entire universe gains entropy as it cools to a uniform temperature. That’s the 2nd Law.”
No. The entire universe gains entropy as it warms–the so-called heat-death. Realize that the universe is also expanding which cools it.
Right, my bad, I got my signs backward. In any case, that’s not going to solve Willis’s problem. He’s trying to claim that a hotter object is gaining energy directly from a colder one. The 2nd Law doesn’t address this directly in the case of a sub-universe-sized system, true, but the same statistical mechanics still apply, just with less certainty. (Not much less in this case – Earth and its atmosphere are pretty macroscopic.)
As you said, two objects of different temperatures placed together will tend to equalize their temperatures, not drift farther apart, due to entropy (via the same statistical mechanics calculations that give rise to the universe-spanning 2nd Law). But that’s not what Willis said in the head post. You wanted him to tell people to stop using the 2nd Law to refute his false claim, but if he instead follows your advice and states more accurately that “although the 2nd Law doesn’t directly refute this statement, it is nevertheless vastly more likely to be wrong than right, due to the same statistical mechanics calculations that the 2nd Law is based on”, do you think that (a) he would understand any of that or (b) anyone else would now give in and let him get away with continuing to be almost certainly wrong with a probability on the scale of the lifetime of the known universe and beyond?
I’m going to reply here, because the troll has ruined the other thread.
Notice in my example, I said “net” heat transfer. That doesn’t mean that the cooler body isn’t transferring heat energy to the warmer body—just not as much.
I know I’m a stupid double-E, but I learned about electromagnetic waves differently than you did apparently. An electromagnetic wave has two vectors: E and H. The E vector is the electric field strength, and the H vector is the magnetic field strength. The units of E are volts/meter and the units of H are amperes/meter. The dot product of E and H is zero. That means the two vectors are perpendicular or ninety degrees apart as they travel through space.
A volt in MKS/SI units is a joule/coulomb. An ampere in MKS/SI units is a coulomb/second.
Where exactly is the “joules” only in the units?
An important operation is the cross product of these two vectors. Specifically, it’s E x H. Multiplying volts/meter by amperes/meter gives watts/meter^2. Wow!—it’s not just joules.
So I shine 1 watt on a 1 by 1 meter square surface. In one second, it’s 1 joule. In one minute, it’s 60 joules. In one hour, it’s 3.6 kilojoules. In one solar day, it’s 86.4 kilojoules. And in one non-leap year, it’s 31.563 megajoules.
Using your “only joules” criteria, let’s say it’s 60 joules. Now explain climate with that value–and good luck.
Did they say that the Stefan-Boltzmann law usually gives a watts/meter^2 value? The correct units for doing this climate nonsense IS watts/meter^2.
Well, Jim, you may have learned some EE, but you haven’t learned enough theoretical physics yet. You can’t just “shine 1 watt on a 1 by 1 meter square surface”. It doesn’t work like that. What temperatures are you dealing with? When you know the temperatures (energy differential), then you can calculate the power developed. Not before. This is what the S-B equation tells us.
And once you have your temperatures, and you calculate the power, the temperatures start to change over time as energy is transferred, and then the power developed will decline accordingly, until equilibrium is reached – which may take a long time of course depending on how your system is set up. (If your system includes the entire universe as a heat sink, then equilibrium won’t occur until the heat death of the entire universe.) Then at equilibrium you will develop 0 Watts.
Remember, you started this part of the thread by trying to defend Willis’s claim that a colder body can develop power (transfer Joules) to a warmer one. That doesn’t happen. And pointing out that the 2nd Law only applies to the entire universe (as written) isn’t going to help you, or him. The same calculations that go into it are also applied at smaller scales, with practically the same result, until you get down to individual atoms. But that’s not what we’re discussing here.
“Well, Jim, you may have learned some EE, but you haven’t learned enough theoretical physics yet.”
Yeah, I’m an engineer that doesn’t know physics. You’re on something, right?
“You can’t just “shine 1 watt on a 1 by 1 meter square surface”. It doesn’t work like that.”
Since when? You’re really off-base with that statement. I can shine a light/radiation source anywhere on any surface.
“When you know the temperatures (energy differential), then you can calculate the power developed. Not before. This is what the S-B equation tells us.”
No it doesn’t. The Stefan-Boltzmann Law is the integral of Planck’s Law. Wein’s displacement Law is the derivative of Planck’s Law. The temperature is only to define the specific curve of the radiating object, and the total radiation emitted. The radiation leaving the object contains no Kelvin term. There is no thermometer in the radiation. The Stefan-Boltzmann constant completely cancels out the Kelvin units.
“And once you have . . . Watts.”
Nonsense! Complete nonsense!
Remember, you started this part . . . that’s not what we’re discussing here.
I remember. I also know some classical thermodynamics. The 2nd Law only applies to isolated systems. The Earth, the Sun, and the Earth’s atmosphere are not isolated systems. If you want to get into statistical mechanics and statistical thermodynamics, then you may have an advantage. I’m not well versed in those subjects.
You are correct that the Earth and its atmosphere are not isolated systems. However, although the 2nd Law as written is descriptive of the entire universe, the probabilistic calculations on which it is based are equally applicable to any subset of it. The only difference is that as the subsets get smaller, the probability of following the expected entropy gradient gets smaller too. But those probabilities aren’t going to decrease to human lifetime scales until you’re talking about a few atoms. On the scale of Earth and its atmosphere, the probabilities involved are indistinguishable from the values that we would calculate for the entire universe. (i.e. either 0 or 1, depending on which direction you’re thinking in)
At night, with the Sun’s hot influence temporarily removed, the Earth’s surface, its atmosphere, and outer space are essentially 3 isolated (from the rest of the universe) thermally interacting bodies. Nothing else is going to add or remove a significant amount of energy to or from these 3.
So your earlier description of the entropy gradient when a warmer object is allowed to exchange energy with a colder one still apply to Earth and its atmosphere (and outer space). And as you said, warmer objects are required by entropy to transfer thermal energy to colder ones, because that is the direction that results in an overall entropy increase, in the absence of other significant influences. Since Earth’s surface is warmer than its atmosphere (which is warmer than outer space), the direction of energy transfer, work, and power is, as you said it must be, Earth -> atmosphere -> space. Not the other way around.
(Yes, because this subset of the universe is less than the entire universe, the probability of energy flowing in the direction we both think it should is not quite 1. It is 1 – ε. But ε is a tiny number indeed. 0 is not a better guess than 1 – ε, if you’re going to bet on it.)
So this obviously contradicts what Willis wrote in the head post. You are agreeing with me, not with him… so why are you blowing raspberries at me? Are you just trolling, or really unable to grasp the significance of what you yourself wrote?
(Your complete inability to grasp what the S-B law means is a separate topic, and isn’t necessary in order to follow what I wrote here, so we’ll leave that aside – you’ve already presented enough correct physics that you obviously agree with me and not with Willis)
Your comment differs from your last one as I seem to have learned more physics in the interim. I must be a fast learner.
Okay, a new presentation. There are three black bodies that I will label “A”, “B”, and “C”. A is colder than B and C is warmer than B. Since they are black bodies, they radiate at all frequencies. It’s possible for A and C to emit photons with identical frequency/energy. And if they arrive at B, then how does B know which is which? Also, if B refuses one of the photons, that would violate the property of a black body.
Now let’s say they are gray bodies (more realistic). With gray bodies you have an emissivity term in the S-B equation. Emissivity is frequency dependent. However, it’s still possible for gray body A and gray body C to emit photons with identical frequency/energy. If those photons arrive at gray body B (and assuming gray body B likes to absorb those specific photons), how does gray body B know which photon is which?
In other words, the 2nd Law does not prevent B from absorbing photons from a colder object.
“So your earlier description of the entropy gradient when a warmer object is allowed to exchange energy with a colder one still apply to Earth and its atmosphere (and outer space).”
I said “net” heat transfer. That doesn’t disallow heat transfers back and forth between the two objects.
“So this obviously contradicts what Willis wrote in the head post.”
No it doesn’t.
“You are agreeing with me, not with him… so why are you blowing raspberries at me?”
I like making noises with my tongue when I read nonsense.
“Are you just trolling, or really unable to grasp the significance of what you yourself wrote?”
Hmmm.
“Your complete inability to grasp what the S-B law means is a separate topic . . . .”
Yeah, it’s my double-E coming out. The units of S-B are watts/meter^2. The cross product units of the E and H vectors is watts/meter^2. I simply don’t understand your silly obsession with joules. Radiation isn’t just joules.
“But ε is a tiny number indeed. 0 is not a better guess than 1 – ε, if you’re going to bet on it.”
Did you know that according to statistical mechanics there’s a non-zero probability that all the air molecules in the room will gather in a corner and leave you gasping for breath?
I need to clarify a couple of things. When I said frequency/energy, I mean either frequency and/or energy not frequency divided by energy.
And the last statement should include the words: “But I wouldn’t bet on it happening anytime soon.”
Your latest scenario went off the rails when you got to this point:
“In other words, the 2nd Law does not prevent B from absorbing photons from a colder object.”
Now you are attempting to conflate classical thermodynamics with quantum physics. No one has managed to square that circle yet, and I doubt you’ll be able to either. I won’t say you can’t, of course – maybe you know more than you’re letting on, and you’re welcome to give it a shot. But it’s going to take a lot more heavy lifting than what you’ve done here so far. There’s a Nobel Prize in it for you, too, if you manage to accomplish it. Possibly more than one.
“Did you know that according to statistical mechanics there’s a non-zero probability that all the air molecules in the room will gather in a corner and leave you gasping for breath?”
I did know that, yes. And my entire point about your attempt to discredit the universal 2nd Law as a way to disprove Willis’s claim is that this probability exists, and is non-zero, but again, as you yourself said, “I wouldn’t bet on it happening anytime soon.” That would be my point exactly.
Your earlier description of the entropic motivation for a hotter object to lose heat to a colder one is indeed a direct contradiction of Willis’s claim. This is what we mean when we say that his scenario is a violation of the 2nd Law. That’s a bit of a shorthand, yes, and so the longer and more precise version would be more like “the odds are against it, and staggeringly so – don’t hold your breath”.
“Your latest scenario went off the rails . . . .”
Raspberries!
Keep on blowing those raspberries, Jim. So much fun, isn’t it? My daughters thought blowing raspberries was hilarious too – when they were about 3.
“Keep on blowing those raspberries, Jim. So much fun, isn’t it? My daughters thought blowing raspberries was hilarious too – when they were about 3.”
KIRK: What are you?
GUARDIAN: I am the Guardian of Forever.
KIRK: Are you machine or being?
GUARDIAN: I am both and neither. I am my own beginning, my own ending.
SPOCK: I see no reason for answers to be couched in riddles.
GUARDIAN: I answer as simply as your level of understanding makes possible.
SPOCK: A time portal, Captain. A gateway to other times and dimensions, if I’m correct.
GUARDIAN: As correct as possible for you. Your science knowledge is obviously primitive.
SPOCK: Really.
KIRK: Annoyed, Spock?
–Star Trek–TOS–The City on the Edge of Forever
Apparently blowing raspberries is all you understand.
“Now you are attempting to conflate classical thermodynamics with quantum physics. No one has managed to square that circle yet, and I doubt you’ll be able to either.”
This is from the back cover of my text on “An Introduction to Statistical Thermodynamics” by Terrell L. Hill:
“The book is divided into four major sections. Part I deals with the principles of quantum statistical mechanics and includes discussions of energy levels, states and eigenfunctions, degeneracy and other topics. Part Il examines systems composed of independent molecules or of other independent subsystems. Topics range from ideal monatomic gas and monatomic crystals to polyatomic gas and configuration of polymer molecules and rubber elasticity. An examination of systems of interacting molecules comprises the nine chapters in Part Ill, reviewing such subjects as lattice statistics, imperfect gases and dilute liquid solutions. Part IV covers quantum statistics and includes sections on Fermi-Dirac and Bose-Einstein statistics, photon gas and free-volume theories of quantum liquids.”
You incorrectly changed the subject so you wouldn’t have to answer my questions. Apparently, you don’t know about Statistical Mechanics and Statistical Thermodynamics, or you wouldn’t have made such an ignorant statement.
Someone several months (maybe years) ago tried to claim that the Sun would “reflect” any radiation from the Earth. Apparently he and others don’t realize that when a surface reflects radiation, it must first absorb it.
I also don’t think you have a clue about the vector cross product.
Which part of your textbook tells you that you can deduce the principles of classical entropy (the four laws of thermodynamics) from the absorption of a quantum of energy transmitted by a single photon?
There is no “entropy” in the universal quantum wavefunction. There is no arrow of time either, which is closely related to entropy. So these two domains are separate, and you cannot mix and match them to try to make a fake point.
You still haven’t explained how your accurate description of energy flow (heat) from a hotter object to a colder one, due to probabilistic maximization of entropy, i.e. what the 2nd Law is based on, fails to disprove Willis’s claim. He claimed exactly the opposite in the head post. (Measurements don’t support his claim either.)
“Which part of your textbook tells you that you can deduce the principles of classical entropy (the four laws of thermodynamics) from the absorption of a quantum of energy transmitted by a single photon?”
It was two photons actually. But we can increase the number of photons to millions from each source if that will make you happy. I’m sure it would be “theoretically” possible to find millions of photons from object C that all wouldn’t necessarily have the same frequency and/or energy and millions of photons from object A that would match up with the same frequency and/or energy—there would be a one-to-one correspondence in each group. Now please tell me how object B could tell the difference between photons from A and photons from C?
I do see your problem. If you answer that object B knows which-is-which, then you’ll have to explain how that magic occurs. If you answer correctly and say object B has no idea, then that destroys your argument. It’s best to try to discount my question and send me off on a “hopefully” wasteful hunting trip.
One of the failures of “classical physics” was trying to explain the heat capacity of diatomic gases. The idea was to assign (1/2)*k*T to each degree of freedom. A monatomic gas has only three degrees of freedom and that means the heat capacity of monatomic gases should be (3/2)*k*T–and it was. For diatomic gases, the degrees of freedom were 7 and so the heat capacity of diatomic should be (7/2)*k*T—and it wasn’t exactly.
The (1/2)*k*T represents the energy assigned to how many particles? I’ll give you a hint: it’s greater than zero and less than two. There’s a relationship between R, the ideal gas constant, and k, Boltzmann’s constant. That relationship is R = N0*k, where N0 is usually referred to as Avogadro’s number. If you see k in an expression, then they are talking about individual particles. The number of particles is usually large, but they don’t have to be.
“There is no “entropy” in the universal quantum wavefunction. There is no arrow of time either, which is closely related to entropy. So these two domains are separate, and you cannot mix and match them to try to make a fake point.”
Yet, there it is in the text book. The fake point is all yours.
“You still haven’t explained how your accurate description of energy flow (heat) from a hotter object to a colder one, due to probabilistic maximization of entropy, i.e. what the 2nd Law is based on, fails to disprove Willis’s claim.”
I said “net” heat flow. You keep leaving that out.
“He claimed exactly the opposite in the head post. (Measurements don’t support his claim either.)”
What measurements? The ones that show “net” heat transfer or the ones that show instantaneous heat transfer? I disagree with Mr. Eschenbach’s claim that heat energy isn’t energy. The units of heat are joules—energy. In the past they used calories and BTU’s, but those are energy units too.
Your word salad is getting far beyond relevance. Maybe you should show some math derived from the universal quantum wave function that proves there is no entropy and no time in our unique universe.
Keep in mind that a photon and quantum of energy are the same thing. As far as I know the universal quantum wavefunction does not disprove that.
Jim, the 2nd law allows a local entropy decrease at the expense of a wider entropy increase. But that’s not what the head diagram is showing, is it? It’s showing the colder atmosphere applying work (Joule heating) and therefore power to the warmer surface, without any other energy input from anywhere else. That is a direct violation of the 2nd law, no two ways about it.
It’s not applying work – the downwelling crap is just shorthand for the insulative (and other) effects of the STUFF between the ground and outer space.
Really it should be modelled as depending on the solar radiation coming in and all the physical properties of the materials involved, instead of a mystery heat source in the sky.
But that would mean spending all that juicy climate change grant money on proper scientific research instead of jet-setting to climate conferences.
Knowing the specific heats and mass of CO2 vs H2O, the latent heat transfer on the diagram appears off to me. N2/O2 seem to be a heat reservoir until the sun falls to the point where they begin to supply energy to more radiative gases.
As I said before, averages just don’t cut it in a complicated system. My thermodynamics professor would have failed us.
That’s not exactly the case. A colder object DOES radiate toward a warmer object. And, it will radiate more and more until equilibrium is reached. But that really only applies to similar, ideal objects. What’s missing are gradients with time, mass, absorption/emission, specific heats, source varying as a sinusoid, conduction, etc.
Radiates energy, yes. Power, no. Please study the relationship between these two concepts very carefully and then get back to us.
I’m not sure what you call “power”, but Watts per square meter, i.e., Joules per second per square meter is a power term to me.
Ideal bodies radiate based only on their temperature. Other bodies do not change this. Granted there are lots of other variables in the real world that complicate the amount of radiation emitted, but radiation is still based on temperature.
Now if you are talking about the NET exchange of power, I’ll agree that a cold body can not generate sufficient power to add additional energy to a hot body.
Ideal bodies radiate energy based on their temperature. This is measured in Joules. Bodies do not radiate power. The conditions for developing power are more complex than that. And there is no such thing as “net” power. There is only power, or no power.
Are you saying that radiation from a colder black body can’t be absorbed by a warmer black body? How would the warmer black body know that the radiation is coming from a colder black body? Obviously theoretical objects, black bodies have three basic properties: they absorb all frequencies, they radiate all frequencies according to Planck’s Law, and they adjust their temperature depending on the total energy received and radiated each instant.
Gases are not black bodies…
I thought the Sun was an extremely close example of a black body.
Strictly speaking, the sun is mostly plasma.
All objects above absolute 0 radiate energy; but developing power is a very specific process that cannot occur from a colder body to a warmer one, due to the 2nd law.
…in an isolated system. A lot of people ignore the most important part of the 2LOT.
Sure. In a non-isolated system, you can calculate the probability that energy will flow against the entropy gradient. What is that probability if we are talking about Earth and Earth’s atmosphere? It’s not 0, but how close to 0 is it?
Energy flows against the entropy gradient even in an isolated system with 100% certainty. It is heat that does not flow against the entropy gradient in an isolated system. Remember, heat is the net flow of energy. Using contemporary 1LOT nomenclature ΔE = Ein – Eout the terms Ein and Eout are the energy transfer components while ΔE is the heat transfer.
Anyway the point is that in non-isolated systems heat can transfer from cold to hot. The most obvious obvious examples are evaporator/condenser/compressor systems like what is found in refrigerators and HVAC systems. This does not violate the 2LOT because the system is not isolated.
Energy does not “flow” against the entropy gradient, ever. Energy is the “potential to do work”. It is present as a field, when dealing with an EM radiation context. The field doesn’t “flow”. And work is only done “down” the entropy gradient. Statistical probability tells you how unlikely it is for the opposite to happen, which depends on the size of the system you are looking at. For human-scale systems, the likelihood is essentially 0.
At no point in an HVAC system is a cold part of the system making a hot part hotter through direct energy transfer. Energy always flows from hot to cold.
Patently False. And you can easily convince yourself of this with nothing more than an IR gun.
Correct. But don’t conflate energy with heat. Energy absolutely moves from cold to hot. It is heat that does not move from cold to hot spontaneously.
Patently False. Heat (the net transfer of energy) is moving from the evaporator (cold) to the condenser (hot). The cold evaporator definitely plays a role in making the condenser hot. This is not a violation of the 2LOT, however, because the system is not acting by its own means (isolated). Work is being performed on the system to make the heat go against the entropy gradient.
Patenly False. First, I think you are conflating energy and heat. Remember energy is the ability to do work. Heat is the net transfer of energy. Look at the 1LOT closely. ΔE = Ein – Eout. In this equation ΔE is heat while Ein and Eout are just energy. If Ein = Eout then ΔE = 0. In that scenario there is no heat transfer despite there being energy transfers.
Look at the radiant heat transfer equation Q = εσ(Th^4 – Tc^4)A . It is derived from the 1LOT and SB law. Start with the 1LOT ΔE = Ein – Eout. Multiple both sides by time (t) and area (A) to get ΔE/tA = Ein/tA – Eout/tA. Rewrite in flux form ΔF = Fin – Fout. Swap in the SB law to get ΔF = εσTa^4 – εσTb^4. Notice that ΔF is the net transfer of flux between bodies A and B so we can rewrite this as Q = εσ(Th^4 – Tc^4)tA where Q is the heat gain, body A is hot with temperature Th, body B is cold with temperature Tc, t is the time of heat movement, and A is the area for the view factor being analyzed.
Notice in the derivation of the radiant heat transfer equation that the heat transfer is from hot to cold despite there being an energy transfer from cold to hot. It is because the energy transfer from hot to cold Ehtoc is greater than the energy transfer from cold to hot Ectoh such that Ehtoc > Ectoh that you have a net transfer (or heat) moving from hot to cold. The fact that heat moves from hot to cold spontaneously does not mean that energy is not transferring across the entropy gradient.
“Energy absolutely moves from cold to hot.”
No it doesn’t.
“Heat (the net transfer of energy) is moving from the evaporator (cold) to the condenser (hot)”
Not directly. The compressor (using energy from elsewhere, such that total entropy will increase when that energy usage is accounted for) creates a series of intermediary entropy gradients. At each step, heat is flowing from a hotter object to a colder one, in accordance with entropy and statistical mechanics (the same calculations that the 2nd Law is based on).
Energy cannot be transferred in two directions along a single entropy gradient at a given time. That’s not how power works. You are inventing imaginary flows of energy while calculating the amount of power being developed. As a mathematical tool, that is fine, if it makes your calculation simpler, which it really doesn’t – but it is not physics.
“with nothing more than an IR gun.” How do you think an IR gun works?
“Heat (the net transfer of energy) is moving from the evaporator (cold) to the condenser (hot)”
You are trying to confuse Joule heating with isentropic compression, and hoping that no-one notices. These are different thermodynamic processes, but you knew that, right? What makes you think that when I said “heat flows only from hot to cold”, that I was talking about any process other than Joule heating? Willis certainly wasn’t, nor was I when I pointed out that he was wrong. Are you deliberately trying to deflect from the central point of the argument?
“Energy does not “flow” against the entropy gradient, ever.”
Raspberries!
Yummy! But what have raspberries got to do with anything?
Truly, you can’t be THAT ignorant!
It is 0. But that is moot because in the sun/atmosphere/surface/space system, which is effectively isolated, the heat flow is sun (hot) ==> surface (warm) ==> atmosphere (cool) ==> space (cold).
If we then put a thermal barrier between the lower and upper atmosphere the heat flow is still sun (hot) ==> surface (warmer) ==> loweratm (warmer) ==> barrier ==> upperatm (cooler) ==> space (cold). The fact that the surface and loweratm warmed while the upperatm cooled does not change the direction of heat flow. What it does is change rate of heat flow.
That description of energy flow based on temperature gradients is correct. But that’s not what Willis wrote, is it? He wrote that energy (transfer of Joules, i.e. Watts) flows from the atmosphere to the surface. It doesn’t.
I believe stevekj has a conceptual problem in expressing what goes on. I hope he is trying to say that a hot body can not absorb and RETAIN energy from a cold body, thereby making it “hotter”.
Absorption and emission are two separate phenomena. If absorption is less than emission, a body may cool at a slower rate, but it will not reverse the gradient.
Isn’t there a Kirchhoff law about emission and absorption?
What’s funny is that if the 2nd Law always applied, then nothing could cool down. Objects that lose heat are also reducing their entropy. Negative heat transfer is a negative entropy change.
Jim may also be having conceptual problems distinguishing between quantum and classical (macroscopic humanly observable) descriptions of a situation. They are not the same. Which one are you dealing with?
Jim knows the difference. Of course, I forgot about adiabatic systems. They can cool down without losing heat.
You should do some research about what you are talking about. It has nothing to do with classical vs quantum. In fact you will find little to nothing about the actual mechanism whereby an atom or molecule absorbs a quanta (photon) of energy. The results of absorption are well described by quantum theory but the actual physical mechanism is still only theoretical. Kinda like entanglement. We know it occurs and can describe the effects. Exactly how it works is only theoretical.
Radiation of energy occurs by electromagnetic waves. The power contained in a EM wave is accurately described using Maxwell’s equations. An EM wave can contain the power of one quanta (photon) to many millions of watts. A single atom or molecule can intercept the EM wave and absorb some of that energy depending on various factors such as frequency. One atom or molecule does not have to absorb the entire power contained in a given wave, it only has to extract the amount of energy needed to move electrons into a new energy level(s). But let there be no doubt power is transmitted based upon the temperature of an object. The object doesn’t need to be the either hot or cold, each emits radiation.
“But let there be no doubt power is transmitted based upon the temperature of an object.”
This is what I keep trying to teach Willis, and he’s not listening…
stevekj,
Willis is both incompetent and ignorant. I see a similarity to Einstein, though – both Einstein and Eschenbach start with the same letter.
That is one similarity, indeed, Mike. I think it ends there though!
I did manage to get him to write that objects emit radiant energy, not radiant power, and that radiant energy is measured in Joules, not Watts. I count this as significant progress. But I don’t think he really understands any of this either, even though he wrote it himself. I keep trying…
Dim Jim,
You really don’t understand, do you?
You wrote –
“Obviously theoretical objects, black bodies have three basic properties: they absorb all frequencies, they radiate all frequencies according to Planck’s Law, and they adjust their temperature depending on the total energy received and radiated each instant.”
Your imagination is no substitute for reality. How hot is a blackbody absorbing the energy from a candle? Two candles? A supernova?
Next thing you’ll be saying that submerging a block of ice an a bowl of water will make the water hotter, because the water is absorbing the IR emitted by the ice!
Ah, reality has reared its ugly head, has it? Pity, that.
Back to the playground with you. Leave physics to the grown-ups.
I’ll add to some things recently said.
Quoting: “DLR exists, it’s measured both manually and automatically all over the
planet every day … Get over it.”
Really? Prove it. No, subtracting radiation intensities doesn’t prove it. Nor
does detecting longwave infrared radiation in the lower atmosphere prove it.
Especially when no correlation is ever made with the intensity of an alleged heat
source higher up (or lower down) in the atmosphere allegedly producing that radiation.
The “Downwelling Longwave Radiation” (“DLR”) they measure is, in fact, overwhelmingly,
blackbody radiation coming from the many transient, colliding, multi-molecular
entities produced in any gas at one-atmosphere pressure at ambient temperatures, where
the rate of collisions, per molecule, is about 10 billion/second (each collision,
especially if it’s a 3-body collision, being a potential source of such radiation,
even if the colliding “molecule” is Argon!). And, anyway, this blackbody radiation
detected doesn’t have anywhere near the energy to heat the surface of the Earth as
another form of “downwelling” radiation does — we call that unmentioned radiation
that quaint term, “sunshine,” though I’m sure the global warming wackadoodles forgot
that it even exists.
Furthermore, the vast preponderance of the heat in the atmosphere at 5000 feet
altitude and above, caused by convection from the surface of the Earth, just ends up
going still higher through more convection, not lower through radiation. This is
especially true of those frequencies of infrared radiation in the atmosphere at that
point, allegedly substantially contributing to its heat content, that are absorbed by
“greenhouse gases” like water vapor and CO2 — the energy of that radiation lasts a
few centimeters before it is reabsorbed and then continues its journey, by convection,
towards outer space.
You’re going to censor me? Yawn, yawn.
David Solan
They’ve never actually measured downwelling longwave infrared power at the surface, at ambient temperature, at night. They publish numbers that make it look like they have, but they haven’t. The actual measurements from the pyrgeometer sensors are negative when looking upward, which is what you would expect, since the surface is warmer than the atmosphere. Then they add fake numbers from mathematical fictions (an incorrect usage of the S-B law) to make it look like there is downwelling power. It’s quite a racket they’ve got going…
First, the total change in greenhouse radiation is more than three times the change due to CO2
Is “greenhouse radiation” here the DLR you were defending just prior to this statement or some other radiation category?
The same. Sorry for the confusion.
I’ll clear it up in the head post, thanks for the heads up.
w.
I note that despite my polite request, some impolite folks want to hijack this thread to proclaim that the “greenhouse effect” isn’t real and downwelling radiation doesn’t exist.
I said I’d snip such comments, but I realized that if I do, the authors will likely style themselves as martyrs rather than just common thread hijackers. So I’m gonna pass.
In hopes of expanding their concepts of the greenhouse effect, however, let me recommend the following posts of mine:
Let me also repeat my request. Please, guys. There are lots of places to debate these questions. But this is not one of them.
Sadly,
w.
Let’s just call it the atmospheric mass effect..
That is, after all, what it is. !
Well, there definitely is a Mass Effect – if you take the average temperature of Earth, Mars and even the greenhouse gas poster child Venus at an altitude where the pressure is half the surface pressure (i.e. half the atmosphere is above you and half below you on a mass basis), you’ll find it equals the SB computed value – regardless of the differences in atmospheric composition.
But all this climate change doomsday nonsense is focused on surface temperatures, or actually the temperature of the atmosphere just above the surface where the weather happens, so I am not sure if the previous observation would negate the idea of CO2 increases leading to higher temperatures near the surface because it slows the heat transfer to outer space.
But there is a mass effect on atmospheric temperature.
Thanks, PCman. You know me, I never believe something until I run the numbers myself.
Now, for the US standard atmosphere, that’s true … but it’s only true if we use the post-albedo value for solar strength (240 W/m2). Can’t use the TOA solar strength of 340W/m2, it gives the wrong answer.
240 W/m2 converts by S-B to a temperature of 255K. And the US standard atmosphere temperature at 506mb, which is about 5.5 km altitude, is 252K.
So far, so good … but then I looked at Venus. The relevant numbers are here.
The problem is that the albedo of Venus is about 0.7. So after albedo, Venus is only getting about 198 W/m2 … and that’s less than the Earth. But we know that Venus is very hot.
Now, 198W/m2 converts by S-B to 243K.
According to the second graph in the link above, surface pressure on Venus is about 1e5 millibars. Half of that pressure is half the mass above and half below, which would be 5e4 millibars … which occurs at an altitude of about 10 km.
But by the first graph in the above link, the temperature at 10 km is on the order of 650K … far from the 243K S-B temperature.
Hmmm … what am I missing here?
w.
Willis see fig 6 of the following article, also in your relevant numbers, I see. Its not the altitude which corresponds to 50% of the atmosphere. It’s the altitude at which the atmospheric temperature is equal to the radiative emissions temp of 243, then add the lapse rate down to surface. For Venus that 243 you calculated is about 60 Km up. The lapse rate is about 10/km depending on source. The atmosphere must be in convective mode for lapse rate to apply since lapse rate depends on the work lost/gained by the parcel of atmosphere as it rises/falls. So 243+ 11 x 60 = PFH…plus there are lots of other things going on in Venus’s atmosphere. Ask StephenWilde.
Sorry, D, but first, what is the “following article” you refer to?
Next, are you saying PCMan is wrong about the 50% above/below?
Next, I don’t understand the procedure. From my link above, I find the altitude of the Venusian atmosphere where the temperature is ~ 255K = 243W/m2. As you note, that’s at about 60 km altitude. And that, in turn, is at a pressure of 200 mbar. So ~ 99.8% of the Venusian atmosphere is below that.
Then you say the lapse rate is about 10K/km, so the surface must be at about 600K + 255K = 855K, which is higher than the top of the range of surface temperatures in the link. Kinda true, given the rough numbers—according to the link the lapse rate is ~ 8K/km.
But … so what? I mean, you could pick any other temperature and repeat the procedure with the same result.
w.
Sorry Willis, I just don’t think your Steel Greenhouse article is correct.
It isn’t.
That was years ago guys why can’t you focus on THIS article instead for any possible weak spots to talk about.
Sorry, Nelson and mkelly, but until you:
1) QUOTE exactly what you think is incorrect in the article, and
2) SHOW (not claim but demonstrate) exactly how and why I’m wrong
your opinion is totally and completely worthless.
Your move.
w.
I have in the past and you never responded so I am not doing again.
A very convenient excuse. Put up or shut up.
w.
Willis,
In your steel sheel article, you show the graphic (attached below I think) with the planet emitting radiation of 235 W m2 at the surface. The steel shell obtains the same radiating temperature and hence emits 235 W m2 back to the surface. Now you add the planet’s surface emitted radiation to the “back radiation” to get 470 W m2 now emitting from the surface. For this to be the case the planet’s surface temperature is much hotter than it was when it emitted at 235 W m2 originally. Here is where you lose me. For two bodies emitting at the same temperature, there is no net heat flow. The steel shell emitting 235 W m2 can not heat the planet’s surface which is also emitting at 235 W m2.
I am currently sitting near an oil-filled electric heater. It temperature at the top is 190 F when on high. As a thought experiment, suppose I wrap the heater in lots and lots of tin foil. Do you believe that the temperature at the top will increase? I don’t, because I have tried the experiment. I can not get the top of the heater any higher than 190 no matter what I try. Of course, this isn’t a perfect experiment. But if what you say is correct, I should be able to see some temperature increase, but I don’t
Thanks, Nelson. You say:
You seem confused about the nature of radiant energy. Here’s a thought experiment.
Suppose we have two stars at the same temperature, but widely separated.
What happens to their temperatures when we bring them close together?
The answer is that the temperature of both stars will increase. They have to because now they are both absorbing more energy from the surroundings than when they were widely separated.
And the same is true for a hotter star and a cooler star. Both will warm.
Ponder my thought experiment for a bit with those examples in mind.
w.
Willis,
You are off with the fairies again. Suppose we have two blocks of ice, each radiating 300 W/m2, widely separated.
You aren’t seriously suggesting that bringing them closer will make their temperature rise?
Try bowls of water – The Eschenbach Miracle Water Heater – boil two bowls of water simultaneously! No energy required!
Oh I see, it’s all in your imagination, is it? No connection to reality!
That explains everything.
Mike – The fact that you cannot distinguish between the case of the stars, with internal power sources (fusion), and blocks of ice or bowls of water without, tells us all we need to know about your lack of understanding of these issues.
Try this simple experiment. In a cool room, lie in bed with enough bedclothes and blankets to keep you comfortably warm. Your body at 37C, with about 100 watts internally generated from your metabolism, you are in steady state condition, outputting that same ~100W to room ambient.
Now bring in another body (hopefully someone you like…) at 37C with about 100W metabolic supply in close contact with you. You will quickly notice the additional warming, and your body will react (e.g. by starting to sweat) in reaction to that warming.
Note that this effect would not occur with two corpses without the metabolic power generation.
Without the steel shell the surface budget is Ein = 235 W/m2 and Eout 235 W/m2 yielding ΔE = 0 W/m2. If you add the steel shell some of the 235 W/m2 will get radiated back to the surface. ΔE > 0 W/m2. The surface now has a positive energy imbalance and must increase its T according to ΔT = ΔE/(m*c).
In your thought experiment remove conduction from the picture by separating the foil from the heater by a couple of millimeters. You will observe (assuming it is properly instrumented) that the top of the heater does indeed warm up.
This experiment was carried in real life and in grand fashion onboard the JWST. After the sunshield was unfolded the Sun facing side began warming while the mirror facing side began cooling. After the new stead-state was achieved the Sun facing side of the shield was warmer than it was before. The JWST is similar to Willis’ steel greenhouse except it had 5 layers (ie shells) instead of 1.
Adding the shell raises the equilibrium temperature of the central sphere, yes, as I explained in my correct-physics description of the scenario lower down in the thread. However, there is no “some of the 235 W/m2 will get radiated back to the surface” because the surface is hotter than the shell. Power doesn’t work that way… no one has ever measured it doing so…
Willis makes the following claim in that article.
Willis goes on in the comments to claim this is fine because the shell has twice the surface area, but he doesn’t take that into account on the input side. My understanding of physics/math has a problem with an object radiating away 235 watts/m2 energy when it’s only receiving 117.5 watts/m2 over the total surface area.
Using that article to now to claim he has already answered all of his critics is quite interesting.
Richard, it’s not clear to me where you see a problem.
When the shell is first added to the system, it intercepts all the radiation from the planet. That’s 235 W/m2 absorbed on the inside of the shell. Then, because the shell has two sides, the shell radiates the energy, half inward and half outward, at half that rate, 117.5 W/m2.
Of course, the system at that point is out of balance. It’s receiving far more energy from its radiative core than it is radiating. So it warms up until the planet is radiating 470 W/m2 to the shell, which in turn is radiating it inwards and outwards at 235 W/m2.
At that point the system is at steady-state and no further warming occurs.
w.
This is where you got confused. Your shell has 2 times the radiating surface vs its absorbing surface. It doesn’t matter what direction they are facing. You have to look at the total energy flow (aka power). In your example, at 235 watts/m2 absorbed by the input surface, you will be able to generate only 117.5 watts across twice the radiating surface.
Think of a big vat with water coming in at rate X. What happens when you have a two drains at the bottom with a capability of rate X? What will their actual drain rate be?
Yes Willis I had the same problem when this first came up. I gave the example of a dichroic mirror which reflects half the incident light and transmits half in a certain frequency band (I used various types of these in my research lab), the same people didn’t believe that either.
Willis, you are right that if you censored people who disagreed with you and pointed out your errors, you would look like a bully and a politician rather than a scientist. So good job on backtracking on that plan.
However, pointing us to your false “Steel Greenhouse” is not going to help. We have pointed out why that is wrong, and you ignored everyone who told you that.
This is because, as you said, you are “not a theory guy”, so you can’t defend any of these words that you write. Instead, you rely on logical fallacies like the Argument from Authority. (“But the climate scientists said so!”) Then when we point these out, you call us names (“Pig wrestling! Lead a horse to water! Pond scum!”) and finally shout “Pass!” until we go away.
You wrote: “DLR exists, it’s measured both manually and automatically all over the planet every day, and it doesn’t violate the 2nd Law”
No, all of that is false, and this is easily proven. Downwelling longwave radiant energy (which as we have already established is measured in Joules, not Watts) exists, yes, and corresponds to atmospheric temperature; but power is a different animal. Power requires (and is a measure of the rate of) work, or some other thermodynamic process such as Joule heating. This can only occur from hotter objects to colder ones, as you know. Therefore, power can only be developed in that direction. Not the other way around. The measurements you refer to produce negative numbers. Then they are falsely adjusted to produce a positive “adjusted reading”. That is not a measurement any more.
We have already established that power (in Watts) is another way of saying Joules per second. But what is a Joule? For someone like you who is “not a theory guy”, it might as well be a Jabberwocky; that means about as much to you. However, there is a bit of a trap here, laid for us by the thermodynamicists themselves. They have used the word Joule to mean two different things. And not just “apples and oranges” different, but “apples and the-process-of-slicing-an-apple-into-small-pieces-to-eat” different. Joules measure both “energy” (the potential to do work) and actual work. Sweet Honey Barbecue, how are we supposed to explain this to a fisherman? I think probably the best bet is to qualify Joules as either Joules-energy or Joules-work. And Watts are Joules-work per second. Not Joules-energy per second, which would make no sense whatsoever – but I assume is probably what is going through (what passes for) your brain when you think of photons floating through space. (Photons are energy, not work, as you know very well by now.)
“Hijack the thread”? No, Willis, you started this post with a series of false physics assumptions. Pointing those out is not “hijacking the thread”. It is directly on thread. Indeed, every time you post this nonsense, we will correct you. I will note that in another post where you didn’t make this particular error, you refused to continue our physics lesson because you preferred it to be in a thread related to a post that directly addressed the issue. That’s this thread. So we are actually here by your request.
And contrary to what you wrote, everything I have written can be verified by any physics textbook or professional physicist. Climate scientists (including Dr. Spencer) need not apply, though. They are not physicists, and are being paid to be wrong. Dr. Spencer refuses to engage in debate too… why is that?
While we are on the subject of your complete inability to engage in an intelligent physics debate, where do you think “everyone who disagrees with me please shut up” fits on this handy debating pyramid?
Steve, immediately above your post I asked for two simple things:
1) QUOTE exactly what you think is incorrect in the Steel Greenhouse analysis, and
2) SHOW (not claim but demonstrate) exactly how and why I’m wrong
If you’d done that, you’d be at the top of the pyramid above.
But you’ve done neither of them. Instead, you’ve simply claimed that I’m wrong and you already showed that in some unspecified place at some unspecified time. Plus you’ve thrown out insults, claimed I’m wrong about some physics somewhere, and closed with … wait for it …
… more insults.
That puts you down below the middle of the pyramid somewhere.
So … the invitation is still there. Quote me and refute me. Leave out all your snide insinuations and unpleasant remarks, they’re not productive. Just quote what in the “Steel Greenhouse” post you claim is wrong, and show us why.
w.
Sure, Willis, no problem. Many others have already done this, and you ignored them, but I will do it again.
Your “Steel Greenhouse” involves a sphere and an enclosing shell. Two objects, separated by a vacuum and surrounded by space at a background temperature of about 3K. No problem yet.
Then you specify that the sphere is provided with an internal power source measured in W/m^2. That is bogus already, because that is not how power sources work. To do this correctly, you would specify that your sphere has a power source measured in Watts. For simplicity, let’s give it 100 Watts of internal power. And we’ll give it a surface area of 1 square meter.
Now, at equilibrium, the sphere needs to dissipate all of those 100 Watts. It has a surface area of 1 square meter, so we can immediately determine that it has to lose 100 W/m^2 in order to maintain its equilibrium temperature.
We can’t calculate that temperature yet, because we don’t know the temperature of the environment (the shell) that is insulating the sphere from the 3K background of deep space. But we’ll get to that.
Next, we have a shell. It is larger than the sphere, to enclose it, so let’s say it has a surface area (more or less the same on both sides) of 2 square meters. It can be a few microns thick, just enough to absorb radiation from the sphere.
But what are the constraints on the shell? We know it is going to be absorbing 100 Watts from the sphere, and in order to maintain equilibrium, it has to dissipate all 100 Watts to deep space itself. It has 2 square meters to do that with, resulting in an area flux of 50 W/m^2, and we know that the temperature of the colder environment (“Tc”) is 3 K. Now we can calculate the shell’s equilibrium temperature (“Th”), using the S-B radiant heat transfer equation Qt =σeA(Th^4 – Tc^4) and solving for Th, with Qt = 50 W/m^2 and A=2 (let’s assume that everything is a blackbody). It comes out to 145 K if I’ve done the math right. That sounds plausible.
Great! Now we can calculate the equilibrium temperature of the central sphere, by plugging in the shell’s 145 K as the environment’s Tc, with the area under consideration “A” as 1 square meter, and Qt is 100 W/m^2. “Th” in this case comes out to 217 K. It has to be hotter than the shell in order to lose heat to it, of course. Also a plausible number. And this is a bit of a simplification because the sphere will probably not have a uniform temperature throughout if it is made of a normal material like steel, but at the surface it will be 217 K.
Now we have calculated the equilibrium state of the system – all three temperatures, and the amount of power being dissipated by each element to the next colder one. (Always 100 Watts, exactly the amount of power being supplied by something inside the sphere, such as possibly a nuclear reaction)
From this, it is easy to observe that the presence of the shell surrounding the sphere affects the equilibrium temperature of the sphere, as we would intuitively guess. Without the shell, the sphere’s equilibrium temperature would be 205K. It is indeed a “greenhouse” (notwithstanding that that is not how terrestrial greenhouses work, so really, “blanket” would be a better term, or even “thermos bottle”, since that is how thermos bottles work). But nowhere do you see any Watts being radiated by the shell towards the sphere. That is a fiction, and a violation of the 2nd law, since the sphere is hotter than the shell, as we have just calculated, and indeed can intuitively guess. So please redo your description while taking this law into account. Thanks!
It all gets down to conservation of energy. Good job of pointing out the main problem with the steel greenhouse.
Thanks, Richard.
And whoever voted “-1”, please explain what you thought was wrong with my description. (Calculations are carried out to approximately +/- 1K.) Thanks!
Well stated stevekj.
It is worth noting that the infamous Trenberth energy budget diagram that appears at the opening of this blog no longer gets a place in the WG1 section of the IPCC reports. In fact Trenberth is no longer referenced. His version of climate fisics is gradually dying. It would have been an embarrassment to radiation physicists like Michael Mishchenko, who GISS eventually employed to get a handle on atmospheric radiative transfer and conseptualise suitable measuring instrumentation.
Ah that is good to know, Rick! The sooner the fake “climate fizix” dies out, the better!
” In fact Trenberth is no longer referenced. His version of climate fisics is gradually dying.”
No. Just updated by more recent observations (more particularly CERES) …..
From IPCC AR6 ….
“Figure 7.2 | Schematic representation of the global mean energy budget of the Earth (upper panel), and its equivalent without considerations of cloud
effects (lower panel). Numbers indicate best estimates for the magnitudes of the globally averaged energy balance components in W m–2 together with their uncertainty ranges in parentheses (5–95% confidence range), representing climate conditions at the beginning of the 21st century. Note that the cloud-free energy budget shown in the lower panel is not the one that Earth would achieve in equilibrium when no clouds could form. It rather represents the global mean fluxes as determined solely by removing the clouds but otherwise retaining the entire atmospheric structure. This enables the quantification of the effects of clouds on the Earth energy budget and corresponds to the way clear-sky fluxes are calculated in climate models. Thus, the cloud-free energy budget is not closed and therefore the sensible and latent heat fluxes are not quantified in the lower panel. Figure adapted from Wild et al. (2015, 2019).
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7366598/#CR58
Did you notice that “back radiation” no longer gets a mention. It is now thermal down surface -whatever that means. It has no physical reality.
Interesting. But it’s still denoted in W/m^2, so the original fiction remains, sadly, even though the wording has changed. (“thermal down-surface??”)
(I’m going to “guess” that by “thermal down surface” they are still referring to the original notion of “downwelling longwave infrared power at the surface coming from the atmosphere”.)
stevekj, There is no fiction here. 5.5e24 joules is the component of the surface Ein originating from the atmosphere.
RickWill, It doesn’t matter what it is called. It’s all the same concept. It is the component of the surface Ein originating from the atmosphere. If back radiation or thermal down surface are offense to you then propose a new term. Just make sure the definition of your new term is the same…that is Ein_fromatm_tosfc.
Why would I propose a new term for something that does not exist in the physical world. There is ONE electromagnetic field and power flux in that field is unidirectional in space and time.
Earth’s surface gains energy from the sun and loses it to the atmosphere and space. But at any point on the surface at any instant in time, the surface is either radiating power or receiving power. It is not doing both simultaneously.
Take a look at this video on destructive interference of coherent light rays:
https://www.youtube.com/watch?v=RRi4dv9KgCg
Now use your concept of radiation to explain why the image goes dark when the two EMR beams interact.
Energy transfer exists in the real world. It is even represented as two terms in the 1LOT ΔE = Ein – Eout.
Yes. Earth’s surface has both Ein and Eout components. It can and does both receive energy and send energy simultaneous.
Patently False. You can use something as trivial as a handheld IR gun to convince yourself of the fact that the ground you walk on is both radiating energy and receiving energy simultaneously.
Remember, according to the 1LOT and SB law the only scenario in which a body is not receiving any flux is when that body is surrounded by an environment at absolute zero (0 K). Even the dead of space is at 3 K which means the Sun at 5778 K is receiving a flux, albeit very small, from space.
It does not convince me because I know the handheld IR gun is battery powered and quite capable of sending EMR from its warmer source to the cooler atmosphere. It is calibrated to the S-B equation for both incoming and outgoing EMR.
It may convince you because you do not understanding the physics of the EM field.
You have still not attempted to explain the destructive interference of two EMR beams as demonstrated in the MIT video using your special brand of EMR energy transfer..
Be careful about what your assumptions are.
Your video should tell you that precise positioning was required for both alignment and phase to get cancelation. With molecular motion you can’t assume that is going to occur for any length of time.
I am not making any assumptions. The reason for the video was to challenged bdgwx’s understanding of EMR and show how EM waves interact.
And I am not concerned with molecular motion because EM power transmits through a vacuum. I am only interested in the electric and magnetic components of the EM field. I am making the point that there can be no concurrent power fluxes in different directions in a single field. And there is only on EM field in the entire universe as we know it. All matter within the field interacting at the phase velocity of light.
They are welcome to label those arrows in Joules, as you described. I would buy that. But not in Watts. That’s not how Watts work.
And as I’ve said before the figures are left in units of W/m2 because 1) in this context it is more relevant to the purpose of the model and 2) the figures can be plugged directly into other models (like the SB law assuming a negligible rectification effect) without any other conversions. If you don’t understand why W/m2 are chosen for these 3 layer energy budget models then you haven’t understood the purpose of these models.
Oh, don’t worry, I understand very well that the purpose of the models is to bamboozle gullible people into surrendering their money and their freedom. And yes, W/m^2 are more suitable to that purpose, because they look scary. But they are not physics, not when aimed from a colder object to a hotter one.
IPCC AR6 WGI Chapter 7 section 7.2 regarding Earth’s Energy Budget and its Changes Through Time on pg. 933 specifically references Trenberth’s energy budget research.
Thanks for finding that and the reference.
Only the wording “back radiation” has disappeared. And replaced with a sillier notion of “thermal down surface”. At least it is no longer anything associated with EMR.
So your criticism boils down to the name given to the concept and not the concept itself?
Not at all. My criticism is that it is claptrap. It bears no relationship to the physical process of energy transfer in the electromagnetic field. There is but one electromagnetic field and power flux in that field is unidirectional in space and time.
I made the point about the name change because the diagram has nothing to do with radiation. It is hypothetical with no physical basis.
We will have to wait till AR7 to see if they persist with this claptrap.
NASA use the gravity waves on water as an analogy for EMR.
https://science.nasa.gov/ems/02_anatomy
Any physicists looking at the Trenberth’s energy budget diagram would shudder with its silly conceptualisation.
Downwelling radiation is not claptrap. It’s a real thing.
Energy budget analysis are just an application of the 1LOT which is also a real thing.
And not only does Trenberth not hold a monopoly on energy budget analysis, but he wasn’t even the first to do so. It was first deployed as a tool in [Dines 1917].
I am not referring to dowelling radiation. Short wave from the hot sun to Earth certainly exists. I am referring to long wave radiation from a cool atmosphere to a warm surface. You cannot have power flux against the gradient of a temperature difference.
The diagram shows. some sort of “thermal” flow from the colder atmosphere to the warmer surface. It cannot exist because there can be no heat transfer from hot to cold and EMR flux is unidirectional. There is no co-existence of back and forth EMR.
There is only ONE EM field and there can only be one power flux in that field at any point in time and space as defined by the Poynting vector – it is a vector – therefore unidirectional.
Mishchenko offers a proof as described here:
https://pdfs.semanticscholar.org/c03b/2b493f57e13d3c3e2b58d17c9656d2dee978.pdf
And if you struggle with Mishchenko’s proof. Then try to use you understanding of EMR to explain what is observed in this video:
That is patently false. There absolutely exists an energy transfer (power flux) from cold bodies to hot bodies. What you cannot have is a heat transfer from cold bodies to hot bodies in an isolated system. Remember, in the 1LOT (ΔE = Ein – Eout) energy is the Ein and Eout terms while heat is the ΔE term.
For example, a blackbody shell around a blackbody sphere at temperature 278 K will transfer energy per unit time and area of 340 W/m2 toward the sphere while the sphere at 289 K will transfer 401 W/m2 toward the shell. The heat transfer is 61 W/m2 is from the warmer sphere to the cooler shell. That does not invalidate the fact that the shell is still transferring 340 W/m2 of energy toward the sphere.
This is the 1LOT in action. If you try to contrive an example where a colder body does not transfer energy to a warmer body then you WILL violate the 1LOT. Note that energy transferring from a cold body to warmer does not violate the 2LOT because it only says that heat cannot move from a cold body to a warm body in an isolated system.
Do not conflate energy (the ability to do work) with heat.(the net transfer energy).
Nope. There is no netting out. You can separate the T^4 terms to do the arithmetic but it does not have any physical meaning. There is only a single power flux of 61W/m^2 from hot to cold.
P= 5.67E-8*(289^4 – 278^4) = 61W/m^2
better describes the physical process. Because there is only one EM field and there can only be one power flux at any point in space and instant in time in that field as described by the Poynting vector..
Use your special brand of EMR transfer to explain what happens in the MIT video above showing the destructive interference of the two EMR waves..
The T^4 terms are radiating potentials. The power flux is the result of different potentials. If there is no potential difference, there is no power flux.
I’m not sure what you’re saying nope to. I’m defending the 1LOT here.
The T^4 terms are separate physical quantities. They come from the 1LOT. Remember, the radiant heat transfer equation is derived from 1LOT by swapping out the Ein and Eout terms with the SB law.
First…the MIT video does not refute the 1LOT or the fact that energy from a cold body can transfer to a hot body. Second…it is irrelevant to this discussion.
That does not refute the fact that energy from a cold body can and almost always does transfer to a hot body even in an isolated system. This is true even where there is no heat transfer such as the case when the T^4 terms are the same.
This is the key to your misunderstanding of the physical process.
You cannot have concurrent EM waves at any point in space.
There is a single EM field with energy interchange between the electric and magnetic components.
If the destructive interaction of EMR is beyond your comprehension then maybe the interference of water gravity waves will give you more insight:
https://www.youtube.com/watch?v=Sug0iBjTmtc
The point here is that there is only one surface that, from the perspective of the waves, is represented by the altitude of the surface over average level in the gravity field and the vertical velocity or momentum of the water particles. There is an energy intercghange between potential and kinetic energy that transfers the power from the hand driven bobs outward through the water. Note how the waves interact and the dominant wave front from the interaction advances outward from the centre of the two bobs. The power transfer is now a single, but more complex, wavefront that combines the power of the interacting waves.
The EM field exists in three dimensions but is a single interchange of electric and magnetic energy. EMR waves from different sources combine to produce single time varying electric field and magnet field that interchange energy in a similar way to as potential and kinetic energy of a gravity surface wave.
All EMR exists in a single EM field. There are no concurrent waves at a single point.
RickWill Reply to bdgwx November 24, 2023 3:01 pm
This is easily disproven. Imagine a room that is empty except for two lights built into opposite walls, and both turned off. Clearly, light is an electromagnetic (EM) wave.
Now, turn on one light. There’s only one EM wave at any point in the room …
… but then, we turn on the other light. When that happens, at any given point inside the room, we have two concurrent independently-sourced electromagnetic waves in the form of light. And that’s not counting reflected EM in the form of some part of the light bouncing off the walls.
But wait, it gets worse. Radio is also an electromagnetic wave … since radio waves penetrate the walls of our houses, at every point in the room we have at least two lights and dozens of AM and FM electromagnetic waves all existing concurrently at every point inside the room.
Now, can we add all those together at any given point to give a NET overall EM field, as you keep saying?
Sure we can … but that doesn’t mean that the two different room lights and the light they are each independently radiating don’t exist. They do, they are very real, and despite your claim to the contrary, they coexist happily as concurrent waves at all points inside the room.
Regards,
w.
That is where standing waves come in. Energy is not destroyed just changed into a different form.
If you think of electric and magnetic fields rotating in different directions combining, sometimes the E field will be at maximum while the B field will be zero and vice versa. The magnitudes of each field will double at the maximum.
If the two waves aren’t exactly the same power, different results occur.
This is why EE’s take vector calculus. To deal with Poynting vectors via Maxwell’s equations
This is where your physical understanding of EM transmission is flawed. Sure all these waves interact but each source experiences the presence of the other at the speed of light and there is only one resultant field. It may carry numerous frequencies but there is only one direction of power flux at any instant at any point in space.
If you struggle with the concept of a single EM field then have a look at the water surface in this video:
https://www.youtube.com/watch?v=Sug0iBjTmtc
See how these waves interact. There is only one surface and the expanding wavefront at distance appears as if it is being generated from a single source between the two sources. The expanding wavefront no longer represents the separate sources.
Being a fisherman you may have observed the interaction of numerous waves – like rolling ocean swell coming into a shore with smaller local wind driven waves diagonal to it with these being reflected from the side of a ship coming back to meet the incoming waves all creating a conundrum of waves that is impossible to untangle by local observation but there remains but ONE surface where potential energy and kinetic energy is being exchanged.
For EMR power transmission, there is but one EM field where electrical and magnetic energy are exchanged.
The other aspect to observe is that the waves from the different oscillating bobs in the water interact at their point of generation as well.
The phase velocity of gravity waves on water is very slow compared with the phase velocity of EM waves so the interaction between sources occurs in much shorter time interval.
The only problem with your description is that there are a number (infinite?) of EM waves being generated all in different directions and frequencies. Two in-phase EM waves in exactly opposite directions will form a standing wave as the E and B fields rotate in different directions. As different directions and frequencies interact, lordy, I’m not sure what you end up with.
At some point inter molecular distances limit how many waves can interact. I can think of other factors also.
There is no matter (read molecules) in the vacuum of space but there is an EM field.
There are countless objects in the universe all emitting EMR all making their unique contribution to the EM field at any point in time and space in the vacuum. The power flux in the field at any instant is unidirectional. Because there is only one EM field.
“There is no matter (read molecules) in the vacuum of space . . . .”
One wonders how stars and planets form in the vacuum of space.
Jim,
By definition, a vacuum is the absence of matter.
From the Oxford dictionary –
noun
Either you can support your silly comment with an authoritative definition which disagrees with mine, or you can look like an ignorant sarcastic Warmist.
Others can decide for themselves.
You don’t even know where the IR energy from a block of ice submerged in water (all in an environment of standard temperature and pressure, of course) goes, do you?
You can’t even find the answer on the internet, can you? Nor can anybody who believes in the GHE!
Go on, put me in my place – tell me what happens to the IR energy from a block of ice submerged in water. How hard can it be?
I’m glad you found something that you can read. Apparently you can’t read or comprehend any comment on this site.
Jim,
Awwww.You’re ignoring me, are you? Or are you just a wee bit slow. Here, again –
Go on, put me in my place – tell me what happens to the IR energy from a block of ice submerged in water. How hard can it be?
There’s another definition for space: (also outer space) the area outside the Earth’s atmosphere where all the other planets and stars are.
You should stop reading only what you want to believe.
You are really an infantile piece of nonsense.
Jim,
A vacuum is the absence of matter.
Maybe you could quote the words with which you disagree, and provide some facts to support your disagreement.
You wrote –
“You should stop reading only what you want to believe.
You are really an infantile piece of nonsense.”
I repeat, what mental defect leads you to believe that I should value your opinion? I don’t, of course. Why should I?
A standing wave from identical EM waves traveling exactly in opposite directions has no direction. That is why it is called “standing”.
If the two waves are not equal, there will be a standing wave and the more powerful EM wave will have its remaining power continue onward. This is more applicable to heat radiation.
If two EM waves are normal to each other, what direction does their combination travel?
“There absolutely exists an energy transfer (power flux) from cold bodies to hot bodies”
No there doesn’t. What experimental evidence have you got in support of this claim? Remember that energy is measured in Joules, so to support this claim, you would have to show a reduction in Joules in the colder body, and an increase in the hotter one. Joules of thermal energy are usually measured in degrees C or F, or Kelvins.
Your steel greenhouse fails as an analogy of Earth’s atmosphere because the inner sphere is not gaining heat from outside the steel shell. You need to think of an analogy that places the heat source outside the shell.
Earth’s surface is mostly warmed by an external heat source well outside the atmosphere. The surface is usually warmer than any part of the atmosphere between it and the sun – the primary heat source.
I will agree with the last sentence in your quote. But I will add the point that, however you want to define the “greenhouse effect” it is unrelated to Earth’s energy balance and climate. The energy balance is fundamentally controlled by convective instability. That drives most of the heat advection as well as limiting the heat input. This is the process that needs to be understood to grasp how energy flows into and out of Earth’s climate system. The “greenhouse effect” is a diversion – clap trap.
There is an interesting paradox developing with regard the “faint sun” during the early evolution of Earth. There is a lot of effort to find proof of CO2 in great abundance during the Archaean eon that is constantly confounded by the evidence and geological proxies that indicate the atmospheric pressure was much lower than present and CO2 was a relatively low proportion. .
“. . . the “faint sun” during the early evolution of Earth.”
So now we are talking about the faint young Sun paradox. You are really getting off the subject.
The paradox is that it does not comply with the notion of a “greenhouse effect”. So it is right on topic because there are still people who believe the GHE influences Earth’s energy balance and climate.
First, I don’t think CO2 controls temperature. Second, if you are talking about C.R Scotese’s graph of CO2 vs. temperature, then it doesn’t seem to be saying what many think it does. Not having his actual data, I pulled values off the graph and ran a statistical correlation analysis. Surprisingly, I got something like .6 or higher–I don’t remember exactly. And lastly, I think the faint young Sun paradox can be explained with orbital mechanics, but that is really off topic.
Jim,
You wrote –
“First, I don’t think CO2 controls temperature.”
Agreed. Self proclaimed climate scientists who publish nonsense like “Atmospheric CO2: Principal Control Knob Governing Earth’s Temperature.” are obviously not authorities to whom you would appeal.
Gavin Schmidt, for example.
Please go away.
Jim,
Awwwww. And I was agreeing with you, and everything!
Now you want me to “go away”?
And if I ignore your desire, and laugh loudly into the bargain?
What then? A tantrum? Call me rude names?
RickWill November 22, 2023 7:09 pm
Been there, done that …
w.
All that indicates is that the atmosphere is going to be cooler than the surface if it has high transparency to short wavelengths and opaque to long wavelengths. However it does not have anything to do with the energy balance on Earth because the transparency and opaqueness of the shell is a function of surface temperature. Earth’s energy uptake is temperature controlled. The energy balance has nothing to do with “greenhouse” gasses and minute variations in transmissivity of the atmosphere. The transparency of the atmosphere drops rapidly once the ocean surface reaches 26C and reaches thermal balance at 30C.
With regard to your model. The power flux from the surface to the shell is properly calculated as:
P= 5.67E-8 * (303*4 -255^4), = 240W/m^2
You can choose to bring the power potentials outside the brackets and do separate sums but it has no physical meaning. The radiated power flux is unidirectional and ALWAYs from higher potential to lower potential.
As an aside, with regard the surface of your sphere, any point on the surface is either warming or cooling from a unidirectional EMR flux or in thermal equilibrium with no EMR flux. It is not taking in some EMR energy and releasing some EMR energy simultaneously.
I do not care how you choose to define the “greenhouse effect” because it has as much significance to Earth’s energy balance as the number of fairies that can dance on a pin head. Its claptrap that gullible soles accept as having physical significance.
RickWill,
it seems that NASA is giving up on “back radiation”.
A quick look shows another fantastic statement –
“Most of the emitted longwave radiation warms the lower atmosphere, which in turn warms our planet’s surface.”
Really? The atmosphere is generally colder than the surface – in sunlight. Are these people quite mad?
Even when exposed to an atmosphere warmer than the surface (surface inversion at night, dry atmosphere, calm conditions, cloudless) – the surface still cools!
Only Warmists like Willis and his ilk could possibly believe such nonsense.
Willis,
At your link, you wrote –
“Finally, Fig. 4 shows the energy balance when the shell is transparent to shortwave (solar) and is opaque to longwave (“greenhouse”) radiation. This, of course, is what the Earth’s atmosphere does.”
Complete and utter nonsense. Quite apart from the fact that you do not even bother specifying what longwave radiation frequencies are, you imply that some amount of unspecified energy emitted by the Earth’s surface s somehow trapped, or somehow retained.
Balderdash, a fantasy – unless you can provide experimental results to back up your nonsensical statement.
The surface temperature falls at night. All energy emitted by the surface escapes to space. I point out that slow cooling does not result in a rise in temperature as far as the surface is concerned.
And, of course, the surface as a whole has cooled from the molten state to its present temperature. All the energy lost has fled to outer space. Vanished. Gone. Disappeared.
Go on, say that the Earth has an “energy balance”, where “energy in” equals “energy out”, or vice versa, or something equally ridiculous.
There Is no GHE. You can’t even describe such a mythical concept, can you?
The Earth has cooled, and continues to do so. You live in a fantasy world – choosing to reject reality.
If you disagree, quote my exact words, and support your disagreement with facts. The contents of your fantasy are not facts.
Ok, I’ve been struggling to get a handle on this for a few months since I read “Nitrous Oxide and Climate” (C. A. de Lange, J. D. Ferguson, W. Happer, and W. A. vanWijngaarden).
According to that paper, upwelling radiation is absorbed within a few meters of the surface by GHG molecules (whether h20, co2, ch3, or no2). Since the LW radiation is at a frequency that is > 12,000 times the diameter of the molecule that makes aense. Once the lw radiation is absorbed then it energizes the molecule with vibtrational, rotational energy. So far every source agrees with that.
Next is somewhat contentious, according to standard AGW theory the LW radiation is re-emmitted and bounces around the atmosphere in a random direction, and the more GHG there are the more it bounces around.
According to Lange, et al, spontaneous re-emission from an excited molecule takes a few tenths of a second, but the Kinetic Theory of Gasses postulates that that at STP a gas molecules are going to collide with a few million other gas molecules a second and will transfer that vibrational energy to the rest of the 99.9% of Non GHG atmosphere (well at this point there is no distinguished difference between types of molecules and how they transfer kinetic energy, i.e. heat among themselves). And eventually the Kinetic energy will be emitted higher up it the atmosphere where there are both much fewer molecules of all types and even fewer H2O molecules. Which is emitted mainly as blackbody radiation that corresponds to a temperature of 255k.
Did I get that right?
This diagram may help.
?fit=1040%2C720&ssl=1
It is based on the optical depth of water vapour in a saturated atmospheric column over a surface temperature at 290K.
The solid black curve is the emissivity from any point in the atmosphere to space. Around 10,000m, the emissivity goes to 1. So all EMR emitted to space from this point makes it to space. The dotted black curve is the absorption. Above 10,000m there is little to now water vapour amor a 290K surface so the absorption becomes zero.
The red curve shows the broad spectrum emitting power at any point in the atmosphere. It is a function of the temperature at that altitude and the emissivity to space. The peak radiating power of 237W/m^2 is effectively 4700m up.
The absorption is 1 below 800m. So there is nominally no EMR making it to space below this altitude.
This is based on average power but water vapour has different absorption for different EMR frequencies so satellites can often “see” the surface for different radiation frequencies. The attached shows the EMR spectral transmission through 5km. The radiating power at 5km is down to 12.4% of the transmitting power at the source.
You left out a few details. You have the basics right.
-Energy is absorbed very low in the atmosphere via surface IR absorption (plus collisions) and directly via conduction.
-The energy is stored as kinetic energy of all the atmospheric gases.
-That energy also moves around. It can be transferred to other molecules via more collisions or you can get IR emitted from GHGs.
-The IR normally travels a short distance before again being absorbed and transferred again to another molecule via a collision.
This ignores convection which also moves energy around, but it’s good to understand this part of the process before adding in convection.
Since the atmosphere gets less dense as you higher in altitude, it cannot hold as much kinetic energy. Where does that energy go? Some of those upward direct IR emissions end up going directly to space even from the lowest altitudes. Energy isn’t just lost “higher up in the atmosphere”, it is lost everywhere.
Energy does flow upward through the atmosphere as well. This is because IR emissions headed upward are travelling through a less dense medium. They take longer before reabsorption than an IR emission downward.
The emission temperature will be the average of all the energy emissions that find there way to space.
“at STP a gas molecules are going to collide with a few million other gas molecules a second and will transfer that vibrational energy to the rest of the 99.9% of Non GHG atmosphere” Make that ‘a few billion’, about a tenth of a nanosecond between collisions.
You don’t get to break the laws of fizzix just because you refuse to listen to them
The laws of climate fizzex will be whatever they want them to be based on the power they wield due to the cozy relationship with authority. I have seen a few spectacular business failures where the opinions of the top executives could never be challenged. The western world is heading towards a great cliff for the same reason..
The power of climate alarmists is currently overwhelming.
Willis spends a good part of his time disproving the “global warming” meme but he is stuck with their dogma; unwilling to learn.
For the radiation at the earth
s surface we have the following numbers (Wild et al. 2019):</span>t make much difference, and Methane stands for 22,9 % of trace gas warming, we get:<span style="color: black;">Solar radiation absorbed: 160 W/m2/decade with an increasing trend.</span>
<span style="color: black;">Longwave cooling from increased temperature: -56W/m2/dec with an increasing trend.</span>
<span style="color: black;">Evaporation, without presentation of trend: -82W/m2/dec</span>
<span style="color: black;">Sensible heat, conduction/ convection from surface: -21W/m2/dec</span>
<span style="color: black;"> </span>
<span style="color: black;">Earth Energy Imbalance measurements tell us that there is a warming of 0,51 W/m2/dec from change in these variables, SWsurf down, LWsurf up, Evaporation, Sensible heat. The components behind these changes are Temperature change, Albedo change, Cloud radiation change, Water vapor Change, and Trace gases change. These are also the feedback components of climate change.</span>
<span style="color: black;"> </span>
<span style="color: black;">Loeb et al, 20 years of energy imbalance from 2000 to 2020:</span>
<span style="color: black;">Temperature surface radiation, Net LW cooling: -0,51 W/m2/dec</span>
<span style="color: black;">Albedo reduction. SW solar warming: 0,19 W/m2/dec</span>
<span style="color: black;">Cloud LW cooling (less clouds) -0,23 W/m2/dec</span>
<span style="color: black;">Cloud SW decreased absorption 0.44 W/m2/dec</span>
<span style="color: black;">Water vapor LW warming 0,33 W/m2/dec</span>
<span style="color: black;">Water vapor SW warming and latent heat. 0,05 W/m2/dec</span>
<span style="color: black;">Trace gas, aerosole LW warming 0,237 W/m2/dec</span>
<span style="color: black;">Trace gas, aerosole SW warming 0,002 W/m2/dec</span>
<span style="color: black;">If we assume that most trace gases and aerosols don
CO2 LW warming 0,185 W/m2/dec
Methane LW warming 0,055 W/m2/dec
With a warming of 0,19 degC/decade since 2000, we get the folowing feedbacks:
Temperature feedback (radiation from warmer surface): -2,68 W/m2/degree
Albedo feedback (Less reflection from atmosphere and surface) 1,00 W/m2/degree
Cloud LW feedback (Less backradiation from decreased clouds) -1,21 W/m2/degree
Cloud SW feedback (Less solar absorption of clouds) 2,31 W/m2/degree
SW water vapor warming feedback/forcing 0,26 W/m2/degree
LW water vapor absorption feedback/forcing 1,74 W/m2/degree
SW trace gas and aerosol warming feedback/forcing 0.01 W/m2/degree
LW trace gas and aerosol absorption feedback/forcing 1,25 W/m2/degree
Methane part of trace gas LW «trapping» «forcing» 0,29 W/m2/degree
CO2 part of trace gas LW «trapping» «forcing» 0,97 W/m2/degree
Sum of all feedbacks and forcings 2,68 W/m2/degree
A very little part of forcings and feedbacks has a warming effect on the atmosphere and earth`s surface (about 2% of total heat uptake, so about 0,01 W/m2/dec). Nearly all the absorbed energy becomes reradiated. But they have some important work to do. They have effects on the lapse rate. And shortwave radiation is warming liquid water and ice in clouds, resulting in evaporation and melting, potential heat and cloud dissipation. This may be the greatest contribution to global brightening, and is not a linear function of trace gases. CO2 stands for less than 20% of all positive forcings/feedbacks. So CO2 make only up a minor direct contribution to global warming.
Sorry for the format problem: Another try.
For the radiation at the earth
s surface we have the following numbers (Wild et al. 2019):</span><span style="color: black;">Solar radiation absorbed: 160 W/m2/decade with an increasing trend.</span>
<span style="color: black;">Longwave cooling from increased temperature: -56W/m2/dec with an increasing trend.</span>
<span style="color: black;">Evaporation, without presentation of trend: -82W/m2/dec</span>
<span style="color: black;">Sensible heat, conduction/ convection from surface: -21W/m2/dec</span>
<span style="color: black;"> </span>
<span style="color: black;">Earth Energy Imbalance measurements tell us that there is a warming of 0,51 W/m2/dec from change in these variables, SWsurf down, LWsurf up, Evaporation, Sensible heat. The components behind these changes are Temperature change, Albedo change, Cloud radiation change, Water vapor Change, and Trace gases change. These are also the feedback components of climate change.</span>
<span style="color: black;">Loeb et al, 20 years of energy imbalance from 2000 to 2020:</span>
<span style="color: black;">Temperature surface radiation, Net LW cooling: -0,51 W/m2/dec</span>
<span style="color: black;">Albedo reduction. SW solar warming: 0,19 W/m2/dec</span>
<span style="color: black;">Cloud LW cooling (less clouds) -0,23 W/m2/dec</span>
<span style="color: black;">Cloud SW decreased absorption 0.44 W/m2/dec</span>
<span style="color: black;">Water vapor LW warming 0,33 W/m2/dec</span>
<span style="color: black;">Water vapor SW warming and latent heat. 0,05 W/m2/dec</span>
<span style="color: black;">Trace gas, aerosole LW warming 0,237 W/m2/dec</span>
<span style="color: black;">Trace gas, aerosole SW warming 0,002 W/m2/dec</span>
<span style="color: black;">If we assume that most trace gases and aerosols dont make much difference, and Methane stands for 22,9 % of trace gas warming, we get:
CO2 LW warming 0,185 W/m2/dec
Methane LW warming 0,055 W/m2/dec
Willis, I’d be curious what you think would happen to Earth’s surface temperature if N2 was eliminated from the atmosphere and it it’s mass was doubled?
Unclear. When you say “it it’s mass was doubled”, you talking about N2 or about the atmosphere or about the earth?
w.
The total mass of N2 in the atmosphere is 3.96 X 10^18 kg. What happens to the Earth’s surface temperature if this mass is taken out of the atmosphere by the elimination of N2. What happens if we double the mass of N2 in the atmosphere?
Please correct me if I’m wrong Nelson but I take it you mean if N2 was removed but the atmosphere remained at the same mass (other atmospheric gases increased in proportion) so that the only variable to be calculated was the loss of N2?
It does not need to be speculated on. There is good evidence indicating atmospheric pressure got as low as 400hPa when the cyanobacteria were in overdrive. That created a true snowball condition. It needs to get to about 500hPa to avoid freezing. That is where convective instability can be generated to produce clear skies.
The atmospheric pressure has been as high as 1100hPa is the past 60Ma and that increased ocean regulating temperature by about 3C; up to 33C rather than the present 30C.
I have not looked at what happens at really high pressure in Earth’s atmosphere.
I believe Venus is where it is is because it did not have a moon producing collision. It rotates very slowly. The air advects longitudinally at real pace; taking 4 days to circulate compared with about 250 days for the surface below. Such high velocity would prevent the same sort of vertical convective instability that puts Earth’s climate in the Goldilocks zone.
Rick, the pressure effect is what I was trying to highlight. Without N2 in the atmosphere, the Earth’s surface would be a colder place. It would be much hotter if the mass of N2 was doubled.
Nelson, I have proven that that is not possible. See my post, “A Matter Of Some Gravity“.
w.
Lots of people disagree with you about how atmospheric mass affects surface temperature.
Leading Heat Transfer Physicists/Geologists Assert The Impact Of CO2 Emissions On Climate Is ‘Negligible’ (notrickszone.com)
Willis, you can’t prove anything about “radiation” until you grasp the difference between “radiant energy” (in Joules) and “radiant power” (in Watts). Remember that you told us that objects emit radiant energy, and this is measured in Joules. But that’s not what your “proof” (or your Steel
Thermos BottleGreenhouse) said…Since you always ask for specific quotes about your errors, that article contains this particular error, among others: “some of that absorbed energy is radiated by the atmosphere back to the surface” followed by “ [the surface] absorbing that energy from the atmosphere“. Except that the surface is warmer than the atmosphere. Can a warmer body absorb energy, thereby heating up, from a colder one? When has that ever happened? Note that “losing energy more slowly” is not the same as “gaining energy”, so try not to fall into that trap (like your two-star example).
Another error you included in this “proof” is this one: “theoretical maximum average temperature for that given level of incoming radiation” Since you have no idea of the difference between radiant energy and radiant power, and you didn’t specify, I’ll have to assume you mean radiant power here, which you usually do (except you call it “flux” for some reason). However, there is no such thing as a theoretical maximum average temperature for a given level of incoming radiant power. That’s not how power works. There is a theoretical maximum average temperature for a given level of incoming radiant energy, though, and it is the same temperature as that of the emitting object (in this case, that would be the Sun). There is also a theoretical average temperature corresponding to the amount of radiant power transferred to a 0 K background environment, which is given by the single-temperature variant of the S-B radiant heat transfer equation.
Neither of these theoretical average temperatures has anything to say about the presence or absence of an atmosphere generating a gravito-thermal effect in a gravitational field, which just means that the temperature at the bottom of the atmosphere is warmer than at the top, due to gravity, even when the atmosphere has no IR-absorbing gases or liquids in it.
If our atmosphere were pure nitrogen, would the average surface temperature change? Probably; the water vapour does a lot of heavy lifting in the planet’s energy distribution department. In what direction would it change? I’m not sure exactly, but I would guess it would get a little colder, without the water vapour blanket. There would still be a thermal gradient in the atmosphere, though, similar to today’s, due to gravity.
Several people have tried to disprove the gravito-thermal effect, and all of them have been hilarious failures. Yours is based on several false physics assumptions, chiefly a complete failure to grasp the nature of either “energy” or “power”. So if you want to convince me that there is no such thing as the G-T E, see if you can explain what is going to happen to the potential energy gradient in the atmosphere if it is not counterbalanced by a kinetic energy gradient to produce an isoenergetic total mass at rest. The kinetic energy gradient, of course, manifests as a thermal gradient, while the potential energy gradient is “invisible”. To put it another way, the gravito-thermal effect is a consequence of the fact that we can feel kinetic energy, but we can’t feel potential energy. And at rest, these two have to balance out. Otherwise something would rise, or fall, until it did. That’s my elevator pitch.
RickWill November 21, 2023 7:01 pm
If there is “good evidence”, then LINK TO IT OR I’LL JUST POINT AND LAUGH. Can’t tell you how tired I am of people making claims without putting forward a single scrap of evidence.
w.
Here you go:
https://www.washington.edu/news/2016/05/09/early-earths-air-weighed-less-than-half-of-todays-atmosphere/
I am certain you can find more references showing similar results if you care to search. There are numerous efforts to reconstruct the early atmosphere and all the recent world point to the pressure being much lower mass then present.
This paper gives limits for the gasses and reasoning for the limits:
https://www.researchgate.net/publication/339533189_The_Archean_atmosphere
So dragonflies with more than a meter wingspan didn’t have a thicker atmosphere to fly in? They couldn’t fly in today’s atmosphere.
Dragon flies were not around in the Archean eon. The highest pressure was likely up around 1100hPa during the Cretaceous Period. Then is when pterosaurs were flying. The largest had a wing span as wide as 11m:
https://www.livescience.com/24071-pterodactyl-pteranodon-flying-dinosaurs.html
I believe the physiology of dinosaurs still remains a mystery. The flight of pterosaurs is still not well defined.
“Dragon flies were not around in the Archean eon.”
Neither were you. We don’t know the past pressures. Please don’t say you know what they were. They are all estimates.
Yes, dinosaur physiology is a mystery. They could be warm blooded, but the latest data indicates that they were dinosaur blooded–somewhere between warm blooded and cold blooded.
You haven’t a single scrap of evidence for your claims of positive DWLWIR power measured at the surface at ambient temperature at night, either, Willis. Are you pointing and laughing at yourself?
Proving that CO2 is not a problem is just wasted effort as the real objective, as stated by a former head of the UNIPCC, is the destruction of capitalism and by inference establishment of worldwide communism – we need to start touting all the benefits to mankind from capitalism and all the destruction that communism has caused
342 incoming
105+237=342 outgoing
(with no error bars on any of the numbers…)
Where is photosynthesis in this balance? Is it really negligible?
Willis,
You said: ” …the change in the clouds over the period has more than offset the change due to CO2 … who knew?”
What have clouds done over the time period 2000-2022?
And relatedly, how well have GCMs done in modeling them?
I have read people talking about an overall decrease in cloudiness even though there has been more precipitation..
Thanks for another interesting article!
Willis,
You wrote “There is a lot of misinformation floating around the web about the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, or a “mirror” that reflects part of the heat back to Earth, or “a pane of glass” that somehow keeps energy from escaping. It is none of these things.”
Maybe you could explain how spheres of different radii, radiating the same W/m2, can radiate identical amounts of energy, because it is simply impossible. The areas of the spheres are different, and if they are radiating the same amount per unit area, the total energy must be different.
If you don’t feel like admitting you were wrong, feel free to abuse me. You have done so in the past, as I recollect. Only joking, I’m sure you just overlooked reality in your enthusiasm.
Mike, I ask again:
1) QUOTE what I said that you think was incorrect, and
2) SHOW (not claim but demonstrate) exactly how and why I’m wrong.
I have no idea why you think I said that “spheres of different radii, radiating the same W/m2, can radiate identical amounts of energy”.
w.
Willis,
Ok, from your silly steel greenhouse link –
“Figure 1. Building a steel greenhouse. (A) Planet without greenhouse. Surface temperature is 235 W/m2 heated from the interior. (B) Planet surrounded by steel greenhouse shell. Shell radiates the same amount to space as the planet without the shell, 235 W/m2. It radiates the same inward, which warms the planet to 470 W/m2 (29 C, 83°F, 302 K). [Clarification added] Note that the distance from the shell to the planet is greatly exaggerated. In reality, it is close enough to the planet that the difference in the areas can be neglected in practice.”
No, the areas cannot be “neglected in practice”, as you would quickly find if you tried to translate your fantasy into reality. No matter how minor the difference in areas (which you now seem to be denying while admitting their existence), your imaginary outer sphere will be radiating energy at slightly less W/m2 than it received – being greater in area.
Slightly cooler than the body heating it. Unfortunately, you seem to be subscribing to the bizarre notion that a warmer body can have its temperature increased because it is in the path of radiation emitted by a cooler one. This is demonstrably silly – submerging a block of ice in a bowl of hot soup does not make the soup hotter, even though the IR radiation from.the ice has to go somewhere. You really don’t know what happens to the radiation from the ice, do you?
Another example is that of a low level temperature inversion at night, wherethe atmosphere is actually hotter than the surface beneath it. The surface still cools in reality, if not in your dreams. John Tyndall explained why, and he still believed in the luminiferous ether.
Mike Flynn November 21, 2023 9:02 pm
Mike, the difference in the answers is about 0.15%. Now, you’re free to re-do the thought experiment and include that. But it will make absolutely no difference to the underlying argument, which is why it can indeed be “neglected in practice”.
But like I said, you can redo the analysis in that analysis, at steady-state the outer shell will radiate at 234.63 W/m2 rather than at 235 W/m2, which makes exactly zero difference to the underlying argument.
Get back to us after you re-run the numbers including the meaningless 0.15% difference, and tell us why we should care about trivialities.
Suppose I stand next to a block of ice. Whoa, I can feel the cold. Now, I put a sheet of plywood between me and the block of ice. I feel warmer … but you say that’s impossible because the plywood is cooler than I am …
Same with the atmosphere. If we had no atmosphere radiating down on us, we’d be much colder. The question is not “does the air make the surface warmer?”
The question is, “does the air leave the surface warmer than if there were no air, just the cold of outer space?”
Same with the plywood. The question is not “Does cool plywood warm my body?” The question is “Am I warmer with the plywood between me and the ice”.
And for those questions, the answer is yes, despite the plywood being far cooler than my 98.6°F, and despite the atmosphere being colder than the surface.
Since the plywood is warmer than the ice, it leaves me warmer than if the plywood didn’t exist. And since the atmosphere is warmer than the alternative, it leaves the surface warmer than if the atmosphere didn’t exist. So yes, cool things can indeed make warm things warmer than they would be if the cool things didn’t exist.
Here’s another way to look at it. Suppose we have a hotter star and a colder star far from each other. What happens to their temperatures when we put the colder star next to the hotter star?
The answer is, both warm up. Both of them are now receiving more energy from the environment than when they were separated. So yes, the colder star definitely leaves the other star warmer than it would be without the nearby colder star.
See my post “Can A Cold Object Warm A Hot Object” for a full discussion.
Regards,
w.
Quantum fizzix of photons hasn’t reached Willis Eschenbach yet
He still thinks he can add radiation willy nilly like beads on his abacus with no regards to wave length and temperature
Mean while, back on planet Earth, submerge a block of ice in a bowl of hot soup. Hmmmm (sarcasm) – didn’t seem to warm the soup, did it?
Pity you can’t explain what happened to the IR emitted by the ice, isn’t it? Maybe you could try to look it up on the internet – how hard could it be?
Oh dear – you write complete nonsense like “Same with the atmosphere. If we had no atmosphere radiating down on us, we’d be much colder. The question is not “does the air make the surface warmer?”
Well, no, Willis. Not if the Sun is shining. High temperatures on the Moon exceed those on Earth – I thought you might have known that. My mistake.
No atmosphere, you see. John Tyndall explains the reasons. Pretty simple, so you should be able to understand it. See, sarcasm backed up with fact – you should try it sometime.
Back to the ice in the soup – you haven’t got the faintest idea where the photons from the ice have gone, and if you can find the answer on the internet, you can’t bring yourself to believe it.
Sad,
Willis,
You wrote “But like I said, you can redo the analysis in that analysis, at steady-state the outer shell will radiate at 234.63 W/m2 rather than at 235 W/m2, which makes exactly zero difference to the underlying argument.”
Well, no, it makes the outer shell cooler. I know you think you can make something hotter using only the radiation from a cooler object, but that is just Warmist cult belief, not fact.
You’ve put your foot in your mouth, and shot yourself in the foot, one might say.
Care to double down, and claim that you can make hot soup hotter by adding ice? What happens to the IR energy emitted by the ice?
Don’t know? Can’t or won’t say?
Pity.
Willis,
You wrote –
“Suppose I stand next to a block of ice. Whoa, I can feel the cold.”
Well, no, you can’t feel “cold radiation”. There are no “cold rays”. You are just making stuff up, now, trying to sound knowledgeable. Everything above absolute zero emits IR, but you cannot feel radiation from anything colder than you.
You are probably confused by cold air currents impacting your skin. At temperatures well below freezing, in still air conditions you won’t feel any “cold radiation” from the surrounding snow field – even naked.
Yes, been there, done that. Your wishful thinking won’t turn fantasy into fact. Now’s your chance to start playing semantic games, and pretend you really meant that you could feel the cold air descending from the ice, around your feet.
Or do you still want to claim that you can feel “cold” radiating from ice?
Go on, put a big block of ice in the Sun, stand upwind from the ice, and listen to me laughing as you say you can “feel the cold.” You really have no clue, do you?
““Suppose I stand next to a block of ice. Whoa, I can feel the cold.”
Well, no, you can’t feel “cold radiation”. There are no “cold rays”. You are just making stuff up, now, trying to sound knowledgeable. Everything above absolute zero emits IR, but you cannot feel radiation from anything colder than you.”
Unscientific garbage, of course you can feel radiation from something colder than you!
The radiation heat transfer equation is dependent on
Th^4-Tc^4. Your body would be Th so next to a block of ice you would definitely feel cold. Just like when standing next to a brick wall which has been heated by the sun during the day you feel the warmth.
Thanks, Phil. I’ve mostly given up on trying to get people like Mike Flynn to read up on basic radiation concepts.
Regarding the question of 2-way energy fluxes, here’s the MIT textbook, still used, 5th Edition.
The textbook is online. It’s worth a read. I added the URL to the top left of the graphic.
Regards,
w.
Tut, tut, Willis.
You claim you can feel the “cold” radiated by ice – you just make this stuff up as you go along. Grab a thermometer and give it a try. You can’t describe the GHE, you can’t explain its role in the fact that the Earth is cooler now than four and a half billion years ago!
Gee, what’s the GHE supposed to do? Make the planet hotter? Colder? Both simultaneously?
As to “thought experiments”, you are no Einstein. You still turn to water and scuttle away rather than trying to explain how an unspecified sphere magically doubles its energy output (from 235 W/m2 to 470 W/m2), and then magically stops heating – just by surrounding it with a cooler spherical shell.
Come on, whine away, refuse to tell anybody where the additional energy comes from. Just act like you have a hair up your nose, and burble about something else.
The Earth has cooled. CO2 does not make thermometers hotter.
Go on, have a go at explaining why the IR radiation from a block of ice (say 235 W/m2) completely surrounded by water, doesn’t raise the temperature of the water at all.
How ignorant do you want to look?
Phil,
Give it a try. Your fantasy is not reflected by reality.
You write –
“Just like when standing next to a brick wall which has been heated by the sun during the day you feel the warmth.”
No, Phil, not just like that at all.
To perceive a sensation of radiated heat, the radiating object must be hotter than you.
Next thing, you’ll be telling me that colder objects radiate “cold rays”! Some eminent people certainly used to believe so, and had experiments which appeared to show the existence of “cold rays”.
Surrounding an object with something colder won’t make its temperature increase. Not the Earth, not Willis’ imaginary sphere, nothing. Unless you have a hidden inexhaustible high temperature heat source lurking in your fantasy, of course. Like Willis.
What about you?
Willis,
Your exact words “I note that despite my polite request, some impolite folks want to hijack this thread to proclaim that the “greenhouse effect” isn’t real and downwelling radiation doesn’t exist.”
Your “polite request’ that no body points out where you are wrong, or challenges anything you say, is just an attempt at browbeating others into agreeing with you – right or wrong. As to your presumably snarky comment about “downwelling radiation”, who has claimed that the gaseous atmosphere doesn’t emit infrared radiation upwards, downwards, sideways + in all directions, in fact. Not me, that’s for sure. I will graciously accept your apology if you were inadvertently referring to me in some patronising and derogatory fashion.
Only joking – I know you were just trying to make me look stupid.
You link to “People Living In Glass Planets” which is equally erroneous as your “steel greenhouse” fantasy.
Are prepared to defend your fantasies, or is everyone just supposed to accept anything you say?
Given that you cannot even describe the GHE in any way which agrees with reality, it’s a bit rich pretending to “explain” this non-existent effect. Shooting the messenger won’t turn fiction into fact.
As Richard Feynman said “It doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are. If it doesn’t agree with experiment, it’s wrong.”
If you can’t actually describe the GHE in any disprovable way, then nobody at all can contradict any fantasy you might advance about it.
You could always start by describing the role of the GHE in the Earth’s surface no longer being molten, in spite of four and a half billion years of continuous sunlight.
I’m all ears to observations and supporting experimental data. You don’t actually have any that fit any GHE description that you can come up with, do you?
Sorry, Mike, but your claim that “You link to “People Living In Glass Planets” which is equally erroneous as your “steel greenhouse” fantasy” is a handwaving joke.
I’ll say it again, real slow this time:
If you wish to show I’m wrong, you need to:
1) QUOTE what I said that you think was incorrect, and 2) SHOW (not claim but demonstrate) exactly how and why I’m wrong.I’ll wait.
w.
Willis,
i can’t copy your silly diagram from the “Steel Greenhouse” link (which shows two spheres of different radii, radiating the same W/m2, emitting the same total energy), but no matter.
You wrote –
”In order to maintain its thermal equilibrium, including the new shell, the whole system must still radiate 235 W/m2 out to space. To do this, the steel shell must warm until it is radiating at 235 watts per square metre. Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet. The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell. This will warm the planetary surface until it reaches a temperature of 470 watts per square metre. In vacuum conditions as described, this would be a perfect greenhouse, with no losses of any kind. Figure 1 shows how it works.”
Where do I start? For example “in order to maintain its thermal equilibrium. . .” There is “thermal equilibrium” applying to the Earth. It has cooled since the surface was molten. Neither remnant heat, radiogenic heat, four and a half billion years of sunlight, not CO2, or any non-describable GHE stopped the Earth radiating more energy than it received. Go on, tell me I’m wrong, and that energy in equalled energy out!
Second, you wrote “To do this, the steel shell must warm until it is radiating at 235 watts per square metre.” Nonsense. If the steel shell has a greater surface area than the underlying sphere, then it will be receiving and radiating somewhat less than 235 W/m2, as its surface area is greater. You just can’t create energy from nothing.
Third, “This will warm the planetary surface until it reaches a temperature of 470 watts per square metre.” Really? A temperature of 470 W/m2? Don’t be silly, temperatures are measured in degrees [of hotness]. I know you really meant something else, but why not say it?
Now consider this – your original imaginary surface was emitting 235 W/m2. You have now increased its emission to 470 W/m2. Oh dear! 470 plus 235 equals 705. And so on. You are off with the fairies.
Maybe you could start by actually describing the GHE. Only joking, you can’t.
I’ll wait while you try to justify your silliness.
Mike Flynn November 21, 2023 8:12 pm
So you can’t figure out how to copy an on-screen graphic, but you’re here to tell us how the climate works. Got it.
As to the different radii, I thought I discussed that. Hang on … yeah, I did. I guess you didn’t read carefully enough. I said:
Now, you’re free to re-do the numbers including the slight difference in areas. It will make a difference of about 0.15% in the numbers, but the underlying principles and conclusions don’t change in the slightest. Be my guest. Or you can do what any real scientist would do in a “first cut” analysis, ignore the trivialities and focus on the real issues.
You continue:
The Earth? Who said anything about the Earth? This is what is called a “thought experiment” about a theoretical planet surrounded by a steel shell and solely heated from the inside by radioactive decay. Does that sound like the earth to you?
And yes, such a system will absolutely reach a steady-state where it radiates the same amount it receives. Physics. Don’t leave home without it.
You continue.
Seriously? OK. The real number is 234.63 instead of 235. Be still, my beating heart …
As I said, for such analyses these kinds of minor effects are generally ignored. But as I also said, you’re free to redo the numbers and include the hundredths of a watt if you wish. It makes no difference at all to the underlying analysis.
Onwards. You say:
That’s your big beef? Immediately below that statement, I said:
I specified that the planet is a blackbody. If it is radiating at 470 W/m2 it has a certain temperature. See, the thing with blackbodies is, that a certain temperature blackbody gives off a certain specified amount radiation.
So you can measure the same blackbody temperature as 470 W/m2, or as 29 C, or as 83°F, or as 302 K. They all specify the exact same temperature
But heck, if folks like you are getting easily confused, I’ll go change it. Hang on … OK, now it says:
Happy now?
So, we’ve dealt with your trivialities, none of which affect the underlying argument in the slightest.
Do you have any real arguments as to how it is wrong?
w.
I cannot eliminate the possibility that this may be the same “Mike :Flynn” that trolls Dr. Spencer’s blog with endless posts about how the Moon does not rotate and other nonsensical claims. If that is the case you won’t even be able to convince him of the 1st law of thermodynamics.
Willis,
“So you can’t figure out how to copy an on-screen graphic, but you’re here to tell us how the climate works. Got it.”
Ooooooh! That’s mature and helpful. Maybe you could tell commenters how to insert graphics into comments. If you wanted to be helpful, of course. Your assumption that you can read my mind is incorrect. Climate is the statistics of historical weather observations. What “work” are you referring to?
“As to the different radii, I thought I discussed that. Hang on … yeah, I did. I guess you didn’t read carefully enough. I said: . . . “
No, Willis, if your outer shell is emitting less W/m2, it is cooler. Cooler, get it? You probably think that you can use a vast amount of radiation from ice, concentrate it, and heat a drop of water. You can’t. You can’t force a hotter body to accept radiation from a colder, and get hotter as a result. That’s just silly.
“The Earth? Who said anything about the Earth? This is what is called a “thought experiment” about a theoretical planet surrounded by a steel shell and solely heated from the inside by radioactive decay. Does that sound like the earth to you?”
Oh goody – I mistakenly assumed were insinuating that your nonsense had something to do with the greenhouse effect. I apologise – anything is possible in a fantasy. I suppose when you wrote “In hopes of expanding their concepts of the greenhouse effect, however, let me recommend the following posts of mine:”, you were referring to an imaginary greenhouse effect, not applicable to the Earth. Why bother, then? All imaginary anyway, is it?
“Seriously? OK. The real number is 234.63 instead of 235. Be still, my beating heart …
As I said, for such analyses these kinds of minor effects are generally ignored.”
Oh, I see. You didn’t really mean to write –
“For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square metre of surface area.”, did you? You meant to write 234.63, but had an attack of forgetfulness?
Here’s the thing, Willis, ice may be a millionth of a Kelvin below freezing, and submerged in water that is a millionth of a Kelvin above it. The radiation from the ice will be greater than 234.63 W/m2, or even 300 W/m2 – but will not raise the temperature at all, will it.
Just like your “steel shell”. If it is a millionth of a Kelvin cooler (and it must be cooler, because you hopefully admit its surface area is greater, and therefore emitting less W/m2, the total energy emission not being more than it received), then it manifestly cannot make anything warmer than itself!
You attempt to ignore your sloppiness by writing –
“This will warm the planetary surface until it reaches a temperature where it radiates at 470 watts per square metre. Happy now?”
Ooooh! That’s harsh Willis, but I’m pleased to see that you are so concerned about my happiness. But seriously, no it won’t. As I pointed out, a cooler body cannot raise the temperature of a warmer one, by it’s own radiation (just in case you start waffling about insulation).
Now obviously, you think you have invented a perpetual motion motion, or you forgot to mention you had a hidden heat source up your sleeve, You have a sphere internally generating and radiating a fixed amount of energy to space. You surround this sphere with a steel shell, the sphere immediately start emitting twice as much energy – and then stops, for some inexplicable reason.
if you started with a sphere radiating, say, 470 W/m2, surrounded it with a steel shell, would it double in temperature? Or would it remain at 470 W/m2 emission?
You claim that once reaching an emission of 470 W/m2, no further temperature increase occurs. This is where the magic occurs, apparently.
Not a bad illusion. I suspect you rewrote your Steel Greenhouse description after I pointed out your basic error previously.
Come on Willis, confess that you believe that you can make a warmer body increase its temperature by surrounding it with a cooler one (no sneaky additional heat source, mind).
Or admit you were wrong.
Or be your usual snarky, browbeating, sarcastic self – pile abuse on me, and scuttle away pretending that you actually know what you are talking about.
Go on, tell everyone what happens to the IR radiation from ice which is submerged in water? It doesn’t appear to heat the water, does it? Only joking, you don’t know and you can’t find out can you?
Mike, What about this….
The Earth gets warmed by the sun. It radiates (cools) that heat at a given but constant rate (remove the sun and it would lose all it’s heat in a matter of hours? days? whatever) Add radiation to Earth from an atmosphere. The Earth is not warmed by the atmosphere (second law) but the rate of cooling is slowed. If the rate of cooling is slowed and the input from the sun remains static, would not the Earth’s temperature rise slightly? It’s temperature is not increased by the cooler atmosphere but by adding to the sun’s input to make a new total.
As a layman I don’t know. I’m trying to understand the whole thing. I would have thought that if you add more heat to an gaseous atmosphere (not a steel cage) it just expands to a new volume (with a greater surface area) and nothing else changes…….
Mike,
There is no sunlight at night. I see no reason to disagree with Baron Fourier, who said that during the night, the surface loses all the heat of the day, plus a little of the Earth’s “remnant heat”.
The Earth’s surface is no longer molten. It has cooled, and people like Willis can’t explain why, so they just pretend it hasn’t happened.
Slow cooling, fast cooling – the word is “cooling”. Falling temperatures. It takes a cultist to claim “slow cooling” is really heating.
If you heat a seemingly unconstrained gaseous atmosphere, it might expand, but on the other hand, surprisingly, maybe not. A “heat dome” consists of hotter but denser air, constrained by colder denser air. But in general, you are right.
It’s interesting that both the hottest and coldest places in the world are also those with the least GHGs in the atmosphere. Inexplicable to GHE believers, of course.
By the way, some real scientists are now accepting that hotter thermometers result from AGH (Anthropogenically generated heat). I’m sure Willis won’t have a bar of it.
Mike Flynn – You miss several key points that demolish your argument.
In the sphere-alone case, the sphere is in thermal contact with the ambient temperature of -270C (3K) of deep space.
In the sphere-plus-shell case, the sphere is in thermal contact with the much higher temperature of the shell. Even though the shell temperature is lower than the sphere temperature, this reduces the resulting net heat flow to ambient. With the separate power source for the sphere (a key point!), this results in a higher steady-state temperature for the sphere.
It is crucial to realize that the (net) HEAT flux is still from the sphere to the shell, so there are no 2nd Law violations here.
In your soup example, the pot of soup with the burner on will have a higher temperature in contact with water ice at 0C (273K) than in contact with liquid nitrogen at -196C (77K). You are erroneously comparing it to the case of being just in contact with room temperature ambient at ~20C (293K).
Following your argument, you obviously believe that the soup in a 20C ambient will be hotter than in a 0C ambient. But 20C is still below hot soup temperature, and you claim it could not “heat” the soup. Kind of goes against your own argument…
Ed,
I didn’t forget anything. Willis demands precise quoting, but makes sure he provides insufficient information.
You write “With the separate power source for the sphere (a key point!), this results in a higher steady-state temperature for the sphere.”. Well, no, not necessarily. Like Willis, you imply, but do not state, that the interior power supply has infinite capacity to maintain any temperature found necessary.
Obviously, the temperature of Willis’ imaginary sphere, with an emissivity of 1, is -19.424 C. The power source has this temperature, and I assume that Willis’ sphere is also isothermal, as he doesn’t state otherwise.
Luckily, you seem to accept that the outer sphere is at a lower temperature than -19.424 C. Given that Willis steadfastly refuses to indicate the temperature or capacity of his imaginary heat source, what is wrong with me setting it at -19.424 C?
Willis just refuses to accept that surrounding a sphere with a cooler shell will not result in the temperature of the sphere rising. Willis states that the sphere is radiating at 235 W/m2, with a an assumed temperature of -19.424 C. He surrounds it with a shell of lower (unstated) temperature. He claims the temperature of the sphere will rise to 29 C, and stay there. I suppose if he repeated the process, and changed nothing, the temperature would double again to 58 C. Another doubling would result in 116 C, quite enough to provide infinite free power from a primitive steam engine.
At this point, Willis will discover that his internal heat source has unlimited capacity, but a temperature of 29 C at most. Flexible chap, Willis, changes his fantasy to suit his purpose.
No, it doesn’t work.
Then you write –
“In your soup example, the pot of soup with the burner on will have a higher temperature in contact with water ice at 0C (273K) than in contact with liquid nitrogen at -196C (77K).”
Eh? Burner? What burner? Who mentioned liquid nitrogen?
I merely pointed out that submerging a block of ice in a bowl of hot soup doesn’t raise its temperature. I hope you are not trying to say it does?
Maybe you are trying to “school” me, by saying that liquid nitrogen is always colder than frozen H2O. Unless the ice is cooled to a lower temperature, of course. Oh, you didn’t realise that?
Ambient temperature is just a red-herring employed by people who believe in the GHE. It doesn’t matter what the ambient temperature is, you cannot make soup hotter by submerging a block of ice in it, even though the IR from the ice must go through the soup. Yes, I’ll admit I assume that hot soup is liquid, and above the freezing point of water.
To get back to the point, you have no idea where the IR photons emitted by the ice go, do you? Neither has Willis. You see, you can’t warm a warmer sphere with a cooler shell. Any more than you you can warm non-frozen water with ice or liquid nitrogen.
Nor can you warm the Earth by surrounding it with an atmosphere, and applying sunlight – even for four and a half billion years! It cools regardless.
Thanks for your input.
Mike Flynn – You obviously have never formally studied thermodynamics or heat transfer, or you wouldn’t be making the silly mistakes you are. So let’s break it down a little more.
For the sphere alone in space, radiating into a 3K environment, the steady-state condition has the surface radiating away the 235 W/m2 that is internally generated. With virtually no power radiated back toward this surface (the 0.000003 W/m2 from space at 3K can be ignored), the surface temperature of a blackbody in this steady-state condition would be -19.42C (253.73K), as you state.
BUT… consider the surface layer of the steel sphere that is radiating away. For it to be in a steady-state condition, it must also be receiving 235 W/m2 by conduction from the inside of the sphere. This requires a temperature gradient, and so a higher internal temperature and a source “capable” of producing these temperatures.
(To put some numbers on it, typical steel conductivity is 45 W/m/K, so the temperature gradient would be [235 W/m2] / [45 W/m/K] = 5.22 K/m.)
So even though the exact nature of the internal power supply is not explicitly stated, to anyone experienced in this type of analysis, it is obvious that this supply is capable of producing temperatures high enough to cover all the scenarios in this problem.
In an engineered system, electrical resistance heat fills this bill. In a planetary system, internal radioactive decay will work.
Now, consider the sphere-plus-shell system. In the steady-state condition, it must radiate away from the outer surface of the shell the same power that is generated internally: (235 * Asphere). This means the radiative power flux density from the outer surface of the shell must be (235 * Asphere / Ashell) W/m2. This means that the steady-state temperature of the (outer surface of the) shell must be at a level that can radiate this much. For Asphere/Ashell ratios near 1.0, this puts the temperature near -19.4C.
But the inner surface of the shell must also be radiating at least this much power inward. So let’s analyze the sphere itself when the shell is in place. In steady-state conditions with (235 * Asphere) W internal power input, there must be (235 * Asphere) W resulting output from the surface of the sphere.
Whether you consider the radiative transfer to be 470 out and 235 in “radiative power fluxes”, or a (470 – 235) unidirectional “heat flux”, the surface temperature of the sphere must be at 28.5C (301.7K) to radiate away enough power to maintain steady state.
You complain that Willis does not specify the starting condition of the shell in his analysis. Again, you just show your inexperience in analyzing this type of problem. In steady-state analysis, it does not matter what the starting condition is – the resulting steady-state will be the same.
Let’s use the case where the initial state of the shell is the same 3K as deep space. The sphere-plus-shell system is still receiving (235 * Asphere) from the internal power source, but is radiating virtually nothing away. So its internal energy, and so temperature is increasing. As the shell temperature increases, it radiates more away to deep space – but also radiates more back to the sphere (or reduces the radiative heat flux from the sphere, if you prefer). This process will continue until the steady-state conditions are reached.
Your example of adding ice to soup is irrelevant here. I was trying to tweak your example to make it relevant to this example, but you missed that. I laughed out loud when you said worrying about ambient temperatures is a “red herring”. Only someone completely inexperienced would say that.
My engineering group does a lot of thermal testing of electronics. Recently we were testing a product. The requirement was that it operate properly (i.e. powered) in an ambient of 45C, with a desire for it to do so in an ambient of 55C. A key requirement was that the surface temperature of any component could not be over 125C. At 45C ambient, the hottest component had a surface temperature of 122C, so we met the requirement. Then we increased the ambient temperature of the thermal test chamber to 55C.
According to you, this increase could not cause an increase in the component temperature, but “strangely”, it did! The component surface temperature increased to 131C, so as a result we could not specify it for operation in a 55C ambient.
Have you ever stopped to think about how putting on a jacket can warm you up in cold weather, even though it has a lower temperature than you do?
Out of time for now, but I’ll see if I can respond to more of your points later.
Ed, you have the patience of a saint.
w.
Ed,
Willis’ request that you quote peoples’ words doesn’t apply to you, either? Oh well.
You wrote “or the sphere alone in space, radiating into a 3K environment, the steady-state condition has the surface radiating away the 235 W/m2 that is internally generated. ” Internally generated with a switchable variable heat source, which doubles its output at Willis’ command – and maintains it.
You write –
“BUT… consider the surface layer of the steel sphere that is radiating away. For it to be in a steady-state condition, it must also be receiving 235 W/m2 by conduction from the inside of the sphere.” Except that the inside of this sphere cannot be receiving 235 W/m2, as Willis finally admitted. The inside of the steel shell has greater area than the surface of the sphere, and consequently less than 235 W/m2 – as Willis reluctantly acknowledged – 234.something W/m2.
This unfortunately for you, means that the steel shell is actually at a slightly lower temperature than the ~19.4 C. Good luck with warming a sphere by surrounding it with a colder body!
After a load of nonsensical waffling, and appeals to your own authority, you write –
“According to you, this increase could not cause an increase in the component temperature, but “strangely”, it did! The component surface temperature increased to 131C, so as a result we could not specify it for operation in a 55C ambient.”
Why according to me? I didn’t say any such thing! I’m well aware of insulation, heatsinks, and the temperature rise associated with feeding more energy into a component than it can dissipate, consistent with not exceeding design temperature. You may not agree that nothing – not remnant heat, not radiogenic heat, not four and a half billion years of continuous sunlight, has prevented the Earth’s surface from cooling from the molten state – but then others would look at you askance, wouldn’t they?
If you disagree with something I say, please quote my exact words, and support your disagreement with facts, or physics, or something reasonable.
By the way, given your expertise, you should be able to demonstrate your vast knowledge, and explain what happens to the IR energy emitted by a block of ice totally immersed in water.
You can’t? Why is that?
Mike Flynn – The extent of your confusion is just astounding! You have said in this thread, and I quote:
“Willis just refuses to accept that surrounding a sphere with a cooler shell will not result in the temperature of the sphere rising.”
“Good luck with warming a sphere by surrounding it with colder body!”
You have several more basically equivalent statements.
And yet when I present to you the results of real-world experiments that did just this — surrounding an object at 122C with a cooler “shell” or “body” (the thermal test chamber) at 55C resulted in the temperature of the object rising to 131C — you don’t have the technical chops to realize that this is a direct contradiction of your arguments!
Instead you protest that you “didn’t say any such thing!” But you absolutely did!
(I should mention that the electrical power input to the device was carefully monitored during this testing, and remained constant. And while the device wasn’t spherical, surely even you are not so confused to think that makes a difference here!)
You say that you are “well aware of insulation, heatsinks, and the temperature rise associated with feeding more energy into a component than it can dissipate.” But you are not “well aware” enough to realize that this is the exact situation of Willis’ “steel greenhouse”.
Oh, and despite your confused protestations, his problem has a CONSTANT power input of (235 * Asphere) watts. He surrounds this powered object with a cooler shell and the temperature increases!
You don’t have the level of understanding to realize the difference between you counter example of adding ice (at 0C) to a pot of soup (above 0C) in a warm room (also above 0C) – which will reduce the soup temperature as you claim — from surrounding a (powered) pot of soup with a shell of ice in a very cold ambient (below 0C) — which will increase the temperature of the soup compared to the case where the soup is in direct thermal contact with the <0C cold ambient.
By the way, Willis, you wrote –
“See, the thing with blackbodies is, that a certain temperature blackbody gives off a certain specified amount radiation.”
Well, no, you are confused. Here is a normal black body definition “A black body or blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.” – Wikipedia.
You are probably talking about “black body radiation” – which is a completely different thing. I have refrained from mentioning this before, because you are even more confused about other things.
You demonstrate your confusion by writing “a certain temperature black body gives off a certain specified amount [of] radiation. Once again, a black body is a theoretical construct which absorbs all incident radiation, be it from a lump of concrete or a supernova. It has no temperature at all. I’m not sure why you believe in a GHE anyway, when you can’t even say what the operation of the GHE achieves.
Certainly not heating the Earth, or making thermometers hotter, or anything of that nature.
Have you found out where the IR energy from a block of ice submerged in water above freezing, goes? No?
Once you do, you might have an epiphany, and abandon the Warmist cult.
Who knows?
Mike Flynn
Reply to
Willis Eschenbach
November 24, 2023, 5:51 pm
This is a perfect example of why I’ve given up discussing science with folks like Mike Flynn. Here’s a typical definition of the S-B Law:
Mike’s claim that a blackbody “has no temperature at all” would be risible if he didn’t actually believe it.
w.
Willis,
No, you didn’t mention a blackbody radiator in relation to the sphere. You said “a perfect black body”, and I took you at your word. Silly me.
Here’s the first blackbody definition that popped up –
“A black body or blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.”
No temperature mentioned, for good reason.
I suppose you now want to claim you really meant to say “black body radiator’ rather than “perfect black body.” Off you go, give it a try.
As I said, you were confused. You tried to change your sphere description to a ball of rock, or something. Still doesn’t work. Unless you explain why, and provide some physical characteristics. You can’t just make stuff up on the fly, unless you want to be laughed at.
You might be gullible, but at least you are ignorant.
Mike, a perfect black body BY DEFINITION is a blackbody radiator. Look at the definition above. It absorbs all incident radiation, and by Kirchoff’s Law, it perforce must be equally able to radiate.
You say “No temperature mentioned, for good reason”. Yes, you are right.
The “good reason”, however, is that the overwhelming majority of folks studying the subject are not as stupid as you, so we don’t need the obvious fact that a blackbody has a temperature pointed out to us.
The extent of your ignorance of these matters is beyond belief.
w.
Willis,
I have looked at “A black body or blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.” again.
Maybe you could point out where Kirchoff’s Law is mentioned, or that a blackbody radiates. It just absorbs all radiation.You are probably referring to “black body radiation”, and I can understand your confusion.
You write “The extent of your ignorance of these matters is beyond belief.” Strange, are you saying that you don’t believe that I am ignorant of these matter, or are you just trying to be gratuitously offensive?
If the latter is true, you obviously need to try harder. I don’t feel offended at all. Am I supposed to?
Carry on.
As a follow up to Willis’s thought experiments here’s mine which can also be carried out as a real experiment.
Take a vacuum flask with a black material at the bottom and a transparent window at the top and evacuate it. Shine ~400 W/m^2 down on the black surface, the surface will warm up until it reaches ~290K at which point approximately 400 W/m^2 will be emitted.
Replace the window with a dichroic mirror which transmits visible and ~50% of IR. In that situation equilibrium will be reached when about 800 W/m^2 is emitted by the black material ( ~345K).
Phil,
You haven’t stated the temperature of the ~400 W/m2 source. 400 W/m2 from sunlight will produce a different result from one side of Leslie’s cube filled with boiling water.
i’m not sure what all this is about. Are you trying to say that immersing a block of ice in a bowl of liquid water will make the water hotter? Or that surrounding a sphere of constant temperature with a colder shell, will double the amount of radiation emitted by the sphere? Obviously, for this to be true a hidden source of heat of adequate temperature and energy output is necessary. Of course, this heat source remains hidden up the sleeve of the conjuror, to be produced on demand, whilst claiming it was in plain sight all the time.
You might agree that ice in a vacuum flask, or water, or hot soup, will eventually reach the temperature of the flask’s surroundings. No GHE, no magic.
Thanks for your input.
“Are you trying to say that immersing a block of ice in a bowl of liquid water will make the water hotter?”
Heat is transferred by conduction, convection, and radiation. Are you trying to tell us that heat transfer by conduction (the primary method of heat transfer in your ice example) disproves heat transfer by radiation?
Jim,
No, I said what I said. Obviously, Willis, request that people quote comments exactly only applies to Willis and his supporters.
i asked Phil “Are you trying to say that immersing a block of ice in a bowl of liquid water will make the water hotter?”
Just agree with me that it won’t, and try and explain why – if you wish, of course.
Then you can try and support Willis’ implication that a GHE exists, even though is not even prepared to state what it does. Nor are you, of course.
Feel free to prove me wrong by describing the GHE in some way that accords with reality.
Thanks for your diversionary input, anyway.
“Just agree with me that it won’t, and try and explain why – if you wish, of course.”
Agree with what? You’re conflating radiation with conduction. Please stop.
“Feel free to prove me wrong by describing the GHE in some way that accords with reality.”
Disproving GHE or EGE is easy. It’s just not possible with your nonsense.
Jim,
“Are you trying to say that immersing a block of ice in a bowl of liquid water will make the water hotter? Just agree with me . . .”
What part of my question do you disagree with? You don’t have to agree, of course. What reasons do you have for thinking that immersing a block of ice in a bowl of liquid water will make the water warmer?
None at all? Maybe you agree with me after all. Wasn’t that hard, was it?
You wrote – “You’re conflating radiation with conduction. Please stop.” You need to seek a refund for your mind reading course. It’s not working. As to “Please stop.”, what if I don’t? Will you have a tantrum, or just hold your breath until you turn blue?
You also wrote (for some unfathomable reason) –
“Disproving GHE or EGE is easy. It’s just not possible with your nonsense.”
How do you “disprove” something that doesn’t exist and has never even been described? Go on, describe the GHE in some way that reflects reality. Only joking, of course you can’t. Nobody can, because it doesn’t exist.
Time for you to rejoin the fairies.
Looney Tunes! I do like Bugs Bunny cartoons.
Jim,
Willis wrote –
“To Avoid Misunderstandings: When you comment please quote the exact words that you are discussing.”
Are you expressing your contempt for Willis, or just trying to annoy me?
I generally decline to feel annoyed by ignorant fantasists, and you give me no reason to make an exception in your case.
Carry on.
Well, you told me to rejoin the fairies. Are you trying to annoy me?
Jim,
Willis wrote –
“To Avoid Misunderstandings: When you comment please quote the exact words that you are discussing.”
Are you expressing your contempt for Willis, or just trying to annoy me?
I generally decline to feel annoyed by ignorant fantasists, and you give me no reason to make an exception in your case.
Carry on.
Say again?
If I must –
Jim,
Willis wrote –
“To Avoid Misunderstandings: When you comment please quote the exact words that you are discussing.”
Are you expressing your contempt for Willis, or just trying to annoy me?
I generally decline to feel annoyed by ignorant fantasists, and you give me no reason to make an exception in your case.
Carry on.
“You haven’t stated the temperature of the ~400 W/m2 source. 400 W/m2 from sunlight will produce a different result from one side of Leslie’s cube filled with boiling water.”
The temperature of the source is irrelevant.
I’m not saying anything about blocks of ice nor spheres..
Of course ice in cirrus clouds will give a warming when compared to outer space at 3K.
Phil,
You wrote –
“The temperature of the source is irrelevant.”
I don’t agree, obviously. For example, concentrating 300 W/m2 from sunlight can melt steel. Concentrating 300 W/m2 from ice cannot even warm water.
Now, a body with a temperature of 290 K and an emissivity of 1 will radiate over 400 W/m2. You claim “Replace the window with a dichroic mirror which transmits visible and ~50% of IR. In that situation equilibrium will be reached when about 800 W/m^2 is emitted by the black material ( ~345K).”
With a source having a temperature of 290 K, you have raised a piece of black material to 345 K! Impossible.
You might want to reconsider your ill-advised claim that “The temperature of the source is irrelevant.”
Or you could fly off at a tangent, trying to avoid the fact that you are not looking too intelligent just now.
The ~400 W/m^2 shone down on the black surface was intended to be visible.
Phil,
You wrote –
“The ~400 W/m^2 shone down on the black surface was intended to be visible.” Do you mean white light or monochromatic light?
You also wrote –
“Replace the window with a dichroic mirror which transmits visible and ~50% of IR. In that situation equilibrium will be reached when about 800 W/m^2 is emitted by the black material ( ~345K).”
Are you using a different light source, and if so, what IR wavelengths are you talking about, and why only allow ~50% of the IR to enter?
I don’t have a problem with a surface reaching a higher temperature if exposed to more energy, and I can’t see why you think this is at all unusual. What is the relevance of the vacuum flask, and the “window”? If you are trying to say that 800 W/m2 of sunlight will make an object hotter than 400 W/m2 of sunlight, I agree.
Ground temperatures vary between about -90 C and +90 C for this reason. Nothing to do with any GHE, if that is what you are implying. Maybe you could say what it is you are trying to say.
““The ~400 W/m^2 shone down on the black surface was intended to be visible.” Do you mean white light or monochromatic light?”
It doesn’t matter it’s irrelevant.
““Replace the window with a dichroic mirror which transmits visible and ~50% of IR. In that situation equilibrium will be reached when about 800 W/m^2 is emitted by the black material ( ~345K).”
Are you using a different light source, and if so, what IR wavelengths are you talking about, and why only allow ~50% of the IR to enter?”
The same light source, the IR wavelengths are the blackbody radiation emitted by the black material. The 50% was used as an illustration, the fraction just changes the equilibrium temperature.
“What is the relevance of the vacuum flask, and the “window”?”
To restrict the heat transfer to radiation.
Phil (and others)
It’s amusing to me to watch Flynn do what he has done for years (decades?) and troll people in his unique and confabulating parrot fashion.
PS he calls himself Swenson on Roy’s
No one has no will ever get through any sense.
Whether he does it knowingly is the wonder.
Anthony,
As I pointed out above, Phil has no clue.
He claims that he can raise the temperature of an object to 345 K, using an object with a temperature of 290 K.
He did write “The temperature of the source is irrelevant.”, after all.
He’ll probably complain that it’s not fair that I use a temperature of 290 K for his source – because it makes him look foolish.
Do you agree with Phil that the temperature of an object is irrelevant when considering heat transfer? For example, ice can emit more than 300 W/m2. Are you really going to say that 300 W/m2 from ice will warm an object to the same temperature as 300 W/m2?
You wrote –
“It’s amusing to me to watch Flynn do what he has done for years (decades?) and troll people in his unique and confabulating parrot fashion.”
Some may value your opinion, I suppose. Do you have an opinion on the role of the GHE in the observed fact that the Earth’s surface is no longer molten? Go on, tell everyone that the temperature of the surface is irrelevant, or that a drop in temperature is called heating!
Yes, I’m laughing at you. Others may be able to keep a straight face. Maybe you can contribute a fact or two – you might be able to wipe the smile off my face, but I doubt it.
Off you go, now. Give it a try, if you like.
QED
QED
Yes, an obvious troll. Thanks for the heads up. I should have figured it out sooner.
Jim,
Obviously I have to ask – can’t you decide things for yourself?
Do you always wait for someone to tell you what you should be thinking, and then just follow along?
Go on, make up your mind for yourself – even if you are a bit slow.
I did make up my mind myself. I was giving you a benefit of the doubt, but obviously you deserve no benefit.
Jim,
in your case, what mental defect makes you think that I value your opinion?
Do you believe you have some awesome superpower that you will unleash upon me, unless I care about your “doubt” – or lack thereof?
If you are striving for humour, you could try your hand as a stand-up comedian. You make me laugh, at least.
Carry on.
Willis demonstrates his ignorance of the conservation laws by writing “And because the atmosphere is transparent, this means that the planet is radiating to space more energy than it receives. This is an obvious violation of conservation of energy, so any theories proposing such a warming must be incorrect.
Q.E.D.
Now, I’m happy for folks to comment on this proof, . . . ”
Actually, the planet does radiate to space more energy than it receives – otherwise the surface would still be molten. Willis is off with the fairies, denying observed reality.
As Baron Fourier (Fourier series and all that) mentioned a couple of hundred years ago “Thus the earth gives out to celestial space all the heat which it receives from the sun, and adds a part of what is peculiar to itself” – radiating to space more energy than it receives (from the Sun, at least).
The gravitothermal effect that Willis was addressing is nonsensical, but Willis’ “disproof” is equally nonsensical. The Earth has indeed cooled, and Baron Fourier explains why. Early attempts to establish the age of the Earth were largely based on Fourier’s ideas and measurements.
Willis appears to be off with fairies, as they say. A bit disconnected from reality at least. The planet has cooled, and continues to do so. It’s a big mainly glowing hot body sitting ini an environment not much above absolute zero, a long, long, way from the nearest decent external heat source. Maybe he is even delusional enough to believe such a body is heating up – just because he says so! Of course, he would never say such a silly thing, would he?
As usual, if you disagree with anything I say, quote my exact words, tell me why you disagree, and provide some facts to support your disagreement. Or just complain about your hurt feelings, and ignore inconvenient facts.
Yes, I am laughing at Willis, just as he threatened to do to other commenters. Fair’s fair.
Mike Flynn November 24, 2023 11:00 pm
Mike, you seem unacquainted with the concept of a “thought experiment”. A “thought experiment” posits various concepts and then sees where they lead. Einstein did it with his thought experiment about people falling through space in an elevator.
As a result, the planet in my experiment is NOT the Earth, so all of your claims about how it is cooling are built on sand.
The planet in my experiment is a block of rock that has been at the same temperature for a million years. Get over it.
w.
Willis,
You wrote –
“Mike, you seem unacquainted with the concept of a “thought experiment”.
A “thought experiment” posits various concepts and then sees where they lead. Einstein did it with his thought experiment about people falling through space in an elevator.
As a result, the planet in my experiment is NOT the Earth, so all of your claims about how it is cooling are built on sand.”
You are no Einstein. Presenting pointless fantasies, and arriving at completely impossible conclusions, just shows that you are waffling on for no reason at all.
You keep trying to weasel out of the fact that you are “explaining” a GHE which you cannot even describe. You wrote –
“The planet in my experiment is a block of rock that has been at the same temperature for a million years.”
Now you have a “planet” which is a “block of rock”? Initially, you wrote –
“For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square metre of surface area. How is it heated from the interior? Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.”
OK, now tell everyone that a “perfect blackbody” is no different to a “block of rock.” Or that “the same temperature for millions of years” magically increases when you surround it with an admittedly colder “steel shell”!
Then, in a perfect fit of avoidance, you decide that your “experiment” is not about the Earth, and magically does not change its surface temperature – until you surround it with a colder environment!
If you are now going to claim that the surface heats up due to heat from the interior, “Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.”, that’s fine. Yes indeed, a heat source is required to raise the temperature of an object. These radioactive elements of yours are magical, maintaining constant heat ouput over millions of years – no radioactive decay to be seen.
You are just confused. Are you claiming that a terrestrial “greenhouse effect” exists, or not? You also wrote –
“As a result, the planet in my experiment is NOT the Earth, so all of your claims about how it is cooling are built on sand.”
The real Earth has actually cooled, unlike your fantasy planet which is either internally heated, not internally heated, a perfect black body or a block of rock, is heated at a constant temperature by magic radioactive elements which maintain a constant heat output over millions of years without any decay at all, but which have. control switch which magically doubles the output of energy on demand, and keeps it at that level.
All this is surrounded by a steel shell which is heated to a slightly lower temperature by the hotter sphere, until the magic heat source doubles its output, whereupon the steel shell doesn’t heat up at all! Unless you put another magic cooler shell around the whole impossible assemblage – in which case you say the temperature of the sphere triples!
If you disagree with anything I say, please quote my exact words, and support your disagreement with fact, not fantasy.
Oh wait, I’m addressing Willis, who doesn’t take any notice of what the tells others to do. Yes, that’s sarcasm – I’m flattering you through imitation, you clown! Happy now?
If you can’t even get your fantasy consistent, why do you think anyone less gullible or ignorant than you should take any notice of you?
If nothing in your fantasy relates to reality, why do you bother even writing it down?
Yes, Willis, you are definitely the gift that keeps giving. A thought experiment requires that you think. Oh well, as you said, nothing to do with the Earth. Nor anything else if you refuse to accept physical laws or mathematics. You don’t have the accept that the Earth once had a molten surface, and has since cooled. You can believe anything you like – that CO2 makes thermometers hotter, perhaps?
Pass. DFTT, especially trolls who believe in magic blackbodies that can absorb any amount of radiation without changing temperature …
w.
Willis,
Here’s another black body definition for you –
A blackbody is defined as an ideal body that allows all incident radiation to pass into it (zero reflectance) and that absorbs internally all the incident radiation (zero transmittance).
Or another –
“A blackbody is a theoretical or model body which absorbs all radiation falling on it, reflecting or transmitting none.”
You believe yourself to be a clever chap – what is the temperature of a blackbody after it absorbs the radiation from a supernova? After it absorbs the radiation from a second, even bigger, supernova?
Here’s your chance to educate me. Come on, is a back body twice as hot after absorbing the radiation from two blocks of ice, rather than one?
You really, really, don’t understand any of this, do you? How big is a black body? How long is a piece of string? Can you describe the GHE?
Questions, questions.
Mike Flynn Reply to Willis Eschenbach November 26, 2023 2:44 pm
===
“You believe yourself to be a clever chap – what is the temperature of a blackbody after it absorbs the radiation from a supernova?”
===
Seriously? There is far, FAR from enough information in your question for it to have an answer.
That’s as brain-dead as asking “What is the temperature of a pot of water after it absorbs the heat from the stove?”
Without knowing the mass of the blackbody, the specific heat of the blackbody, the starting temperature, and the total number of joules absorbed from the supernova, the question is meaningless.
But hey, asking a meaningless unanswerable question and then bragging that someone can’t answer it, that’s totally on-brand for you …
w.
Willis,
OK, Willis, what is the mass of a blackbody?
“A blackbody is defined as an ideal body that allows all incident radiation to pass into it (zero reflectance) and that absorbs internally all the incident radiation (zero transmittance).”
It seems to have no mass. Kirchhoff in 1860 introduced the theoretical concept of a perfect black body with a completely absorbing surface layer of infinitely small thickness.
Was this concept wrong? Would a perfect black body of infinitely small mass suffice to satisfy the requirements of the definition I quoted above? Maybe you can specify the density of a theoretical black body, and show experimental support. Not terribly bright, Willis. Do you make this stuff up as you go along?
You wrote –
“Without knowing the mass of the blackbody, the specific heat of the blackbody, the starting temperature, and the total number of joules absorbed from the supernova, the question is meaningless.”
I see. That is no doubt why you changed your Steel Greenhouse planet from a “perfect blackbody” to “The planet in my experiment is a block of rock that has been at the same temperature for a million years. Get over it.”
Go on Willis, keep promoting fizzix.
While you are at it, maybe you could provide some figures for your Steel Greenhouse “block of rock”. Radius of the sphere, internal radius of the steel shell, maximum power output of the magic radioactive elements which provide millions of years of unvarying power, without decaying.
You might also mention whether there is any other hidden heat source – someone supporting you mentioned a “background temperature”, and you didn’t disagree.
You wrote –
“But hey, asking a meaningless unanswerable question and then bragging that someone can’t answer it, that’s totally on-brand for you …” Did I hurt your feelings? Oh dear!
Are you now admitting that you can’t tell me where the IR radiation from a block of ice submerged in water, and emitting 300 W/m2 goes? Or do you know, but are just being unhelpful. If you think that pointing out your lack of knowledge is bragging, then I accept your encomium.
Off you go now, feel free to sulk. Why should I care about the feelings of such as you?
I believe that when the earth is closest to the sun during the year, the solar radiation is 90 W/m2 greater than six months later, that is 90 times greater than the 1 W/m2 of DLR being discussed here. Is this correct? Is this effect identifiable in the temperature record?
Pat Smith November 25, 2023, 9:49 am
While the instantaneous radiation varies, a curious process occurs. When the earth is near the sun, it gets more solar energy, but it’s moving faster. So it spends less time in the warm zone. And the opposite occurs when it’s far from the sun.
And since both light and gravity decrease as the square of the distance, these two effects basically counteract each other completely and as a result, there’s little difference in the amount of energy available annually to each hemisphere.
w.
Willis,
Thanks for informing us that “since both light and gravity decrease as the square of the distance, these two effects basically counteract each other completely”
That’s good to know. What do you mean by “basically counteract each other”? I assume you don’t mean that light and gravity counteract each other, completely or otherwise. Maybe it is clear to you, and my comprehension is flawed. Maybe you could explain what you mean in some way that it makes sense – we don’t all share your Einsteinian level of intellectual attainment.
Are you trying to imply that the effect of 90 W/m2 of solar radiation cannot be distinguished from 1 W/m2 – in daily temperature records?
When you write “there’s little difference in the amount of energy available annually to each hemisphere.”, presumably you have taken joy in examining the copious records available, and based your statement accordingly. What does “little difference” mean? Can you put a number on it?
Only joking, I know you can’t. I’m trying to rise to your level of sarcasm, Am I succeeding?
Mike, I’d said that regarding the amount of energy absorbed when the earth is nearer and further from the sun, the effect of greater sunlight when the earth is close to the sun is counteracted by the fact that it’s moving faster, so it spends less time there. These two effects cancel each other out.
Now, you didn’t understand that. No probs, lots of stuff I don’t understand.
But instead of acting like a decent human being and politely saying “Willis, what did you mean by …?”, you once again opened the valve on your endless torrent of personal attacks.
Why? Do you think your ugly abuse will make me MORE or LESS likely to respond to your lack of understanding?
Next, in your usual nasty fashion, you say:
Where to start … first, there are not “copious records” of gridded TOA solar input. The CERES data is the best we have. And yes, I’ve run the numbers. What were the results? From memory, little difference. Hang on. You ask if I can put a number on it. Of course I can. I do my best to avoid unsupported statements …
… OK, since you are obviously incapable of actually obtaining the data and doing the computations required, I’ve “put a number on it” as you say. As you may or may not know, the total solar irradiance (TSI) at the top of the atmosphere (TOA) is about 340 watts per square meter (W/m2).
In the CERES data, the difference in the amount of solar energy striking the two hemispheres is 0.64W/m2. (339.86 W/m2 vs. 340.5 W/m2) This is about a 0.2% percent difference. This is from uncertainty and instrument error, as the physics of equal hemispheric radiation is clear.
Finally, could you dial back on the aggro? Standing on your tiptoes in a vain attempt to bite my ankles is not a good look on you …
w.
Willis,
You wrote – “Now, you didn’t understand that. No probs, lots of stuff I don’t understand.”
Presumably, you are talking about Kepler’s Laws of Planetary Motion. Oh well, if you believe I don’t understand those laws, bully for you. Maybe somebody believes you can read minds, but not me.
You could always provide some facts to support your opinions about what I understand, and what I don’t, but you really can’t be bothered, can you? Or is there another reason? So I’ll ask politely “Willis, what did you mean by ‘Now, you didn’t understand that’ “?
You wrote “Why? Do you think your ugly abuse will make me MORE or LESS likely to respond to your lack of understanding?” What ugly abuse are you talking about? Why do you bother feeling “abused”? Are you affected by the opinion of an anonymous total stranger? Do you imagine that I have the power to somehow affect your life?
Or are you claiming that your “feelings” have been “hurt”? Give me a reason why I should care whether you respond to me or not. It’s your choice – you don’t seem to like me much, and I can’t be bothered feeling bullied, annoyed, insulted, or anything of that nature due to the utterings of someone who doesn’t like me, anyway.
You go on “Next, in your usual nasty fashion, you say: . . . “. If you “feel” I am “nasty”, that’s up to you. I have no control over your feelings, do I?
You carried on –
“Where to start … first, there are not “copious records” of gridded TOA solar input.” I never mentioned “gridded TOA solar input” – you are confused. I was merely following up on Pat Smith’s enquiry about “temperature records”. I believe there are copious records of temperatures of various sorts, but if you can provide that there are not, fair enough.
And then – “OK, since you are obviously incapable of actually obtaining the data and doing the computations required, . . . “.Are you mind reading again? If your “joy is to graph different relationships and then see what I can learn and find out from the graph.. . . ” is restricted to CERES, I apologise. Just giving you the opportunity to get some joy. My bad.
After writing “As you may or may not know, the total solar irradiance (TSI) at the top of the atmosphere (TOA) is about 340 watts per square meter (W/m2).” Gee, you acknowledge that I might or not know . . . , and maybe you didn’t really mean to write that TSI at TOA Is “about” 340 W/m2.
Maybe you were talking about a different TSI than the one from Wikipedia – “Total solar irradiance (TSI) is a measure of the solar power over all wavelengths per unit area incident on the Earth’s upper atmosphere. It is measured perpendicular to the incoming sunlight.[3]”
NASA doesn’t agree with your “about” 340 W/m2 either, but you might accuse me of being “nasty” if I ask you if you really meant average TSI or something equally pointless.
Getting close to the end, you write “This is from uncertainty and instrument error, as the physics of equal hemispheric radiation is clear.” Oh, I see, discard actual measurements, because they must be wrong! No, Willis, surface temperatures do not necessarily bear any relationship to the amount of solar radiation at TOA. For example, ground temperatures due to unconcentrated sunlight vary roughly between 90 C and -90 C. Of course , for GHE believers, this causes a small problem – the hottest ground temperatures occur where the atmosphere contains the least GHGs! You knew that of course, and will incorporate into your GHE description in due course.
There I go again, laughing at you. You can no more describe the GHE in any way that reflects reality, than fly to the Moon by flapping your arms, can you?
Finally (phew!!) –
You wrote –
“Finally, could you dial back on the aggro? Standing on your tiptoes in a vain attempt to bite my ankles is not a good look on you …”
Willis, what do you mean by that? Dial? Aggro? Tiptoes? Vain? Biting ankles? If you choose to feel offended, insulted, bullied, aggravated, whatever, be my guest. Your emotions are your affair – you can blame me for your inability to control yourself, but it’s not going to affect me in the slightest. Why should it?
Not a good look on me? What particular form of mental affliction leads you to think that I value your opinion?
Oh dear, Willis, you have a high opinion of your opinions, don’t you?
Carry on.
The strange steel greenhouse of Willis Eschenbach.
For any persistent onlookers –
Willis attempts to emulate Einstein, but doesn’t put enough thought into his “thought experiment”.
Willis has a large rock, which has a radioactive heat source within it, keeping the sphere at a constant temperature of 19.424 C for millions of years. The radioactive elements have an infinite half life, keeping the heat output constant.
I spent five minutes or so to consider Willis’ “Steel Greenhouse” – a link appears in Willis’ article above.
The information provided by W is –
Willis leaves much unsaid, presumably so that he can complain bitterly about not having said it.
My assumptions are –
Willis has not provided any information to the contrary.
However, based on the information provided by Willis, I calculated –
It will be noted that the steel shell is slightly colder than the sphere.
For the sake of this exercise, the area of the sphere is set at 1 m2. For a sphere of any other size, apply a multiplier of the number of square meters desired. This makes the interior surface of the steel shell 1.001577 m2, if the rocky sphere has an area of 1 m2.
Now the heat source is obviously 235 W, as the emitted energy is determined by Willis to be 235 W/m2. The temperature of the heat source is also determined by Willis as -19.424 C, as the sphere has an emissivity of 1.
The surrounding environment is nominally 0 K as there is no other heat source specified.
The steel shell absorbs 235 W of power from the sphere, which is spread over a slightly larger area, resulting in a temperature of -19.524 C (calculated from the emitted radiation – 243.63 W/m2 specified by W. The steel shell is slightly colder than the rocky sphere, as it must be.
Willis claims that the surface of the sphere will rise in temperature, due to the radiation from the colder steel shell surrounding the sphere.
Unfortunately, the heat source is limited to 235 W, and has a temperature of -19.424 C. The surface of the sphere cannot make the sphere any hotter than it already is anyway, being colder than the sphere.
Willis also claims that the steel shell emits 235 W/m2 from its exterior surface. This is impossible, as the shell is only receiving 235 W from the surface of the sphere, which is distributed over a minimum of 1.001577 m2 of steel shell. If the steel shell is infinitely thin, it can radiate as much as it receives – only 243.63 W/m2 – set by Willis earlier. If the shell has any thickness, the radiation per square meter from the exterior of the shell decreases further, as does the temperature.
The steel greenhouse illusion is not so much a “thought experiment”, as a “thoughtless fantasy”.
I am surprised that this particular illusion has lasted, unchallenged, as long as it has. There are obviously many gullible GHE supporters, who are quite capable of accepting any nonsense which is foisted upon them.
The “steel greenhouse” problem is quite typical of problems given to undergraduates in an introductory thermal sciences course. One of the reasons I have been participating in this thread is that I am about to start teaching a technical course at this level at a major university, and I want to observe common misconceptions in problem solving.
So let’s look at your analysis of the steel greenhouse. First, some minor points:
You don’t have to “assume” an emissivity of 1.0. The problem states that these are blackbodies, which by definition have an emissivity (as well as an absorptivity) of 1.0. As a teacher, this is a red flag to me, that the student is already missing some key concepts.
The total fixation on the very slightly larger radius of the shell compared to the sphere is excusable for a beginning student, but working through the problem should demonstrate that it is not important.
Also, the fixation on the premise of the problem, that the power source is constant and has allowed the sphere to come into steady state conditions, is unrealistic because sources like radioactive decay can change over millions and billions of years, shows that the student is unable to identify the important issues in the problem – another red flag!
Let’s get on to the analysis here:
You get the right result for the sphere alone in a 0K environment, that it must radiate away (235 * Asphere) watts to this environment to balance the (235 * Asphere) watts input from the internal supply. You also correctly compute the temperature of the blackbody surface necessary to radiate away this power: -19.424C.
You also get the right result (allowing for your typo – should be 234.63, not 243.63) for the combined sphere-and-shell system in a 0K environment, where the system must still radiate away (235 * Asphere) watts to maintain steady-state conditions. Accounting for the slightly larger radius of the shell, the shell must radiate a power flux density of (235 * Asphere / Ashell) W/m2, yielding a slightly lower temperature of -19.524C. (Wow, 1/10 degree cooler – big difference!)
But then you go completely off the rails:
You say: “The temperature of the heat source is also determined by Willis as -19.424 C”. That is a meaningless statement, and Willis “determined” no such thing!
But your key mistake is when you say: “The surface of the [shell] cannot make the sphere any hotter than it already is anyway, being colder than the sphere.” You acknowledge that the inner surface of the shell is radiating almost 235 W/m2 to the sphere, but you argue that this does nothing!
This shows that you have a complete misunderstanding of both the 1st and 2nd Laws of Thermodynamics! For the 1st Law (conservation of energy), you “disappear” the inward radiative power, in violation of the 1st Law. Another way of looking at this is computing the energy balance for the sphere itself when surrounded by the shell.
The sphere has a power input from the source of (235 * Asphere) watts. Under these conditions, it outputs radiative power of (235 * Asphere) watts, but inputs radiative power of (234.63 * Asphere) watts. So there is a power imbalance of ((235 – 235 + 234.63) * Asphere) = (234.63 * Asphere) watts, which will lead to an increase in the internal energy, and therefore the temperature, of the sphere. This will continue until the sphere reaches a temperature where its radiative output causes it to be in power balance again.
But, you protest, the radiation input is from the lower-temperature shell! The 2nd Law forbids this, doesn’t it? Absolutely not! The full process of the (inseparable) “radiative exchange” between the sphere and the shell provides (0.37 * Asphere) watts from the warmer sphere to the cooler shell, consistent with the 2nd Law.
And this has always been the understanding of the 2nd Law. Clausius, the “father of the 2nd Law”, explicitly stated this. The “heat flow” is from hotter to colder, even as radiative power flux goes in both directions.
Previously in this thread, I provided experimental evidence of this phenomenon to show that a colder “shell” (the thermal test chamber) could cause in increase in temperature of a hotter powered object (the operating electronic component). You acknowledged that this was a realistic scenario, but for some reason you cannot realize that this is the very same phenomenon as the “steel greenhouse”.
If I were grading your analysis in an introductory course, I would give you about 20% credit (not a passing score).
Thanks, Ed, you have endless patience. And your analysis is spot on.
w.
Ed,
You have been sharing Willis’ Kool-Aid.
Sorry about the typo, yes, the steel shell is indeed radiating less than 235 W/m2. It must, having a larger surface area. You agree that this results in the shell being cooler, but still claim it can warm a hotter object by its own radiation. That makes you delusional, of course. Try warming water with ice that is only 0.1 C colder than the water.
You say that I argue that a the radiation from a colder body cannot raise the temperature of a hotter one. No argument. It’s a fact. As to Clausius, who died in 1888, physics has moved along a little since then, whether you accept it or not. Clausius was right about “heat” moving from hot to cold, but of course wasn’t silly enough to believe that submerging a block of ice in a bowl of water would result in the water getting hotter, no matter how close the temperature of the ice to the water.
The ice may be emitting 300 W/m2 or so – but you haven’t got the faintest idea where that IR radiation goes, have you? Still so sure that it is exchanging radiation with the water? Oh dear, looks like you forgot to engage your brain before hammering away in the keyboard.
You wrote –
“The sphere has a power input from the source of (235 * Asphere) watts. Under these conditions, it outputs radiative power of (235 * Asphere) watts, but inputs radiative power of (234.63 * Asphere) watts. So there is a power imbalance of ((235 – 235 + 234.63) * Asphere) = (234.63 * Asphere) watts, which will lead to an increase in the internal energy, and therefore the temperature, of the sphere. This will continue until the sphere reaches a temperature where its radiative output causes it to be in power balance again.”
As I stated, for a sphere of 1 m2, the power supply must be 235 W – no more, no less. Otherwise, the sphere will be radiating more or less than 235 W!m2.
235 W output. You demonstrate Willis, level of delusion, by writing “will lead to an increase in the internal energy, and therefore the temperature, of the sphere.”
No, the power supply is 235 W – no more no less. No matter how you believe you can get it to increase – you can’t. Supplying 235 W to a sphere of 1 m2 results in a radiation of 235 W/m2. You believe you can increase the output of this power supply by surrounding it with a steel shell. Off you go, increase the power of your electric motor or your car engine by surrounding it with a steel shell. Oh your electric motor and your car engine are already mostly surrounded with steel shells? How silly of me, that’s obviously what makes them so powerful!
That’s sarcasm, but you may be too dim to realise it.
You say: “The temperature of the heat source is also determined by Willis as -19.424 C”. That is a meaningless statement, and Willis “determined” no such thing!”
I assumed the sphere to have emissivity of 1. Willis set the flux at 235 W/m2. If a temperature of -19.424 does not produce this flux, correct me. It would make no difference anyway. The steel shell is always colder than the sphere. The highest temperature it can ever reach is if it is in contact with the sphere. The further away from the sphere, the lower its temperature must be. You might believe that Willis or yourself can suspend the operation of physical laws at will, but you can’t. Willis even referred to the inverse square law in another comment, so possibly you are slightly more ignorant than Willis.
You also wrote –
“Previously in this thread, I provided experimental evidence of this phenomenon to show that a colder “shell” (the thermal test chamber) could cause in increase in temperature of a hotter powered object (the operating electronic component).”
You were no doubt referring to the heat transfer phenomenon called “insulation”. You didn’t mention any details of course, because this would destroy your conjuring trick. Maybe you are unaware of firefighters – who wear heavily insulated clothing to keep cool. Or cold rooms, where the greater the R-value of the insulation, the colder the contents remain (given the same power consumed by the heat pump)!
You wrote “If I were grading your analysis in an introductory course, I would give you about 20% credit (not a passing score). ” What mental disturbance leads you to think anyone would value the opinion of a fantasist, who believes he can magically turn 235 W into any number of Watts he likes? No you can’t.
Willis wrote “Thanks ed [ . . . ] and your analysis is spot on.” Dumb and dumber.
Come on ed, demonstrate the unswerving nature of your fantasy – tell me where the radiation from ice goes, when it is submerged in water! Maybe you could ask that other deluded fantasist Willis?
Thanks for the laughs – no GHE, no magical conversion of 235 W into 470 W (or anything else).
Try harder next time – maybe you can generate even more laughter.
Off you go.
Mike – You just continue digging yourself deeper and deeper!
You say: “You agree that this results in the shell being cooler, but still claim that it can warm a hotter object by its own radiation.”
NO!!! This is your key misconception! The “hotter object” (the sphere) is “warmed” by its internal power supply. That is, this supply provides a power input to the sphere. The shell reduces the radiative power loss of the sphere to ambient.
In the first case, the sphere is radiating directly to the ~0K ambient and there is virtually no radiation coming back from ambient. (In the real world, we have ~0.000003 W/m2 received from deep space.) So the radiative HEAT transfer is 235 W/m2.
In the second case, the sphere is radiating to a shell that is only slightly colder than it is. In the specific case we are discussing, the radiative HEAT transfer from the sphere is only 0.37 W/m2. With 235 W/m2 input to the sphere, and 0.37 W/m2 output, do you really believe the sphere is not getting hotter?
Let’s say you are depositing $235 per week into your bank account, and withdrawing $0.37 per week. By your logic, your bank balance will not change. I want to be your banker!!!
You keep bringing up the irrelevant example of the block of ice submerged in a pot of soup, and asking what happens to the IR photons from the ice. First of all, with the density of molecules in solid and liquid being ~1000 times greater than in gas, excited molecules that in a gas would emit a photon before colliding with a less excited molecule would instead collide with a less excited molecule before emitting a photon.
That is why we treat these heat transfer problems as conduction phenomena, not as radiation phenomena. Somebody who actually understood heat transfer would realize this immediately.
But I suppose the odd IR photon from the ice could make it to the liquid in the soup. Since water is highly opaque to IR radiation, these photons would be absorbed in the first micron or so of the soup, and that absorption would add to the energy of the soup. Note that, as I have acknowledged upthread, there is greater power transfer from the liquid soup to the ice, so the ice would have a cooling effect on the soup.
(But let’s say it is a radiative transfer problem with the ice emitting 300 W/m2 to the liquid soup. The soup might be at a temperature that is emitting 400 W/m2 toward the ice. Both substances absorb virtually all IR radiation, so there is a net HEAT transfer of 100 W/m2 from soup to ice. If isolated from the rest of the universe, this would result in cooling of the soup.)
But that is not at all the problem we are discussing here, and you make yourself foolish not being able to realize the difference. A relevant example would be a pot of soup with a power input from a stove in a sub-zero ambient environment. If you surround the pot with a shell of ice (at 0C), preventing direct thermal contact with the colder ambient, the presence of the ice shell will indeed result in a higher soup temperature than without the presence of the ice shell.
Related question: Why do you think the Inuits use igloos for warmth? Have you told them that igloos don’t do any good?
You deride my real-world example of the thermal test chamber as just being “insulation”. WTF do you think a shell between a warm sphere and super-cold ambient is? Can you seriously not see that these are the same types of systems???
Now, I have answered your (irrelevant) question about ice in soup. But you have dodged my (completely relevant) question about the radiation from the inside surface of the shell. So I ask it again: What happens to the energy radiated from the inner surface of the shell when it reaches the blackbody surface of the sphere? Answers must be consistent with the 1st Law of Thermodynamics!
You have the patience of Job.
The Mike-troll is probably a warmest masquerading as a denier. It obviously doesn’t know basic physics; it doesn’t know how to read a dictionary; and it projects its ignorance on others.
I would stop feeding the troll.
Jim,
You don’t know where the IR energy from ice totally submerged in water goes, any more than the rest of the dummies. The odd IR photon? Really? Heat travels from the hotter to the colder – The fantasists agree that the shell is colder than the sphere, even when 100% of the power emitted by the sphere’s power supply is reaching the shell. Ah, the magic of the GHE. Who needs reality when you can fantasise?
Of course, you have no intention of actually quoting anything I say, have you?
Watch dumb and dumber make complete fools of themselves – or head for the hills.
Let’s make a list of assumptions so all are discussing the same thing.
From:
https://www.e-education.psu.edu/png520/m16_p3.html
Let’s apply this to the problem.
Yes. The sphere radiates based on its temperature,
regardless of whether it radiates to space or to the
shell.
The shell will absorb the radiation from the sphere
and warm. As heat is transfered from the sphere to
the shell, the shell warms until thermodynamic
equilibrium is achieved.
S/B for two bodies I = σ(T1 – T2) and I = 0 is a
legitimate answer. Therefore, T1 = T2.
2. At thermodynamic equilibrium, both the shell and the
sphere will be emmiting the same amount of radiation
since they are at the same temperature.
As we move away from ideal conditions one must assign various values to all the parameters and develop gradients that define the transfer of heat throughout the bodies.
.
Jim,
You wrote –
Not according to NASA, which says “Thermodynamic equilibrium leads to the large scale definition of temperature. When two objects are in thermal equilibrium they are said to have the same temperature.”
And of course, the sphere and the shell must be at different temperatures, as I explained previously. Everybody else seems to agree.
I know you want to believe that two bodies emitting the same total amount of energy are at the same temperature, but in the words of the song “It ain’t necessarily so”. Would you like me provide more real examples, rather than the obvious one of Willis’ fantasy sphere and a shell at some distance from it? If I do, will you abase yourself before me, bang your head on the floor, and repeat 1000 times “I am not worthy”? Poking yourself in the eye with a hot needle is optional.
By the way, your first “agreed” assumption is disputed by Willis, at least, who wrote “The planet in my experiment is a block of rock that has been at the same temperature for a million years. Get over it.”
You will have to get used to Willis shifting the goal posts as he goes along.
Carry on.
Please note the following condition I explained:
This says the per square meter differences are negligible. Sure, the shell may receive 0.00001 less watts/ sq meter, but that is negligible.
If the shell is substantially larger,where r^2 losses are important, then the per square meter absorbed will be less. However, the total power radiated will remain the same, albeit at a slightly lower temperature.
I wanted to make the point that when discussing thermodynamics, assumptions are important. If you don’t outline them in great detail, then arguments can not proceed properly. As a matter of physics in the real world, average values don’t cut it. Defining gradients is required using physical factors like mass, specific heats, emissivity/absorption, heat reservoir actions, conductivity, etc., are all components.
Jim,
You can’t just make physical laws vanish because you don’t like them. If the sphere is smaller in area than the shell, the shell is at a lower temperature. It doesn’t matter whether it’s a little or a lot – it’s less.
You wrote –
“Sure, the shell may receive 0.00001 less watts/ sq meter, but that is negligible.”. Well, Willis specified a figure. Go on, tell me the limit for “negligibility”. When does a cooler object get the ability to raise the temperature of a hotter one?
You really are an idiot, aren’t you?
Next thing, you’ll be claiming you can warm water with ice that is only 0.00001 K colder!
Dimwit.
Ed,
You wrote. –
“With 235 W/m2 input to the sphere, and 0.37 W/m2 output, do you really believe the sphere is not getting hotter?”
Hang on just a second there, pardner! Have you had a momentary mental aberration? 235 W input to a square meter of the sphere’s surface, and you think you can reduce the output to 0.37 W/m2? The total output is now 0.37 W, from an input of 235 W?
Where did the other 234.63 W go? Don’t be really stupid and talk about subtracting and adding fluxes – you have one heat source – 235 W. You can reflect, concentrate, add, subtract until you have confused yourself, as well as the rest of the dimwits who believe you.
Now, contrary to your kindergarten understanding, all objects above absolute zero emit IR. You cannot stop any of them from emitting, and the intensity at which they emit is proportional to the fourth power of their absolute temperature. Here’s a snippet from a university (not a kindergarten ) –
“The Stefan-Boltzmann law, a fundamental law of physics, explains the relationship between an object’s temperature and the amount of radiation that it emits. This law (expressed mathematically as E = σT4) states that all objects with temperatures above absolute zero (0K or -273°C or -459°F) emit radiation at a rate proportional to the fourth power of their absolute temperature.”
Putting a mirror in front of an object, and reflecting its radiation back at it will alter neither the amount of radiation it emits, nor raise its temperature. Try it. I’m right, and you’re fantasising.
You really don’t understand any of this, do you? I’ll let you and Willis make yourselves look as stupid as you like. You don’t need my assistance.
Come on, tell everyone how you managed to reduce the wattage of the power supply, from 235 W to 0.37 W. I suppose you are going to say that you can supply 235 W to a 1 m2 surface with an emissivity of 1, but it gets annoyed, and its emissivity changes, only allowing 0.37 W out! Yes, Willis stated the surface has been radiating 235 W/m2 for millions of years, and has an emissivity of 1 – you can tell me what temperature is needed to provide this radiative intensity, because you don’t seem to believe me.
Over to you.
Ed,
You asked –
“What happens to the energy radiated from the inner surface of the shell when it reaches the blackbody surface of the sphere? Answers must be consistent with the 1st Law of Thermodynamics!”
Well, the shell is colder than the sphere, as you say.
The system consists of a sphere, surrounded by a shell, in an infinite vacuum.
The first law of thermodynamics can be expressed thus –
The first law of thermodynamics states that, when energy passes into or out of a system (as work, heat, or matter), the system’s internal energy changes in accordance with the law of conservation of energy.
Now, somebody stated that the shell radiates 235 W/m2 into the infinite vacuum. It can’t, because it only receives 235 W, and has an area slightly greater than 1 m2. Willis confirms this by stating that the inner surface of the sphere is radiating 234.63 W/m2, so I hope you are not going to claim that the outside of the shell (which cannot have a lesser area than the inside) emits more energy than it receives!
So the power supply is providing 235 W, the outer shell is radiating 235 W to the infinite vacuum. I’ll let you beclown yourself attempting to answer your own gotcha. Bear in mind that heat travels from hot to cold (heat being measured in degrees of hotness – temperature if you like), and you agree that the sphere’s surface is 0.1 C hotter than the sphere, and the power supply cannot provide more than 235 W. You don’t like me stating the temperature of the sphere as -19.424 C – maybe you can provide your reasons for not believing the Stefan-Boltzmann law – which gives me my answer. Feel free to check my calculations.
Given a sphere with an unchanging temperature, surrounding it with a lower temperature shell won’t increase its temperature. There is only one heat source, 235 W, sufficient to raise the temperature of the sphere to -19.424.
Run away, learn some physics, quote my exact words, then provide facts to support any disagreement you may have.
Whoops, a typo. The sphere’s surface is hotter than the shell, obviously, although some dummies might not realise this.
Hope I didn’t overload your tiny brain. Apologies if I did.
ed,
You wrote –
“Now, I have answered your (irrelevant) question about ice in soup.”
Unfortunately, your answer is no answer at all. As I stated, the ice is emitting 300 W/m2 of IR radiation. Quickly now, demonstrate your cleverness by calculating the number of photons emitted by the ice every second. To make it easy for you, the ice is pure H2O, has an emissivity of 0.97, and has a mass of 0.3 kg., and is spherical in shape.
If you want to use a different mass for ease of calculation, be my guest. The surrounding environment is a vacuum, although the ice will emit the same number of photons per second, whether you like it or not, and regardless of its environment.
When you have established this fairly large number, you might wish to reconsider your statement “But I suppose the odd IR photon from the ice could make it to the liquid in the soup.” You suppose, do you? How odd is that particular photon?
OK. So subtract “the odd photon” from number you just calculated. Where do the rest of the photons go? Come on, you’ve just calculated how many were emitted – they don’t just vanish, do they?
You will have noticed your calculation doesn’t depend on whether the object in question is surrounded by gas, liquid, solid, or a vacuum – nothing at all. It depends only on temperature.
You didn’t know that? Gee, what a surprise!
Off you go now.
Mike – I don’t know whether to laugh or cry. You just keep doubling down on your errors!
You asked about the heat transfer for the case of ” submerging a block of ice in a bowl of water”. To anyone who has actually studied heat transfer, this is a CONDUCTION problem, not a RADIATION problem. This basic concept is totally unknown to you.
Because you have obviously never actually studied heat transfer, I took the time to explain the reason for this distinction to you. (In solids and liquids, heat transfer by direct molecular contact dominates.) But this explanation went completely over your head, obviously due to your lack of proper background.
It is important to realize (although I don’t expect you to), that for these conduction problems, the radiative emissivity/absorptivity of the objects in contact DOES NOT MATTER. Which says, the number of photons emitted at whatever conductive boundary you are considering IS NOT IMPORTANT. (If you think it matters, ask yourself what happens to the IR photons emitted a millimeter below the surface of the block of ice.)
Now you say, “The surrounding environment is a vacuum, although the ice will emit the same number of photons per second, whether you like it or not, and regardless of its environment.”
Ahh, so now you have added a vacuum and think it doesn’t make any difference for the heat transfer! Another own goal for Mike!
First, let’s back to your question about a block of ice submerged in liquid soup, but somehow with a layer of vacuum in between preventing any conduction between them. Now we’ve got a radiation heat transfer problem! You ask what happens to the ~300W/m2 of IR radiation emitted by the ice when it reaches the (water-based) soup.
Well, assuming the IR radiative properties of the soup are not significantly different from pure water, the very thin surface layer of the water will absorb about 97% of this radiative power, and this will increase the thermal energy of the soup.
However, the soup will emit more power toward the ice, which will absorb 97% of that per your specification, so the NET HEAT transfer is from the soup to the ice, consistent with the 2nd Law. If these two objects are isolated from the rest of the universe, the temperature of the soup will decrease. But in a more complex system, this problem is underdetermined.
So I repeat my request for you to answer (and not dodge as you have been doing): what happens to the radiative energy emitted by the inner surface of the shell when it reaches the sphere?
You have been treating the radiative heat transfer problem of the “steel greenhouse” as if it were a conductive heat transfer problem. That is, you get the same result as if the shell were in full conductive contact with the sphere.
To anyone with the least familiarity with actual heat transfer, that is a patently ridiculous idea. (Ever heard of a Thermos bottle???) But you cling to that delusion!
Part of the problem with these “examples” is that they are dealing with averages, I.e., static conditions. That is basically impossible.
If a body is heated to a temperature and the heat is removed, it immediately begins to cool. You need a gradient that describes the loss. IOW, the soup cools on its own. The ice cube absorbs heat and warms thereby melting. Again, you need a gradient. All forms of heat, conduction, convection, and radiation for a body will combine to an overall gradient.
Heat gain/loss is not static. Static only works for static conditions. It is useful for basic learning about heat but sooner or later one must proceed into calculus based math to describe a dynamic system.
Hi Jim,
All thermodynamics courses start with steady-state examples. If the students can’t get them correct, there is no point in moving on to more complex dynamic cases.
Ed,
You seem to have passed up the opportunity to demonstrate your knowledge.
As I stated, the ice is emitting 300 W/m2 of IR radiation. Quickly now, demonstrate your cleverness by calculating the number of photons emitted by the ice every second. To make it easy for you, the ice is pure H2O, has an emissivity of 0.97, and has a mass of 0.3 kg., and is spherical in shape.
If you want to use a different mass for ease of calculation, be my guest. The surrounding environment is a vacuum, although the ice will emit the same number of photons per second, whether you like it or not, and regardless of its environment.
When you have established this fairly large number, you might wish to reconsider your statement “But I suppose the odd IR photon from the ice could make it to the liquid in the soup.” You suppose, do you? How odd is that particular photon?
OK. So subtract “the odd photon” from number you just calculated. Where do the rest of the photons go? Come on, you’ve just calculated how many were emitted – they don’t just vanish, do they?
You will have noticed your calculation doesn’t depend on whether the object in question is surrounded by gas, liquid, solid, or a vacuum – nothing at all. It depends only on temperature.
You didn’t know that? Gee, what a surprise!
Off you go now.
Jim,
So thermal equilibrium doesn’t exist? Two bricks on a table, 20 C environment – all at 20 C, emitting IR. Photons all the same
energy – yes, I know you don’t believe it, but it’s true. Wavelength inversely proportional to temperature (short version).
Off you go now, use your knowledge of “calculus based math”? to calculate the temperature change as the bricks are moved closer to each other. Only joking – you are probably stupid enough to start “adding fluxes” just because the emitting surfaces are closer to each other!
You really are off with the fairies, aren’t you?
Jim,
One thing at a time.
“Well, assuming the IR radiative properties of the soup are not significantly different from pure water, the very thin surface layer of the water will absorb about 97% of this radiative power, and this will increase the thermal energy of the soup.”
I see. The thermal energy of the soup rises, but the temperature falls. Is that it? Or does “very thin surface layer” heat up, but the rest of the soup cools?
Wriggle some more.
Don’t try and be silly by saying something like “But in a more complex system, this problem is underdetermined.”
It’s pretty simple – submerge a block of ice in soup. The soup cools, in spite of the fact that your claim that “this will increase the thermal energy of the soup.”
Are you really trying to convince yourself that increasing the thermal energy will result in cooling? What happens when you reduce thermal energy? Heating?
Nope, you can’t heat soup wth ice. So where does the radiation from the ice go? Maybe, like Einstein, you don’t accept quantum physics. Good for you! Reality doesn’t care what you think.
Mike – Now you’re just being ridiculous!
I very clearly stated: “However, the soup will emit more power toward the ice, which will absorb 97% of that per your specification, so the NET HEAT transfer is from the soup to the ice, consistent with the 2nd Law. If these two objects are isolated from the rest of the universe, the temperature of the soup will decrease.”
But apparently, this is too difficult a concept for you. By the time my kids were 6, they understood that if they spent more than they got in allowance, the amount in their piggy bank would decrease, even though they continued to get their allowance. But considering two “flows” is apparently beyond your cognitive capabilities.
I note that you continue to dodge the question of what happens to the IR photons emitted toward the warmer object, whether shell to sphere, or ice to soup. I think it’s because you have no answer. Prove me wrong!
“Ever heard of a Thermos bottle???”
Yes, actually. Two surface coated maximally reflective surfaces separated by a vacuum. Each one emitting the same amount of radiation towards the other. No flux addition, unfortunately. The contents stabilise at the temperature of the environment surrounding the flask – whether initially hotter or colder.
Maybe you are talking about some other Thermos flask?
Maybe you could explain why a surface powered by a fixed power source emitting 275 W/m2 opposing another surface reflecting 275 W/m2, separated by a vacuum, is different to a temperature source of – 19 C in a vacuum flask?
Do you really believe the flask will heat up?
Enough NET HEAT, or do you need some “calculus based math”?
Carry on,
Ed,
You wrote –
“You asked about the heat transfer for the case of ” submerging a block of ice in a bowl of water””.
Well no, I didn’t and you can’t quote me saying any such thing. You just make up what you thought or hoped I said, so you could answer yourself, trying to sound intelligent.
You might have noticed Willis” request to quote exact words to avoid understanding, and ignored it totally just to give Willis a good serve of gratuitous offence. I don’t know, and I don’t claim to reads minds, unlike some of the delusional dimwits here.
Keep on fabricating, and pretending you can read minds. I don’t value your opinions, anyway, unless supported by facts.
Others are free to form their own views.
I quoted you directly, even using copy and paste! You’ll try any ridiculous stunt to avoid admitting you were wrong!
Ed,
Well, no, you didn’t.
i didn’t use the words “heat transfer” at all, because it’s a silly term. “Heat” is not transferred. It is not “caloric”, and it cannot be poured, trapped or stored. Heat transfer is a description used by people who find it useful to try to explain the practical consequences of physical laws that they do not understand.
So I didn’t ask you about “the heat transfer” of anything.
Try harder next time. Maybe quote my words, and don’t add your own editorialising, in an attempt to convince others of the reasons I used those words. Why not just quote me, and let others form their own opinions?
Seriously, Mike? That’s what you’re going to hang your hat on?
What were we discussing about “submerging a block of ice in a bowl of water”, the decorations on the bowl? The (apparent) fact that you don’t like this widely used and widely understood term doesn’t mean it wasn’t the accurate subject description of our discussion.
Every university engineering school I have ever seen offers courses with the “silly” title of “heat transfer”. The MIT textbook referenced in this thread is titled “A Heat Transfer Textbook”. Yet they don’t teach thte caloric theory.
Just another transparently desperate attempt to divert attention from your losing arguments!
First, my thanks to Ed Bo for his endless polite attempts to educate Mike Flynn and others regarding basic thermodynamics. Ed, your patience is an inspiration and your understanding of the subject is excellent. I know your understanding is good because your explanations are simple and clear.
Next, the main problem seems to be people who do not distinguish between flows of energy and flows of heat. A secondary problem is people who do not distinguish radiative heat flow from conductive heat flow.
Heat is the spontaneous net energy flow from a warmer location to a cooler location. It only goes in one direction, from hot to cold.
Radiative energy, on the other hand, goes in any direction.
Now, there’s an incorrect belief going around that two objects can’t both be radiating towards each other. But we know that if we turn on two flashlights, they each can illuminate the other.
But hey, no need to believe me. Here’s a diagram from an MIT textbook on this exact subject. The textbook is online, you can read the section. I’ve put the URL in the top left corner. Here’s the link.
In this diagram, the letter “Q” indicates an ongoing flow of energy. Two specific flows are identified, Q1 to 2, and Q2 to 1. These are the flows between the two objects. Note that just as with two flashlights shining on each other, these are real, measurable, observable physical flows of energy.
Note, however, that THESE ARE NOT FLOWS OF HEAT! They are flows of radiant energy. Heat is the NET flow of energy between the two, and it only and always runs from hot to cold.
If you are still confused, let me suggest that you look at the diagram. Study it carefully. Read every word. Then look at it again, and read and study it again.
Then go to the URL, get the textbook, and read the chapter.
These are not radical new ideas I’m putting out. It’s bog-standard thermodynamics.
Onwards, ever onwards,
w.
PS—Could folks please dial back on the aggro? First, neither Ed Bo nor I are fools or whatever names we’re being called.
Second, the moment someone starts throwing mud, everyone knows the mud-thrower is out of actual scientific information and doesn’t trust the strength of their own arguments.
Because if you have the facts on your side, you have no need to make such attacks …
TL;DR version?
Throwing mud tells people you know you’re losing the debate. Would not recommend.
Willis,
I’ll take one step at a time. Your reference states –
“All bodies constantly emit energy by a process of electromagnetic radiation.”
Wrong. A body at absolute zero emits no energy at all. Did the author really mean to say what I claim – “All matter above absolute zero emits infrared radiation”, but was either ignorant, sloppy or intentionally misleading?
You wrote “First, neither Ed Bo nor I are fools or whatever names we’re being called.”
Good. Then you both should be able to justify your faith in an author who thinks that bodies at absolute zero emit energy by electromagnetic radiation. I assume you can specify the “whatever names you are being called” – otherwise you might be claiming that you are not wise, intelligent, knowledgeable and so on.
By the way, saying something like “Throwing mud tells people you know you’re losing the debate. Would not recommend.” Seems foolish to me, Why do you think that anybody should take notice of you? What “debate” are you talking about?
One thing at a time. I said “All matter above absolute zero emits infrared radiation.” Do you agree or not? If not, why not?
What would you call someone who claims that bodies at absolute zero emit any electromagnetic radiation at all? Willis Eschenbach? Ed Bo?
Over to you.
Mike,
If you actually understood this stuff – which you obviously don’t – you would know that absolute zero cannot be reached. So there can be no bodies AT absolute zero.
If all bodies above absolute zero emit radiation, and all bodies are above absolute zero, then it follows very simply that “all bodies emit radiation”.
So your feeble attempt at nitpicking pedantry (which wouldn’t be important even if true), fails – as usual!
You’re not even a very good troll!
No, Ed, I am aware that absolute zero is a limit. Theoretical. Just like a theoretical black body which absorbs all radiation which falls on it – even the radiation from a super-nova. I am also aware that such a black body does not exist in reality, nor an infinite plane, nor a friction free bearing, nor a sphere with a fixed temperature that can be heated by a colder body surrounding it, in an environment with no temperature at all – a vacuum. You will accept the existence of a vacuum which consists of nothing at all, I presume? An infinite vacuum, even, which doesn’t exist?
Now, you might nit-pick and play semantic games here, if I say all “matter” rather than all “bodies”, above absolute zero emit infrared radiation, then gases are included. However, you will have noticed that some people claim that noble gases neither emit nor absorb infrared radiation.
However, if you want to stick with reality, and avoid the theoretical, I would be happy. I agree with you that it is impossible for real matter to be neither absorbing nor emitting photons. Oh, wait, you probably don’t accept that photons are ejected at the speed of light from an electron without accelerating! Or do you? Only joking, you don’t even know what I’m talking about, do you?
Tell me (if you don’t mind) – do gases absorb and emit IR? Even monoatomic gases?
One thing at a time.
Willis claims you are not a fool. What objective evidence can you provide to support his statement? You refuse to accept the theoretical concept of matter at absolute zero being at absolute zero because it has no temperature! Emitting no infrared radiation.
Sounds foolish to me – you accept a body with an emissivity of 1, which is impossible for a real body, but not a body which emits no infrared radiation at all – likewise unreal.
Again,
Tell me (if you don’t mind) – do gases absorb and emit IR? Even monoatomic gases?
Ed,
“You’re not even a very good troll!”
You must be living in a fantasy. Show me where I claimed to be a troll of any sort.
I am curious as to the particular mental impairment which leads you to think that I should value the opinion of a dimwit who claims I’m a troll, and then claims I’m not a very good one!
Here’s the thing, Ed. Some delusional character claimed that a sphere with an emissivity of 1, radiating 235 W/m2 (a temperature of -19.424 C, according to the Stefan-Boltzmann Law), in an infinite vacuum, could have its temperature increased by a steel shell – initially radiating nothing at all – absolute zero.
If the temperature of -19.424 C terrifies you for some reason, substitute your own. Now tell me why surrounding a body with a fixed temperature with a colder body (as many of them as you like, made of whatever you like), will change the temperature of the sphere – which has a constant temperature, and is the sole heat source. The delusional character claimed the sphere’s temperature had not varied for millions of years.
Come on, demonstrate that you are not the fool you appear. If you want to, of course. Others can decide for themselves if you are as foolish as I think. They may think you are just ignorant, slow, or deluded. I have no way of knowing, do I?
Mike – Still dodging and weaving, trying any diversion to avoid the issue at hand. Not the behavior of someone with a winning case.
I will repeat my question to you once again: What happens to the IR photons (and the energy they carry) from the inner surface of the cool shell when they reach the warmer sphere? This should not be a problem for a super expert like you! Come on, enlighten us!
On the other hand, you are someone who:
Does not understand the difference between radiation heat transfer and conduction heat transfer.
Does (or at least did) not understand the emissivity of a blackbody.
Does not understand the difference in analysis of a system with a separate power input and one without.
Cannot analyze a system with more than two parts correctly.
Was not aware, even after it was pointed out to him, that absolute zero CANNOT be reached. (Even if it could, it had nothing to do with the point being made — bringing it up was a feeble attempt at distraction.)
(I’ll stop there — this is tedious…)
So maybe I shouldn’t expect enlightenment from you.
Ed,
You wrote –
“I will repeat my question to you once again: What happens to the IR photons (and the energy they carry) from the inner surface of the cool shell when they reach the warmer sphere?”
Precisely the same things that happen to the photons from a black of ice submerged in water. If you don’t believe that matter emits IR even when surrounded by other matter, try surrounding water with ice – or maybe just put the ice above, below, to the side or whatever? Willis’ silly textbook even says that the “cold” emitted IR feels “cold”! Are you silly enough to believe in “cold rays”? What about “cold photons”?
Does ice emit “cold photons” which are absorbed by the soup, making it colder? I don’t think so!
Back to your silly attempt to get me to tell you what you don’t know. Admit you don’t know, grovel in mortification before me, repeating “I am not worthy” a hundred times, and convince me you have made sincere efforts to find the answer for yourself, but have been unsuccessful. I might consider referring you to a suitable explanation by someone whose authority you might accept.
As you wrote – “This should not be a problem for a super expert like you! Come on, enlighten us!”
if you believe I’m a super expert, good for you! I appreciate the flattery. But why should I enlighten you? Are you now appealing to my authority because you realise I am right? You are right when you write – “So maybe I shouldn’t expect enlightenment from you.”
Not unless you convince me that you deserve my enlightenment!
in the meantime, try and figure how wrapping a sphere with a constant temperature with a shell initially radiating nothing at all (you don’t accept it, but that’s a temperature of absolute zero) will make the sphere hotter.
If you believe the shell magically raises its temperature to less than that of the sphere instantaneously, that’s fine, but energy from the sphere is only travelling at the speed of light, so the shell is initially at – absolute zero! No temperature at all!
So now you have a shell, colder than the sphere because it is warmed by the sphere, and cannot be at a higher temperature than the sphere, even in direct contact (it could only be as hot, never hotter), and you claim that adding more colder material will raise the temperature of the fixed temperature heat source?
Don’t blame me if you are unable to accept the fairly simple proposition put by Clausius “Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time”
Maybe you also believe the somewhat delusional Willis when he stated that bringing two heat sources closer together would raise the temperature of both! At thermal equilibrium, two objects have exactly the same temperature. Willis (and others) would have you believe that decreasing the distance between two objects at equilibrium makes them both hotter!
You also wrote –
“Still dodging and weaving, trying any diversion to avoid the issue at hand. Not the behavior of someone with a winning case.”
Tell me again why I should value your opinion of my “behaviour”? You might be concerned about “winning” or “loading” a “case”, but not me. Physical facts don’t change, regardless of what you or I “think”.
Keep believing your fantasy. Waste as much time as you like trying to turn your fantasy into measurable reality. Let me know how you get on. Only joking, you have no intention of actually performing a rigorous experiment, have you?
Damn! Another typo – should be “block” instead of “black”!
Sometimes difficult to laugh and type accurately simultaneously. I don’t know why.
Mike, this time in answer to my question, you answer: “Precisely the same things that happen to the photons from a [block] of ice submerged in water.”
Which is no answer at all, because you have never said what happens to those photons. You have said they are emitted, but not what happens when they reach a blackbody surface.
So you continue to dodge and weave — anything to avoid actually answering the question!
Amusingly, you object to an example in the MIT heat transfer textbook. It states that “your face will feel cooler” when it is facing an open freezer, but far enough away that it cannot be from conductive or convective transfer.
Have you ever tried this? It’s very easy to verify that this is true. And it’s not from any “cold rays”. The inside of the freezer will only radiate about 225 W/m2, compared to about 400 W/m2 from room temperature surfaces. With your face radiating about 500 W/m2, the net radiative transfer from your face almost triples, from ~100 W/m2 to ~275 W/m2. And yes, you can feel this!
As to Clausius, you obviously have not read much. Here’s a key quote:
” What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one. ”
DIE MECHANISCHE WÄRMETHEORIE (1887)
You just continue to demonstrate your complete inability to deal competently. I cease to be surprised.
Ed,
You wrote –
“Mike, this time in answer to my question, you answer: “Precisely the same things that happen to the photons from a [block] of ice submerged in water.”
Which is no answer at all, because you have never said what happens to those photons. You have said they are emitted, but not what happens when they reach a blackbody surface.”
If you know the answer to the question you posed, you are just asking a pointless silly gotcha – trolling.
If you don’t know the answer, look it up.
If you want me to provide the answer for you, I have set the conditions.
You don’t seem to like me quoting Clausius “Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time”
Do you wish to dispute this indisputable fact? No?
OK.
Ooooooh, still don’t know what happens to the photons from ice? Still expect me to tell you? No – I’ve changed my mind.
You wrote “You just continue to demonstrate your complete inability to deal competently. I cease to be surprised.”
Am I supposed to value the opinion of such as you? You can’t even say where the photons from a block of ice go! Go on, tell me how you make a constant temperature sphere hotter by surrounding it with a shell initially at absolute zero. Do you need a bigger shell, perhaps? Maybe made of papier mache, or unobtainium?
Yes, I am laughing at you. You can laugh at yourself too, if you like. Not many adverse effects from laughter, are there?