Thursday, October 05, 2017

The Green Plate Effect


An evergreen of denial is that a colder object can never make a warmer object hotter.  That's the Second Law of Thermodynamics, so according to the Agendaists, the Greenhouse Effect, with greenhouse gases playing the role of the colder object, is rubbish.  They neglect the fact that heating and cooling are dynamic processes and thermodynamics is not.

Eli, of course, is a dynamic bunny and knows how to add and subtract. Divide is also possible.  What is happening is that one does not have just a hot body and a cold body, but a really hot body, the sun, constantly heating a colder (much), but still warm body the Earth, which then radiates the same amount of energy to space.

In elevator speak, Tyndall put it

[T]he atmosphere admits of the entrance of the solar heat, but checks its exit; and the result is a tendency to accumulate heat at the surface of the planet.
Eli had a different but not as elegant elevator tweet

Today on twitter, Eli stepped through the simple math and he thought it would be a good thing to put the thread on this blog for future reference.  We start with a simple case, imagine the Earth is just a plate in space with sunlight shining on it.   Maybe 400 W/m^2


The sun warms the plate, but as the plate warms it radiates until the radiated heat matches the heat being absorbed from the sun


Using the Stefan Boltzman Law you can calculate the temperature of the plate when it reaches equilibrium (400 W/m2) = 2 σ Teq4   where  σ is the Stefan Boltzmann constant 5.67 x 10-8 W/(m2 K4), factor of 2 for a two sided plate per m2. Run the numbers Teq=244 K.

Now lets add another plate. We'll color this plate green for greenhouse. It is heated by the first at a rate of 200 W/m2



But after a while, it too has to heat up and reach an equilibrium temperature. . . so as a first guess something like


That's wrong though because there are 400 W/m^2 going into the two plate system and 300 coming out.  At equilibrium an equal amount of energy has to be going in as coming out  So what happens??

The entire system has to heat up to reach the equilibrium condition.  T1 and T2 are the equilibrium temps of the plates.



Looking at the two plate system, the energy going in is 400  W/m2 and the energy going out is  σT14 +  σT24    Since these will be equal at equilibrium

400  W/m2  = σ T14 +  σ T24 

And there also has to be an equilibrium for the energy going in and out of the green plate

σ T14 =  2 σ T24

The bunnies can rearrange the second equation to get

σ T24 =  1/2 σ T14

and substitute for σ T2 back into the first equation 

400  W/m2  = σ T14 +  1/2 σ T14
or
400  W/m2  = 3/2 σ T14 

Solving for T1 the answer is T1 = 262 K.

Without the greenhouse plate it was 244 K.  

Introduction of the second plate raised the equilibrium temperature of the first by 18 K. 

The Green Plate Effect

Show this to the next fool with an agenda who thinks that the Green Plate Effect violates the Second Law of Thermodynamics


517 comments:

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Betty Pound said...
This comment has been removed by a blog administrator.
EliRabett said...

Clausius rises from the grave to point out that he said:

This principle on which the whole following development is based is : heat can never pass from a cooler to a warmer body, unless another change associated herewith simultaneously happens.

NOT

"Heat is the amount of energy that flows spontaneously from a warmer object to a cooler one"

So he kindly requests that BP not lie by omission

Clausius goes on to point out that

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.

This pretty much shows that BP is making a habit of lying by omission

Tip o the ear to Bindidon @ Dr. Roys..

Betty Pound said...

Thanks for falling into my trap.

"The blue plate will receive 400 W/m2 from the sun and 133 W/m2 from the green plate or 533 W/m2 in total"

The green plate is not an energy source. You recycled 50% of blue's emission back to blue. The 133 is not new raw energy that you can add.

If the blue sends 267 to green and receives 133 from green, then you really sent 267-133=133 of thermal/heat energy to green.

Thus you have a situation where blue emits 267 to space and 133 to green. This violates:

"No, that means that every square meter of the surface of the blue plate will emit 267 W/M2"

You see? Your own blue plate is not at thermal equilibrium.

Further you made the mistake of assuming heat has a preference for going to colder object (space). This is a fallacy.

The blue emitting 267 (heat) to space and 133 (heat) to green, will in no time (via conduction) turn to 200 to space, 200 to green.

There is no reason for 67 W/m2 of thernal energy not to flow to 133 W/m2 side.

"The emission to space will be 267 (BP)+ 133 (GP) W/m2 = 400 W/m2 EXACTLY WHAT IS COMING IN FROM THE SUN"

Uhuh, there you go again foolishly believing that 267 can only heat space, and not 133.

Eli failed physics. He simply spouts rhetoric to make his phony model work.

Betty Pound said...

"unless another change associated herewith simultaneously happens."

Source? Are you talking about Work? Irrelevant to discussion.


"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"

Eli debunks himself.

"warm body radiates heat"

"cold body radiates"

Eli didn't notice the lack of heat in the second case, and thereby completely discredits himself.

"simultaneous double heat exchange"

Can you show me the raw German text of this? Looks like bad translation.

Eli uses an ancient translated text, rather than modern textbooks. How pathetic.

Chris A. said...

"No. If the blackbody is at higher energy or temperature, the "absorbed" energy does nothing."

Not true. It affects the rate at which this black body cools.

"The blackbody will not even absorb lower energy photons."

Not true. A black body absorbs every photon.

"The lower energy photons will just destructively interfere with the same lower frequency photons emitted by that blackbody."

Not true. Photons simply cannot interfere destructively, like acoustic waves can do.

"That's why you can't add 133 W/m^2 to an object emitting 267 W/m^2."

Not true. Because all your former assumptions leading to this are already wrong.

You are making strong claims. The standard literature of physics claim otherwise.
Why are you convinced to know it better than Clausius, Planck and the entire rest of physicians who published on the subject of radiative heat transfer since the subject came up?

Have you ever read their works?

Chris A. said...

"If the blue sends 267 to green and receives 133 from green, then you really sent 267-133=133 of thermal/heat energy to green."

Wow. At least you state something that is perfectly true. The amount of energy that can be characterized as "heat" is 267 -133.

Since you also stated your very self, that heat does not equal energy, that fact tells nothing about the total amount of energy of the fluxes.

If you substitute the numbers by 1267 - 1133, the very same transfer of heat occures. The difference would only be in the absolute temperature of the plates.

You can do the same with 100.267 - 100.133. The fluxes will be much denser, the heat transfered remains the same anyway.

Betty Pound said...

I can't find an English translation of "Clausius - Die mechanische wärmetheorie"

When I search for Eli's Clausius quote, I get nothing but debate pages between the usual suspects.

I even suspect that it's a fabrication, but I could be wrong.

Tell me the exact page the quote is on.

Betty Pound said...

So the blue plate emits 267 heat to space and 133 heat to green.

Then it's not possible for blue to be at 262K (emit 267 heat in all directions).

The blue plate is therefore claimed to not be at equilibrium. That's not possible. It will come to thermal equilibrium (267+133)/2=200, i.e. 244K.

The idea that the blue plate has a preference for heating space is perverse. The 267 will end up being 200 to space, and 67 to 133 side.

EliRabett said...

BP writes:

Eli debunks himself.
"warm body radiates heat"
"cold body radiates"
Eli didn't notice the lack of heat in the second case, and thereby completely discredits himself.

Can you show me the raw German text of this? Looks like bad translation.
Eli uses an ancient translated text, rather than modern textbooks. How pathetic.

----------------------------

This is, of course amusing on several grounds. First Eli provided links to the original German, second, if BP looked at the original German

Was ferner die in gewoehnlicher Weise stattfindende Waermestrahlung anbetrifft, so ist es freilich bekannt, dass nicht nur der warme Koerper dem kalten, sondern auch umgekehrt der kalte Koerper dem warmen Waerme zustrahlt, aber das Gesammtresultat dieses gleichzeitig stattfindenden doppelten Waermeaustausches besteht, wie man als erfahrungsmaessig feststehend ansehen kann, immer darin, dass der kaeltere Koerper auf Kosten des waermeren einen Zuwachs an Waerme erfaehrt.

The phrase "aber das Gesammtresultat dieses gleichzeitig stattfindenden doppelten Waermeaustausches besteht" is correctly translated by "but the total result of this simultaneous double heat exchange is".

German speakers will note that the phrase "sondern auch umgekehrt der kalte Koerper dem warmen Waerme zustrahlt," in English is "but also in the opposite way the colder body radiates heat to the warmer (one)".

Wanna try again?

Betty Pound said...

"A black body absorbs every photon."

But not every photon raises blackbody temperature. Only photons of a higher frequency that the BB lacks will heat the BB.

Microscopically, the lower frequency photon is destroyed.

http://curious.astro.cornell.edu/about-us/137-physics/general-physics/particles-and-quantum-physics/805-how-are-photons-created-and-destroyed-advanced

"Photons simply cannot interfere destructively, like acoustic waves can do."

https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Spectroscopy/Fundamentals_of_Spectroscopy/Electromagnetic_Radiation

Ctrl-F "interference"

You are confidently arguing from ignorance.

EliRabett said...


https://ia800309.us.archive.org/0/items/diemechanischewr00clau/diemechanischewr00clau.pdf
page 315 of the text, page 335 of the pdf for the adobe acrobat file

Really, BP, you are hopeless.

Betty Pound said...

'
"warm body radiates heat"
"cold body radiates"

This is, of course amusing on several grounds.'

Yeah because it debunks you.

Eli is a complete jerk. He provides no Page Number, but expects me to find relevant passage from a ~1000 page book.

"warm body radiates heat"
"cold body radiates"

and

"double heat exchange"

is a CONTRADICTION.

Eli can't find a modern textbook so he cherrypicks an ancient confusing doublespeak to make his case.

This is what passes as scholarship for Eli.

Tell me what page it's on!

Betty Pound said...

Good. I'll look up your ancient source of wisdom.

Meanwhile you respond to my response to your previous comment:

Thanks for falling into my trap.

"The blue plate will receive 400 W/m2 from the sun and 133 W/m2 from the green plate or 533 W/m2 in total"

The green plate is not an energy source. You recycled 50% of blue's emission back to blue. The 133 is not new raw energy that you can add.

If the blue sends 267 to green and receives 133 from green, then you really sent 267-133=133 of thermal/heat energy to green.

Thus you have a situation where blue emits 267 to space and 133 to green. This violates:

"No, that means that every square meter of the surface of the blue plate will emit 267 W/M2"

You see? Your own blue plate is not at thermal equilibrium.

Further you made the mistake of assuming heat has a preference for going to colder object (space). This is a fallacy.

The blue emitting 267 (heat) to space and 133 (heat) to green, will in no time (via conduction) turn to 200 to space, 200 to green.

There is no reason for 67 W/m2 of thernal energy not to flow to 133 W/m2 side.

"The emission to space will be 267 (BP)+ 133 (GP) W/m2 = 400 W/m2 EXACTLY WHAT IS COMING IN FROM THE SUN"

Uhuh, there you go again foolishly believing that 267 can only heat space, and not 133.

Eli failed physics. He simply spouts rhetoric to make his phony model work.

Chris A. said...

Energy emitted to space cannot be characterized as heat, because you can heat up only matter, not space. Temperature is a bulk property of atoms or molecules.

Blue emits 267 each direction, of that 133 directed to green are heat transfer, the rest is not.

And yes, the entire system never equilibrates, because it is not adiabatic, not even quasi-adiabatic and can therefore not equilibrate thermally.

Systems with a heat source do never equilibrate thermally, because that would violate the 1st law as also your calculations do show when assuming blue and green to be both at 244K.

As a native german speaker I assure you that Eli's translation is right.
But note that Clausius did not distinguish between "heat" and "energy" as we do today! He used the term "heatflow" for every direction of energy exchange.

EliRabett said...

BP: But not every photon raises blackbody temperature. Only photons of a higher frequency that the BB lacks will heat the BB.

Microscopically, the lower frequency photon is destroyed.


Eli strongly suspects that BP who popped into existence when this post was posted is strongly invested in conservation of energy denial. Photons carry energy. If they are destroyed by being absorbed they transfer the energy to the body that they hit.

The link that BP cites refers to a photon being absorbed or emitted by a single atom. First, a black body does not have a frequency.

Let's stop there cause it's BP all the way down.

Betty Pound said...

Checks out. Eli managed to find ONE EXAMPLE of confusing doublespeak in his ancient scrolls of wisdom.

He doesn't realize that one man's text doesn't define physics. Why can't he use something without doublespeak or something modern?

https://en.m.wikipedia.org/wiki/Heat

"Heat is the amount of energy that flows spontaneously from a warmer object to a cooler one."

Cold can't heat hot by definition. So there is no double heat exchange, only double energy exchange.

Clausius' text is outdated.
I hope Eli doesn't believe in the phlogiston theory.

EliRabett said...

Eli will disagree somewhat with Christian. One can under some circumstances, and they include this problem, describe a photon gas or an electromagnetic field that has both a pressure and a temperature and an equation of state.

EliRabett said...


BP, if the one guy is Clausius it pretty defines the physics of the situation. Given the choice between Wikipedia and Clausius, 97 out of 100 physicists would choose Clausius. Here is another example.

https://books.google.pl/books?id=WhrezCOLYmwC&pg=PA724&lpg=PA724&dq=exchange+of+thermal+radiation+hotter+colder&source=bl&ots=d3AYNKCs05&sig=qvXakV_zEW9Unjz75eJhlCdbn9k&hl=en&sa=X&ved=0ahUKEwilj_msj4XXAhWMB5oKHdU9De8Q6AEIaTAL#v=onepage&q=exchange%20of%20thermal%20radiation%20hotter%20colder&f=false

Betty Pound said...

LOL! your own source engages in ridiculous doublespeak.

"the hotter body radiates to the colder body and the colder body radiates heat to the hotter body"

What? LOL, that's exactly the opposite of reality.

"the net outcome ... colder body would be heated"

LOL. What a contradiction from te previous statement of cold radiating heat, while hot merely radiating.

Hahahaa, Eli relies on doublespeak cranks.

Eli thinks wikipedia made up their own definition.

"Heat is the amount of energy that flows spontaneously from a warmer object to a cooler one. More generally, heat arises from many microscopic-scale changes to the objects, and can be defined as the amount of transferred energy excluding both macroscopic work and transfer of part of the object itself.[1][2][3][4][5][6]

^ Born, M. (1949), p. 31.
^ Pippard, A.B. (1957/1966), p. 16.
^ Landau, L., Lifshitz, E.M. (1958/1969), p. 43
^ Callen, H.B. (1960/1985), pp. 18–19.
^ Reif, F. (1965), pp. 67, 73.
^ Bailyn, M. (1994), p. 82.

"Transfers of energy as heat between two bodies. Referring to conduction, Partington writes: "If a hot body is brought in conducting contact with a cold body, the temperature of the hot body falls and that of the cold body rises, and it is said that a quantity of heat has passed from the hot body to the cold body."

Pretty straightforward stuff, therefore Eli needs to cite confused sources.

Betty Pound said...

The body has higher energy photons. What happens when a lower energy photons strikes an electron at a higher level?

According to Eli, the photon tells the electron what?

a) go to a higher level?
impossible

b) go to that lower level?
impossible

Eli refused to do microscopic analysis and sticks to vague macro "absorption".

Betty Pound said...

"First, a black body does not have a frequency."

What a retard. A blackbody has a whole ensemble of frequencies and intensities of frequencies. That's what a BB curve shows.

Eli is attempting a strawman here. Very pathetic.

Betty Pound said...

"
BP: "He turned the sun's 400 W/m2 to 2*267 =533 W/m2 using his own logic of adding flux densities."

No, Eli calculated that when the green plate is present the blue plate will be at 262 K and emitting 267 W/m2 from each side for a total of 533 W/m2"

Look at this guy. He says "No", but then repeats the same thing I said.

He violates conservation of energy.

Notice he's avoiding the rest of that discussion.

Bob Loblaw said...

Eli:

Is there someone you are working with on this that is doing a psychology thesis? The extent of BP's denial of physics is truly astounding. Whatever she is paying for therapy and pharmaceuticals, it isn't enough.

Betty Pound said...

Eli, please respond to my previous reply to your reply
-----

Thanks for falling into my trap.

"The blue plate will receive 400 W/m2 from the sun and 133 W/m2 from the green plate or 533 W/m2 in total"

The green plate is not an energy source. You recycled 50% of blue's emission back to blue. The 133 is not new raw energy that you can add.

If the blue sends 267 to green and receives 133 from green, then you really sent 267-133=133 of thermal/heat energy to green.

Thus you have a situation where blue emits 267 heat to space and 133 heat to green. This violates:

"No, that means that every square meter of the surface of the blue plate will emit 267 W/M2"

You see? Your own blue plate is not at thermal equilibrium, and it's also not uniformly 262K.

Further you made the mistake of assuming heat has a preference for going to colder object (space). This is a fallacy.

The blue emitting 267 (heat) to space and 133 (heat) to green, will in no time (via conduction) turn to 200 to space, 200 to green.

There is no reason for 67 W/m2 of thermal energy not to flow to 133 W/m2 side.

"The emission to space will be 267 (BP)+ 133 (GP) W/m2 = 400 W/m2 EXACTLY WHAT IS COMING IN FROM THE SUN"

Uhuh, there you go again foolishly believing that 267 can only heat space, and not 133.

Eli failed physics. He simply spouts rhetoric to make his phony model work.

Quokka said...

Betty:

"And yet you claimed the blue emits 267 heat to space and 267-133=133 to the green plate. An object can't radiate 267 and 133 of heat from different sides, and be at the same temperature."

No, I have always claimed, as has Eli, that the blue plate is radiating 267 W/m2 on both sides. It is also absorbing 400 from the space side, so has a net energy flow (heat) of 400-267 = 133 inwards on that side. And also absorbing 133 from the green plate side, so a net energy flow (heat) of 267-133 = 133 outwards on that side. It is truly amazing that after all this time, you don't even understand the model you are supposedly critiquing.

"The lower energy photons will just destructively interfere with the same lower frequency photons emitted by that blackbody.
That's why you can't add 133 W/m^2 to an object emitting 267 W/m^2."

Got a reference for that? (No, of course not.) This is wrong, and this can be demonstrated with a test case that is even simpler than Eli's example.

Imagine a spherical blackbody floating in 0K space, with 1 m^2 surface area, at a temperature of 262K. It is therefore emitting 267W, and cooling down. Now imagine the same sphere in space where there is background radiation of 220K (equivalent to 133 W/m^2). This sphere will also cool down, but not as quickly as the other one. This matches our everyday experience; we know that objects cool down when you put them in the fridge, cool down even faster when you put them in the freezer, and faster still if you drop them into liquid nitrogen. But both spheres are emitting 267W (per S-B law). The only way the 2nd sphere can be cooling down slower is if it absorbing (and converting into thermal energy) the 133W radiation it is bathed in. Using the definition of heat as net energy flow, the 1st sphere is losing heat at 267W, and the 2nd sphere is losing heat at 133W. But for this to happen, the sphere *must* absorb the background radiation as thermal energy.

Betty Pound said...

Quokka keeps claiming that objects prefer to heat space rather than ANYthing colder. He has blue heating space 267, and blue heating green 133. In his fantasy land, 67 of 267 will not go to the 133 side. What an odd thing.

"Imagine a spherical blackbody floating in 0K space, with 1 m^2 surface area, at a temperature of 262K. It is therefore emitting 267W, and cooling down. Now imagine the same sphere in space where there is background radiation of 220K (equivalent to 133 W/m^2). This sphere will also cool down, but not as quickly as the other one."

Wrong. First case, sphere needs to lose 262K.

Your second "ambient 220K environment" case, the sphere needs to lose 42K max. Not faster.

Of course such a 220K ambient environment must have molecules, not space, so the 262K sphere will heat the 220K until thermal equilibrium is reached.

Betty Pound said...

* "Not faster" =

42K to lose is faster than 262K to lose. The very opposite of what you claimed.

Betty Pound said...
This comment has been removed by a blog administrator.
Betty Pound said...
This comment has been removed by a blog administrator.
Betty Pound said...

except 0K.

Quokka said...

Just to clarify, in my previous example, I was referring only to the instantaneous rate of heat loss at the moment at which the spheres had a temperature of 262K (not time taken to reach equilibrium). And that is, obviously, faster in a colder environment. Since both spheres are radiating the same amount of energy, the sphere in the warmer ambient environment must be absorbing energy from that environment to explain the lower rate of heat loss.

Betty Pound said...

Quokka,
A 262K object instantly brought to space will cool at exactly the same rate as two objects at 262K and 220K brought instantly into space.

Heat flows spontaneously from hot to cold.

Heat will flow from 262K to 220K (and space) to space.

EliRabett said...


BP you really need some course work. Not every transition driven by absorption or emission of a photon involves changing the quantum state of an atom. There are, for example vibrational, rotational, phonon etc transitions. When you try and impose your model of only electronic transitions on every problem you simply show that you have no idea of what is happening.

Also trying to describe the Green Plate Effect as involving only the Green and Blue plate continues your tradition of lying by omission. The Green Plate Effect involves the 400 W/m2 heat source which for convenience Eli called the Sun, the Green and Blue Plates and the heat bath that the Green and Blue Plates radiate to which again, for convenience, Eli called space.

EliRabett said...


BP is on strict probation. Posts containing any denigration of others will be deleted.

Betty Pound said...

"Not every transition driven by absorption or emission of a photon involves changing the quantum state of an atom."

Exactly. Thanks for proving 133 does not heat 267.

Those photons are not "absorbed" into molecules, they are just destroyed.

"There are, for example vibrational, rotational, phonon etc transitions."

How do they apply in your example? Does that make 133 heat 267?

You heard it here folks, Eli claims the sun prevents 267 W/m2 from heating 133 W/m2. Why? Magic.

Sorry folks, equilibrium is simply not allowed in any situation when the sun is out. LOL.

Eli, you will keep evading my arguments, right? You simply have no reasonable responses, so you throw meaningless distractions.

Betty Pound said...

Quokka,
sorry Eli deleted my long posts to you.

Eli,
Don't get mad. It blows your cover.

Anonymous said...

"Those photons are not "absorbed" into molecules, they are just destroyed."

Ehm, in what alternative universe do you live? Photons that just disappear into nothingness?

Quokka said...

"Quokka keeps claiming that objects prefer to heat space rather than ANYthing colder. He has blue heating space 267, and blue heating green 133. In his fantasy land, 67 of 267 will not go to the 133 side. What an odd thing."

Utterly nonsensical. I've never said anything about preferences of objects. What I've said is that objects radiate according to the S-B law, therefore the blue plate will radiate 267W/m2 to both sides.

"A 262K object instantly brought to space will cool at exactly the same rate as two objects at 262K and 220K brought instantly into space."

I can't even tell what this two objects scenario is. Are you claiming that a 262K object and a 220K object will cool at the same rate? Because that's obviously wrong; the 262K object will radiate more energy and hence cool down faster. But this scenario (2 objects at different temps) is not the one I was discussing (2 objects at the same temp in different ambient environments). At that instant in time, the object in the warmer environment is cooling more slowly even though they both radiate the same. Nothing you've said contradicts or even addresses this point.

Chris A. said...
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Chris A. said...

Betty,

I will give it a try to show you a real world experiment, for maybe you will believe experimental proof where you don't believe the textbook knowledge I threw at you.

It's about this statement of mine: "Photons simply cannot interfere destructively, like acoustic waves can do."

With interference of photons of given wavelength you can create locations, where the probability to find a photon becomes zero. That does not mean literally the destruction of any photon by interference is possible. It will simply be that two "destructively" interfering photons will have zero probability to be located at a certain location at a point of time, but simultaneously will both be at a different location at the same point of time ( interfering constructively ).

You don't believe me? Look at this real world experiment: https://www.youtube.com/watch?v=RRi4dv9KgCg

That has to be, because the conservation of energy would otherwise be violated. Energies can never be combined to any amount smaller than their sum, therefore no two energies can "destruct" each other literally. Since photons can be viewed as "carriers of energy", they are therefore not allowed to destruct each other. The conservation law forbids this.

-----

To Quokkas example: That one I found not very useful ( sorry @Quokka ) to teach Betty, because here special measures of simplicity are necessary.

Better it should be like this in my opinion:

Betty, please answer the following request about the time two systems need to equilibrate thermally.

1) A 1 m^2 plate of a black body ( lets say 1mm of thickness ) is heated up to an equilibrium temperature of 244K. Then it is set out into 0K space. It will equilibrate to 0K over a time t_1.

2) An identical plate is heated to 244K, too. After at least t_1 has elapsed, it is set out in 0K space in proximity to the plate from 1). The plates are perfectly parallel ordered. Lets assume them to be 1cm away from each other. The second plate will equilibrate to 0K over a time t_2.

Question: Is t_1 = t_2 Yes/No ???

Bernard J. said...

"Eli is a complete jerk. He provides no Page Number, but expects me to find relevant passage from a ~1000 page book."

It amuses me that someone who pretends to overthrow physics can't even muster the nous to do Ctrl-C/Ctrl-F/Ctrl-V/Enter...

The other thing that amuses me is that if one stands on a corner in the Interweb for long enough someone will come along with the same thoughts as oneself. Bob Loblaw (22/10/17 5:57 PM ) raises something that I was idly wondering, and I even pondered a title - "Flatlander psychology in physics denial: motivated reasoning and the constitutional inability to understand objectively-demonstrable logic in physics and mathematics."

The last thing that amuses me in this ongoing train-wreck of 'greenhouse' gas denial is BP's response to Quokka's lovely example - apparently black bodies surrounded by background thermal radiation cool down more slowly not by transiently absorbing some of the radiation from the background, but by magically radiating each photon more slowly. There are entire new fields possible if this is true...

One cannot help but wonder what neurones are FUBARed in BP's brain that render him incapable of grokking the subject. Is it as simple as not having an operant definition of 'net' vs 'gross'? Is it a mortal fear of acknowledging that humans are indeed warming to planet to a harmful level? Is it something else? Absent the minefield of the ethics complexities involved, this area would be a rich vein of research for the curious psychologist in which to dive with gay abandon...

Betty Pound said...

You have no idea what you're talking about, you're just distracting from the fact that an electron can not emit 267 W/m2 at the same time as it absorbs 133 W/m2.

Stop evading and distracting.

Chris A. said...

Betty,

no, you are distracting.

Your argument is without meaning. Even if you were right, it would be irrelevant, for we are not energizing a single electron.

There is experimental evidence, the most accepted evidence in pysics, that photons are not destroyed by any kind of interference, not even by the most destructive there is.

So they do reach the surface and there they can only be reflected or absorbed.

Do something productive and answer the two questions:

Does a 244K warm plate in 0K space equilibrate within the same time when alone compared to when a second identical plate of 0K is present?

Will the two be heated to the same temperature when brought near a star, one plate in the shadow of the other?

Two yes or no questions which should be answered in seconds by a physics expert.

Don't distract, answer!

Betty Pound said...

So, your plate does not have molecules with electrons? Tell me more.

Betty Pound said...

Eli's diagram shows blue heating space 267, and blue heating green 266-133=133. That's unbalanced.

You actually believe that a hotter object cools faster than a colder object?

"the 262K object will radiate more energy and hence cool down faster."

That's hilarious. What law of physics is that?

EliRabett said...

BP: So your plate does not have molecules with electrons?

It has nuclei consisting of protons and neutrons as well as electrons. Solids have quantum states related collective excitations.

You could look up band structure.

Chris A. said...

Betty,

let us do quid pro quo: I answer your question, you answer mine.

Yes, the plates I have in mind consist of molecules that have electrons.

Now you! Two simple yes/no questions in my last comment.

Betty Pound said...

That's not the full question/comment you must answer.

Betty Pound said...

You keep evading what I asked you to address.

Please send me a link explaining how protons and neutrons process incoming electromagnetic radiation (photons). Show me how it responds to spectra.

Timothy Folkerts said...

Quokka: the 262K object will radiate more energy and hence cool down faster.

Betty: That's hilarious. What law of physics is that?

Ummm ... that would be the Stefan-Boltzmann law that is the basis of this whole discussion! The power loss is proportional to T^4, so an object that is twice as hot loses power 16x faster and hence cools 16x faster (assuming the heat capacity is constant, which it should approximately be).

Or in this case, a 262K object will radiate 2x as much as a 220 K object, and hence cool at twice the rate. In the time it takes a 220 K object to cool 1 K, a 262 K object will cool by 2 K.

Timothy Folkerts said...

"Please send me a link explaining how protons and neutrons process incoming electromagnetic radiation (photons)."

This is BASIC molecular physics! Google "vibrational modes" to find a plethera of links that explain how the vibrations of atoms in molecules can absorb and create IR photons. Not individual electrons -- vibrations of the atoms (which you will recall includes protons and neutrons).

Betty Pound said...

And can these particles vibrate at two different frequencies at the same place at the same time?

No? Then you still need to average.

Everyone distracts from my main point.

Betty Pound said...

Since they cool at the same rate, the 220K will reach 0K first.

Even a child can figure out that 262 has to go through 261,260,...,220. But not these clowns.

jgnfld said...

Most children know that rate is an instantaneous quantity. As such, the rate of cooling for the 262K object is going to be much faster than the rate of cooling for the 220K object when both are placed into space at the same time as was specified.

They must separated by a great enough distance they do not affect each other, of course. If that were the case, as most of us know, it would change things a bit as the the original 262K object when it reached 220K would not be in the same radiative environment as the original 220K object at the time of its placement.

Quokka said...

Quokka: "the 262K object will radiate more energy and hence cool down faster [than a 220K sphere]."

Betty: "That's hilarious. What law of physics is that?"

Stefan-Boltzmann, and conservation of energy. The 262K sphere will emit 267W. The 220K sphere will emit 133W. After 1 second, 262K will have lost 267 joules, 220K will have lost 133 joules. Because that energy comes from the thermal energy in the objects, 262K will have cooled down more. Which of those figures do you disagree with? What do you think will have happened after 1 second? Show your work.

Betty Pound said...

People are actually arguing that 262K cools faster to 0K than 220K. Wow!

Betty Pound said...

No, because the 262K object has to go through the same rate of cooling as the 220K object, when it is at 220K.

Anonymous said...

"And can these particles vibrate at two different frequencies at the same place at the same time?"

IN principle, yes. Vibrations are, of course, a matter of relative motion of atoms in comparison to other nearby atoms. As soon as you have three atoms interacting, you can have at least three fundamental vibrations, each with their own frequency.

To take a simple example: a linear CO2 molecule can have a bending vibration (the molecule becomes non-linear) occurring at the same time as a stretching vibration (bond lengths changing).

Still, even if this were NOT the case, it would be irrelevant for the examples here, as we're dealing with many trillions of atoms, which are not all in the same state at the same time.

I also noticed you still haven't explained what happened to that photon that is just "destroyed".

Betty Pound said...

Great: 1 second. You claimed 262K cools faster to zero than 220K.

Betty Pound said...

Can you read?

"same place at the same time"

Do any of your examples meet that critetia.

I'm still waiting for Eli to explain how a plate absorbs/emits 267 W/m2 and absorbs/emits 133 W/m2 , without this creating an average equilibrium at 200 W/m2. Eli wants to pretend it's still at 267 W/m2.

Quokka said...

Betty: "People are actually arguing that 262K cools faster to 0K than 220K. Wow!"

That is a lie. I specifically clarified in a follow-up message (22/10/17 9:47 PM) that I was talking about the instantaneous rate of cooling, *not* the time to cool to ambient. (I think it was fairly obvious in the context of my original post that I was referring to "rate of cooling", but added the clarification just in case any readers were stupid enough to not comprehend that)

Now that you know that rate of cooling was being referred, please address the example with that understanding.

Betty Pound said...

"I was talking about the instantaneous rate of cooling, *not* the time to cool to ambient."

That makes no sense in the context of your example.

Quokka: "Imagine a spherical blackbody floating in 0K space, with 1 m^2 surface area, at a temperature of 262K. It is therefore emitting 267W, and cooling down. Now imagine the same sphere in space where there is background radiation of 220K (equivalent to 133 W/m^2). This sphere will also cool down, but not as quickly as the other one."

a) 262K to 0K
b) 262K to 220K

You are saying that scenario (a) is completed faster than (b).

I laughed at this and still do.

Did I communicate your example correctly?

Assuming yes, your (b) case becomes a heat flow equation, and is equivalent to cooling from 42K to 0K, which is something that would happen later on as a small part of the (a) case.

Which is why it's funny to think (a) happens before (b).

Betty Pound said...

c) 262K and 220K to 0K

is equivalent to

d) 244K to 0K

Anonymous said...

"Do any of your examples meet that critetia."

Yes. But it still doesn't matter in this context. We're not talking about a single particle absorbing/emitting, but a whole plate of 1 m^2. Even at one atom thickness, we're talking about billions and billions and billions of atoms.

Also, still no answer where that photon went? Just "pooff", disappeared?

Betty Pound said...

Quokka, you see, space can't be heated by 262K plate, because space has no particles to heat, so the photons keep traveling to infinity in an empty universe.

However, the 262K plate can readily heat the 220K plate, and it does, even if they are both cooling.

Eli's GPT is bunk. Equilibrium is reached at 244K.

Anonymous said...

"Which is why it's funny to think (a) happens before (b)"

But this is not what Quokka is thinking, so you are laughing about something you made up yourself.

What Quokka is comparing is a scenario where you start with 267 dollar and give away a certain amount on day one, and a second scenario where you give away the same amount on day one, *but also get some money back*. The amount of money you lost after the first day is less in scenario 2 than in scenario 1.

The problem is that you deny scenario 2 can take place, ased on some weird sentience of photons that if they go to a place that is warmer than where they came from, they, "pooff", disappear into thin air.

Betty Pound said...

the "weird sentience" of photons is their frequency. colder objects lack higher frequency photons that warmer objects have. So cold won't heat warm AND cold will not prevent warm from heating it (sharing its higher frequency photons).

The 220K does not slow down 262K's cooling (or its own heating from 262K).

The two will cool as if 244K alone was used.

Chris A. said...

Betty,

since I found only that one question adressed at me

"So, your plate does not have molecules with electrons? Tell me more."

I felt to have it answered. Yet you are not satisfied.

So, friendly as I am, I try to anticipate what questions you could have in mind additionally and answer those:

- The plates I have in mind are made of molecules that have electrons.

- No electron can have more than one state at the same time.

- No molecule can have more than one specific vibrational mode at the same time.

Does this satisfy your requests?

-----

So now, quid pro quo Betty, quid pro quo!

1) Does a 244K warm plate in 0K space equilibrate within the same time when alone compared to when a second identical plate of 0K is present?
( The two are ordered parallel to each other and in proximity to each other ).

2) Will the two be heated to the same temperature when brought near a star, one plate in the shadow of the other?

-----

And you could be so nice to answer Quokkas question, which you tried so hard to misundertand. I try to formulate it anew, so you do understand maybe:

Is the energy loss of cooling by radiation faster at higher temperatures than at lower ones?
That means the higher the temperature, the more Joules per second ( J/s ) are radiatet away. And that means further, deduced logically out of that plus the fact of a nearly constant heat capacity, that cooling of 10K of temperature needs different times depending on the actual temperature.
262K to 252K cools in less time than 244K to 234K. Both cool in much less time than 10K to 0K.
That is what Quokka claimed and he asked if you agree with that.

3) Do you agree, or don't you?

Here are 3 questions now adressed at you. Don't twist them, don't distract, simply answer!

And for this is quid pro quo for me, feel free to ask 3 questions back.

Anonymous said...

"the "weird sentience" of photons is their frequency. colder objects lack higher frequency photons that warmer objects have. So cold won't heat warm AND cold will not prevent warm from heating it (sharing its higher frequency photons)."

What do those "low frequency photons" do, when they reach the warmer object, Betty? Just, "pooff", disappear into nothingness?

I asked you a few times already, but you refuse to answer this question.

jgnfld said...

The radiative cooling function is rather complex, but crucially in the case of cooling to 0K the time integral depends on 1/T^3.

Sadly, (1/262^3 + 1/220^3) ≠ 2*1/244^3.

You are wrong.

Timothy Folkerts said...

Betty says: "People are actually arguing that 262K cools faster to 0K than 220K. Wow."

Betty is actually arguing that 'cools to 0K' is a thing. Wow.

Objects will cool asymptotically TOWARD zero (or whatever the ambient temperature happens to be). This means that any calculation of time to cool "to 0K" would be infinity, and all objects take infinitely long.

In cases like this, the standard approach is to calculate the time to reach some fraction of the original value (often 1/2 or 1/e of the original). This is known as the "time constant" for the situation.

And the time constant for the hotter system is shorter than for the cooler system.

Quokka said...

"You are saying that scenario (a) is completed faster than (b). I laughed at this and still do. Did I communicate your example correctly?"

No. I have already told you, twice, including in the message that was in response to, that I am referring to the rate of change of temperature when I talked about how quickly objects cool down. So that would be another lie. You may have misunderstood it to mean the total elapsed time until it cools to ambient, but I don't think that is the most natural interpretation, as evidenced by the fact that no-one else here seemed to have any problem understanding what I meant.

EliRabett said...


BP:And can these particles vibrate at two different frequencies at the same place at the same time

Yes, the transitions are called combination and overtone bands. That's just for single molecules.

Anybunny familiar with musical instruments should know this. No instrument produces a pure tone but there are always harmonics that provide richness to the sound and are really combinations of vibrational tones.

These are seen for CO2 and H2O, for example in the near IR part of the solar spectrum as absorptions.

One of the problems of dealing with the Pounds is that they don't know very much about the subject they are pontificating on but think they do and telling them about it has no effect while it often simply confuses readers especially those who also lack the background because the Pounds are so certain about their confusion.

The entire digression about electrons is a good example of this

EliRabett said...


For Christian (and BP)

CO2, for example has three vibrational modes. The vibrational state of the molecule is labelled by the number of quanta in each mode (n1, n2, n3) each of which can be any whole number. The energy of that state is E = hv1n1 + hv2n2 + hv3n3 + anharmonic terms. The higher up you go the larger the anharmonicity.

https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Vibrational_Modes/Combination_Bands%2C_Overtones_and_Fermi_Resonances

and links therein

The situation in a solid "involves motion that is a superposition of vibrations of atoms around their equilibrium sites due to the interaction with neighbor atoms. A collective vibration of atoms in the crystal forms a wave of allowed wavelength and amplitude."

See http://www.chembio.uoguelph.ca/educmat/chm729/Phonons/intro.htm
and links for a start.

EliRabett said...

Since this is getting confused. The rate at which a plate emits energy

k = σ T^4 J s^-1 m^-2 so that means

a hotter body will lose energy (cool) at a faster rate than a colder one.

BP is confusing this with the TIME is will take for isolated plates at different temperatures to cool to 0K.

Chris A. said...

Eli,

thank you for your last two posts. But to me it seems as if it seemed to you I would not know this.

The problem is, I try to lure out the false concept of BP step by step, which includes asking and answering questions that might make no sense for a bystander who already knows the end of the road.

BP cannot see that there is any road at all ( or maybe she can, for she avoids certain questions ).

Timothy Folkerts said...

Betty says: "colder objects lack higher frequency photons"
Just for kicks, tell us what you think the highest frequency would be for photons from a 220K object. Which specific frequencies do you think are lacking?

Timothy Folkerts said...

As a side note, readers may be interested in the "Dunning-Kruger Effect". It seems especially apropos here.

"In the field of psychology, the Dunning–Kruger effect is a cognitive bias wherein people of low ability suffer from illusory superiority, mistakenly assessing their cognitive ability as greater than it is. The cognitive bias of illusory superiority derives from the metacognitive inability of low-ability persons to recognize their own ineptitude."
https://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect

Betty Pound said...

So you admit that 262K and 220K take infinity to cool to zero. Ergo, they have the same rate of cooling. The fact that 262K starts off cooling faster is negligible to infinity.

Betty Pound said...

Well whatever you think you meant is an irrelevant metric to the problem you posed.

Betty Pound said...

No, Eli, waves can merge but you can't have two different frequencies in the same place and time. Plot such a thing on a graph. You violate the function.

Betty Pound said...

Yes, Mr. Obvious.
Now tell us the harmonics of molecules sending/receiving 267 and 133 W/m2 at the same time.

Timothy Folkerts said...

Betty, the fundamental modes for CO2 (and many other molecules) are listed here (along with pretty animated diagrams). http://www.chemtube3d.com/vibrationsCO2.htm

Each frequency can be easily calculated from the wave numbers listed, along with the corresponding energy & wavelength. All integer multiples of these energies are also possible (the "harmonics").

CO2 at any temperature can vibrate in the various modes at various amplitudes simultaneously. (Higher temperature molecules are simple more likely to be in the higher energy modes.)

Timothy Folkerts said...

"If you don't know what frequencies a 220K object lacks compared to 262K, then you are unfit to judge your betters."

Oh Betty! Please enlighten us! Spread some of your god-like knowledge to us so that we too may know the truth!

Chris A. said...

"The blue emitting 267 (heat) to space and 133 (heat) to green, will in no time (via conduction) turn to 200 to space, 200 to green.

There is no reason for 67 W/m2 of thermal energy not to flow to 133 W/m2 side."

That is so funny, for you are the one who insists on the strict distinction between the terms "heat" and "energy". But you don't do it yourself.

From plate to space: Always energy, never heat. What is in space to be heated? Nothing, it's space.

From plate b to plate g ( and vice versa ): Only heat, if no energy flux of the same or higher density comes from opposite direction. But always energy.

Therefore, heat is always energy, but energy not always heat, according to the distinction you insisted on.

To know which part of which energy flux is heat, you must book keep the energy fluxes.

267 from blue to space is not heat, but energy. 133 from green to space is not heat, but energy. 267 from blue to green is energy, but only 133 of that is heat. 133 from green to blue is not heat, but energy.

If both plates emit energy at 200 on each surface, there is no heat at all. The very problem: Blue than receives 600 energy ( 400 sun, 200 green ), green receives 200 ( from blue ).
To emit 200 on each Surface, 200 at blue are destroyed and 200 at green are created.

Impossible.

Anonymous said...

I am always curious to know what motivates someone like Betty Pound Sand. I mean it is clear to anyone who has ever taken a course in thermodyanmics, stat mech or quantum mechanics that she is utterly full of fetid dingo's kidneys. It should be clear to Betty by now that many of the folks with whom he/she/it is arguing now DO know what they are talking about. And yet, Betty persists in pretending everyone else is an idiot and only she really understands physics.

It reminds me of the joke about the man who answers a call from his wife while driving:
Wife: "Oh, George. Be careful. I just heard on the news that there's some idiot driving the wrong way on the freeway."
George: "One! There's thousands of them!

So, is Betty:
A) Dumber than owl pellets
B) A bot (Da, Comrade)
C) A gifted comedian involved in a piece of performance art.

Chris A. said...

"Christian,
By your own logic, Eli's diagram would make you state that
Blue receives 267*2+133=666 energy."

Definetely not. How do you come to this idea?

For Eli's diagram with "my logic" blue receives 533 ( 400 from sun, 133 from green ). This adds up to 267*2.
Why have you added 133 a second time out of nowhere? ( your statement: 267*2+133=666 )
Correct is (400+133)/2 = 267, which is emitted by each surface of blue.

The easiest way to see what works and what not, is looking to it from the most basic laws of physics. If we assume equilibrium with the incoming energy of the sun, then it has to be:

a) Energy IN = Energy OUT for the entire system
b) Energy IN = Energy OUT for every single plate

A state satisfying a) AND b) ( not only one of them ) would be equilibrated to the sun's input, correct?

So, according to this logic Eli's scenario satisfies a) AND b) and is therefore describing a state equilibrated to the sun's input without violating any laws of physics.

Your scenario on the contrary ( both plates at 244K ) satisfies a), but not b) and is therefore either a transient state, or simply impossible.
Since the hypothetical plates have zero thickness, everything happens instantaneously, so there are no transient states at all.

You have 600 IN <-> 400 OUT for blue, and 200 IN <-> 400 OUT for green. Impossible, for this violates 1st law. And you can find these values by simple book keeping of the quantities, no calculation at all is needed.

It's amazing how you dance around this most simple fact, as if it is produced by magic, while you talk physics.

Explain how your scenario, although the plates are energetically unbalanced, can be equilibrium! Find the missing of balance by such a simple thing as book keeping energy quantities.

Don't dare to distract with any distinction between heat and energy, for this is meaningless for simply counting the total values of the energy fluxes.

Quokka said...

Betty: "The correct corollary of the bullshit you performed for my example is: 267*2+133=666. Adding two outputs and an input."

Enquiring minds want to know: where/when was this, exactly, so we can evaluate this claim. Give a quote, or the timestamp of the comment.

"Enjoy your cancer."

Class act, as always.

Chris A. said...

"The correct corollary of the bullshit you performed for my example is:
267*2+133=666
Adding two outputs and an input."

Can you show the sentence or section of my post, from which you deduce that absurdity?
I can only guess you mean this, where I have described the energy fluxes according to your distinction between heat and energy:

"267 from blue to space is not heat, but energy. 133 from green to space is not heat, but energy. 267 from blue to green is energy, but only 133 of that is heat. 133 from green to blue is not heat, but energy."

Of this you seem to read the last sentence, as if I was claiming 267 energy for each surface and additionally 133 heat for the right surface would be emitted by blue, for you state: 267*2+133=666
I claimed nowhere that heat is something that comes additionally to energy. On the contrary I claimed, literally: "To know which part of which energy flux is heat, you must book keep the energy fluxes."

[ emphasize added on the word "part", you know why? ]

Thus I claimed that heat flows are part of the energy fluxes, which is perfectly true if one distincts between energy and heat like you insisted on yourself.
That means further, that the 133 heat are part of the 267 energy that is emitted from the right surface of blue. "A part of", not "added to".
I do claim explicitely, that an equilibrium state with the incoming energy of the sun has by most basic physics to satisfy a) AND b):

a) Energy IN = Energy OUT for the entire system
b) Energy IN = Energy OUT for every single plate

You won't deny this, will you? And by that very basic claim one can easily compare Eli's scenario and your scenario, if they satisfy a) AND b) or not.

Let's do this ( again ):

- Betty Pounds scenario: Equilibrium with evenly distributed temperature, thus each plate at 244K. This results in an emittance of 200 W/m^2 for every surface of the system.

- Eli's scenario: A state equilibratet to the sun's input, temperatures are unevenly distributet. Blue plate emits 267 W/m^2 from each surface. Green plate emits 133 W/m^2 from each surface.

---

Test for claim a):

-In Betty's scenario the system receives 400 from the sun and emits 200 left to space, 200 right to space. 2*200=400

- In Elis scenario the system receives 400 from the sun and emits 267 left to space, 133 right to space. 267+133=400

Conclusion: Both scenarios satisfy claim a). No system has in total a gain or loss of energy.

---

Test for claim b):

- In Betty's scenario, the blue plate receives 400 from the left ( by the sun ) and 200 from the right ( by green ). It emits 200 to the left and 200 to the right.
IN(400 + 200) > OUT(200 + 200)
The green plate receives 200 from the left ( by blue ) and 0 from the right ( empty space ). It emits 200 to the left and 200 to the right.
IN(200 + 0) < OUT(200 + 200)

- In Eli's scenario , the blue plate receives 400 from the left ( by the sun ) and 133 from the right ( by green ). It emits 267 to the left and 267 to the right.
IN(400 + 133) = OUT(267 + 267)
The green plate receives 267 from the left ( by blue ) and 0 from the right ( empty space ). It emits 133 to the left and 133 to the right.
IN(267 + 0) = OUT(133 + 133)

The conclusion of this very simple book keeping of energy quantities: Eli's scenario satisfies claim a) AND b), whereas Betty's scenario satisfies claim a) only. It breaks claim b), because somehow energy is lost at blue while energy is createt at green.

---

Betty, your scenario is stated to be physically impossible by a book keeping operation, that every 4th grade pupil could have performed.

It's truely amazing ( and funny by the way ), that the very reason for your failure, namely the destruction and creation of energy in your scenario, is explicitely the very error you insinuated Eli would be making.

jgnfld said...

"Why have you added 133 a second time out of nowhere? ( your statement: 267*2+133=666 )"

Betts is rather enamored with perpetual energy generation.

EliRabett said...


BP:No, Eli, waves can merge but you can't have two different frequencies in the same place and time. Plot such a thing on a graph. You violate the function.

It's called Fourier analysis

BP: Now tell us the harmonics of molecules sending/receiving 267 and 133 W/m2 at the same time.

Wins the prize for confusing a with b

A single photon has a single frequency.

A group of photons can contain multiple frequencies.

This is called a spectrum.

If the photons are emitted by a source at 262 K, the distribution of photon frequencies will be one thing and if they are emitted from a source at 220 K they will be another.

If the source of emission is a solid at 262 K then the total emission flux will be 267 W/m2.

If the source of emission is a soli at 220 K then the flux will be 133 W/m2

Betty Pound said...
This comment has been removed by a blog administrator.
Betty Pound said...

The blue plate is a laptop, and this is what Eli adds to it:

http://www.howtohacks.com/wp-content/uploads/2011/08/unlimited-laptop-power-cable.png

EliRabett said...

Eli charges his smartphone from his laptop on occasion

Chris A. said...

Did I miss anything of substance? Though deleted, on the start page I find a comment of Betty starting with "Christian, you still confuse heat and energy, you..."

To that: If you are not claiming that heat is no energy once put in by the sun, you have nothing of logical substance to say against my book keeping. But if you do, you left the realms of physics. Simple as that. Feel free to decide of what sort your failure is!

Betty Pound said...

Eli, and charging your smartphone simultaneously charges your laptop too, right? Just like the blue plate "charges" the green plate and itself.

Christian, you're out of luck. Eli is embarassed by my response to you.

Betty Pound said...

Christian, why would I repeat the several paragraphs I wrote, when Eli will just delete them for contrived self-serving reasons, instead of the actual reason: his embarassment.

Looks like you will be stuck in fantasy land forever, just like the little child you are.

Chris A. said...

Maybe the relation of offensive terms and reinvented physics was too unbalanced, if you simply book keep those?

It is a pity: I had been curious about the new creative approach you made to tell how creation and destruction of energy fit perfectly well within the frames of physics by just the right distinction between heat and energy, and - I'm nearly sure - the ongoing neglecting of this little thing called "constant input" coming from the sun.

You have invented at least two parts of physics yet unknown to all who study the subject: adiabatic enclosures containing energy sources and destruction of photons by interference. What else you might have in store to revolutionize the ancient knowledge of standard physics? We will never know, for Eli suppresses this superior knowledge out of pure shame...

jgnfld said...

Re. "Christian, why would I repeat the several paragraphs I wrote, when Eli will just delete them for contrived self-serving reasons, instead of the actual reason: his embarassment (sic)," about all that comes to mind when dealing with a perpetual energy crank is:

Dennis: Ah, now we see the violence inherent in the system!...Oh! Come and see the violence inherent in the system! Help, help, I'm being repressed!...Ooh, what a giveaway! Did you hear that? Did you hear that, eh? That's what I'm on about! Did you see him repressing me? You saw it, didn't you?

I guess you got to be "scientist" in your anarcho-syndicalist commune for a bit. But isn't your term supposed to be limited to just one week?

Do you remember I predicted you'd behave like this several hundred posts ago? Cranks never fail to act like cranks.

EliRabett said...


Hey BP cool your jets or Eli will start deleting posts where you complain about your posts that call others names being deleted. Rabett Run is not a bullshit friendly neighborhood. Don't even think of whining.

EliRabett said...

BP:Eli, and charging your smartphone simultaneously charges your laptop too, right? Just like the blue plate "charges" the green plate and itself.

Well yeah, if the laptop is plugged into the wall.

lifeisthermal said...

"They neglect the fact that heating and cooling are dynamic processes and thermodynamics is not."

I came this far, and then I stopped.

Thermo-*dynamics* is apparently not dynamic according to Eli. Speaking volumes about climate science

lifeisthermal said...

Try this:

TSI(1360.8W)=W(4g²)+Q(4*sb*256⁴)

First law, you lose.

Anonymous said...

If with g you mean gravity, then the units don't fit.

That usually indicates a major problem...

Timothy Folkerts said...

First law is ΔU = Q + W

Since TSI (radiation from the sun) is Q, not ΔU, then the comment makes no sense.

Anonymous said...

If a colder object is unable to transfer energy to a hotter object/plate/box....

Then Maxwell's Demon must be between them blocking all the colder/slower/lower frequency energy from getting into the hotter box/plate/object?
izen

Betty Pound said...

Eli thinks his laptop charges faster when he has a smartphone charging from it.

Eli can't complete his "fourier analysis" because it demolishes his stupid arguments.

Kevin O'Neill said...

BP: Do you want Fourier's analysis? I don't think so. As Spencer Weart writes in 'The Discovery of Global Warming" -

"Beginning with work by Joseph Fourier in the 1820s, scientists had understood that gases in the atmosphere might trap the heat received from the Sun. As Fourier put it, energy in the form of visible light from the Sun easily penetrates the atmosphere to reach the surface and heat it up, but heat cannot so easily escape back into space. For the air absorbs invisible heat rays (“infrared radiation”) rising from the surface. The warmed air radiates some of the energy back down to the surface, helping it stay warm. This was the effect that would later be called, by an inaccurate analogy, the "greenhouse effect." The equations and data available to 19th-century scientists were far too poor to allow an accurate calculation. Yet the physics was straightforward enough to show that a bare, airless rock at the Earth's distance from the Sun should be far colder than the Earth actually is.


There's Fourier's analysis. 200 years ago. Without calculator or computer. Funny how some people are still stuck on 'problems' that were solved 200 years ago.

Wombat said...

I really don’t know whether to congratulate those commenters who clearly do understand physics for their persistence and ongoing civility or to marvel at their naivety for thinking it is possible to explain anything to the likes of Betty. Surely once she broke out into her initial burst of obscenities it was time to declare victory and ignore her. We have better things to do with our lives.

Quokka said...

Greetings, fellow Aussie marsupial. I don't think anyone was under any illusions about the likelihood of Betty seeing reason. It was more about making it painfully obvious why he was wrong, and that he didn't have a clue what he was talking about. And Betty's rantings helped make that case.

Betty Pound said...

Kevin, please quote Fourier himself, not some crank who fabricates history. What did Fourier say AFTER the layers-of-GLASS experiment?

P.S. That's not even what fourier analysis is about, dumbass.

lifeisthermal said...

"Now lets add another plate. We'll color this plate green for greenhouse. It is heated by the first at a rate of 200 W/m2"

This means that the added plate has a temperature of 0K.

If the rate of heat transfer to the added plate is 200W/m^2 from the emitting plate at a temperature equal to 200W/m^2, the added plate must have a temperature of 0K.

Heat is the transfer of energy according to differences in emissive power.
Of course, the added plate will not have a temperature at 0K. Which means that the transfer cannot be 200W/m^2. The transfer of heat, depends on the emissive power of the added plate. The transfer depends on the potential in the differences.

Both plates at equal temperature=no transfer.

Then, why would there be a transfer from low T to high T if the added plate drops in temperature?

This would mean that decreasing the energy in one body, causes an increased transfer of energy from that body. Which is a ridiculous statement.

Adding a plate at lower temperature, causes an addition of transferred energy, ON TOP of the emission of energy according to the temperature of the hotter emitter.

Adding a colder body, only causes increased draining of heat from the source. The hot body, the heat source of the cold body, must now supply them both with energy to sustain constant emission simultaneously.
The transfer sustains emission from the cold body. They now share the energy of the heat flow. Of course, this causes the warm body to drop in temperature.

Example: heat a rock by hanging it over a fire. Then put it in a bucket of water. Does the temperature increase when adding the water? Or does it drop?

Adding a heat absorbing mass to a hot body, always causes a drop in temperature. Only heat and work can increase temperature.

lifeisthermal said...

Blogger Christian Anders said...


" explain one thing: How does clothing warm the human body in winter? The clothing is always colder than the body."

This will hurt a bit.

The answer is:

BY PREVENTING ABSORPTION OF HEAT IN THE SURROUNDINGS!!!

The opposite of the greenhouse effect, which increases absorption in the cold surroundings. Which means, gh-gases cools the planet.

https://en.wikipedia.org/wiki/Thermal_insulation

"Heat flow is an inevitable consequence of contact between objects of different temperature. Thermal insulation provides a region of insulation in which thermal conduction is reduced or thermal radiation is reflected rather than absorbed by the lower-temperature body."

lifeisthermal said...

OpenID marcoclimate said...
"If with g you mean gravity, then the units don't fit.

That usually indicates a major problem...

28/10/17 11:10 AM"

Units are N/m^2. Same as Newtons G.

Do you mean that g can only act in a point, and not on a surface?

Retarded.

Anonymous said...

You indicate that the unit on the left hand side is Watt. Watt is Nm/s. That's not the same as N/m^2.

NotBrainDeadYet said...

Not sure if any thinking persons will ever read this - why would they visit this website, but:
Numbering the pictures from above downwards as 1, 2, 3 etc., madness arises in the comment to picture 4.
In picture 4, there is 400 going into the blue plate, and 400 leaving it, and 200 going into the green plate, and 200 leaving it, thus thermal equilibrium. There is 400 going into the two plate system, and also 400 leaving it, 200 from the blue plate, and 200 from the green plate. It follows from the 2nd law of therodynamics that the green plate, warmed by the blue plate, has a lower temperature - and a lower radiation intensity, and cannot heat the blue plate.
It is this 2nd law of thermodynamics that "professor" Eli is trying to disprove, and he does so by assuming - as a foundation for his "analysis", that the law doesn't work.
And people are buying it? I'm amazed anything in the US works!

Max said...

Green Plate effect is fantastic phyisics. Let's have the two plates in close contact, and make the experiment. Obviously, when they are in contact, they reach the same temperature. Then detach them - all of the sudden the blue one will have a nice temperature boost !
Great entertainment for future generations to learn their grandfathers were so incredibly dumb.

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