## 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

Timothy Folkerts said...

"Why would half the incoming radiation decide to go back?"

That is the wrong way to think about it. *ALL* the incoming radiation gets absorbed, making the atoms and jiggle more an more; making the plate get warmer and warmer. This energy looses any identity it had as "incoming radiation" and becomes simply "thermal energy of the plate".

Now the plate has some thermal energy, allowing it to radiate. The warmer it gets, the more it radiates. Since we set the plate up to be thin and have good thermal conductivity within it, it warms up the same on either side. As it continues to warm, it radiated more and more until the same amount radiates as gets absorbed. If it absorbs 400 W (on one face) it will warm enough to emit 400 W of its own energy (ie 200 W from each of the two sides).

Timothy Folkerts said...

Betty says: "Whatever energy the plate received can not be boosted by recycling that energy off of something else to boost itself higher."

OK .. here are three idealized scenarios for you, Betty. Please calculate your answer for each. A 1 m^2 "blue plate" receives a continuous 400 W of heat via electrical power (rather than sunlight, just for simplicity). This blue plate is surrounded by either:
1) deep space @ 3 K
2) a thin, thermally conducting, blackbody (emissivity = 1) sphere of radius 1 m with no independent power supply (other than what it might get from the blue plate).
3) a thin, thermally conducting, perfectly reflecting (emissivity = 0) sphere of radius 1 m with no independent power supply (other than what it might get from the blue plate).

Show that the temperature of the "blue plate" in cases (2) and (3) is indeed "boosted" by the energy "recycled" from the unpowered spherical shell.

Extra credit, find a general answer for any emissivity between 0 < ε < 1.

Betty Pound said...

400 entered the blue plate, and 200 left immediately. You turned 200 into 267.

You can't add a large ice cube's 315 W/m^2 radiation to your body's 98.6F worth of radiation and expect it to raise your temperature, but you do anyway.

Betty Pound said...

What does clothing raise your body temperature to? 100.6F? 110.6F?

"Alternatively you demand, that the 2nd law has to be applied differently depending on the kind of heat transfer ( conduction, convection, radiation )."

Actually they do. My argument was that 2 plates is no different from 1 fat plate (assuming no vertical heat loss).

Betty Pound said...

The blue plate will heat the green plate until both are at equilibrium. Since the sun is a point source, we assume there's leakage on the front side. Warming energy can never flow back to where it came from.

Betty Pound said...

There is no back-jiggling. A less jiggling objecr will not amplify a more jiggling object. Hence Eli's radiative flows are bunk.

Betty Pound said...

Eli's argument:

1 plate) 244K
2 plates) 262K 220K

The colder plate boosted the first plate's temperature.

WRONG.

Betty Pound said...

An object emitting 267 W/m2 will heat an object emitting 133 W/m^2 until reaching equilibrium - 200 W/m^2.

Heat flow is spontaneous and unstoppable.

Betty Pound said...

I did the calculations and presented them.

Betty Pound said...

* Heating energy can never flow back to where it came from to do more heating.

John Garland said...

Ya' gotta admit, Betty's ignorance knows no bounds!

This is rhetoric.

Quote accurate rhetoric!

Betty Pound said...

That's funny coming from a guy who thinks a warmer object will not heat a colder object.

John Garland said...

"I did the calculations and presented them."

And your "calculations" have consistently assumed a perpetual energy machine.

But don't let little things like perpetual motion get in the way of a great argument! You are obviously speaking to a higher truth than simple physics!

John Garland said...

EliRabett said...

First, many thanks for those who carried the carrot, in particular John, Christian and Timothy.

As to Ross, all Eli can say is you need to redo your calculation. There are some issues which may not be obvious, such as the Planck law output is in units of Watts per meter squared per steradian, but its all there at spectralcalc and if you are using wavenumbers you have to convert everything to inverse meters.

Since the blue plate radiates to space the output per unit area has to be multiplied by 2, while when the green plate is present the emission to space is from one side of the blue and one side of the green, but it all adds up correctly.

If Eli has done it right anybunny interested in the details can look at

Calculation is in wavenumbers. Temperatures are on line 2, sums of the output on line 11.

Blue plate output when alone (BPA) is 399.74 from both sides
Blue plate output when green plate is around is 266.91 W/m2
Green plate output when blue plat is around is 133.06 W/m2
Sum of Blue + Green output (B+G) when both present is 399.97

There is a figure showing the frequency dependence of each

Note that there BPA is peaked slightly to the low frequency side of B+G as shown in the difference.

Right now this is view only.

Betty Pound said...

What perpetual motion you lying piece of human filth?

The plates equilibrate to 200 W/m2 (244K) each.

Betty Pound said...

Eli, why can't you learn buster bunny?

Warmer heats colder. The blue heats the green until they reach equilibrium at 244K - 200 W/m2.

Betty Pound said...

Betty Pound said...

How dumb does a person have to be to believe that a colder object will prevent a hotter object from cooling (to it)? Really dumb.

EliRabett said...

Ah yes, having been shown the light, Betty tries the bluster. Eli may go to the zapper.

Quokka said...

The absurdity of Betty on the green plate which receives 200 W/m2 from the blue plate to its left, radiates 200 back to the blue plate, and also radiates 200 rightwards to space is best illustrated by these two quotes:

"The 200 from blue to green, and 200 from green to blue cancel each other out ..."

"The moment green loses energy to space, the blue transfers energy to green, to maintain equilibrium."

So the 200W/m2 is expected to replace both the 200 radiated back to the blue plate, *and* the 200 radiated outwards to space. This is as nuts as putting \$1000 in your bank account, and then claiming you can cover two separate \$1000 checks drawn against that account.

Christian Anders said...

Betty Pound,

before starting with what might become a longer comment, please be so kind to answer one question: If one slows cooling of a body while not slowing the energy supply of the very same body simultaneously, this body will end up with higher steady-state temperature than before. Would you agree?

I will come back to this at the end of my comment.

Christian Anders said...

And now to the comment.

You are misinterpreting things. Then you challenge the physics of your own misinterpretation. Two examples:

1.) "Objects can't warm themselves by their own reflected sun-warmed radiation."

True. But irrelevant, for no one has ever claimed what you indicate with the phrase "themselves". It never was about objects doing anything "themselves" all the time - that is your misinterpretation.

2.) "Whatever energy the plate received can not be boosted by recycling that energy off of something else to boost itself higher."

True. But irrelevant, for no one has ever claimed what you indicate with the phrases "received" ( past tense ) and "itself". It never was about boosting anything higher that has been received in a past period of time and it never was about - to say that again - something doing anything by "itself".

Please tell me what the worth of your criticism is, when you are not criticising what the thought experiment in the article really is about?

In contrast to your interpretation the claim was, that a constant energy supply from a source can heat a plate more effectively, as soon as a second plate is brought in proximity to the first one.

Nowhere is claimed, that there is something happening by itself or that some amount of energy from an earlier period of time is boosted by some mechanism. That is only what you make of it.

One can see the errors you make in your interpretation by the above statements, which are literally what you have written. Additionally, by reading some of your earlier comments, you seem to assume the system will work to get into equilibrium over time. That's also wrong. A system that constantly receives energy on the one hand and looses constantly energy on the other hand will get into steady-state. Thus your assumption, the two plates had to become equal in temperature over time is physically wrong.

Adding this up I state three mistakes you make by interpreting the thought experiment of the two plates:

1.) You interprete that something is doing heating "by itself" and thus must be physically impossible.

- Correction: Nothing is going on by itself, everything is driven by a constant energy supply from an external source.

2.) You interprete the constant energy supply in a way that you separate it to periods of time. That is the only explanation of how you can get the idea of "Whatever energy the plate received can not be boosted [...]"

- Correction: There is no "boost" of energy the first plate has once received from the energy source. There is instead a constant supply of energy from the source, bringing up the internal energy of the plate until steady-state is reached.

3.) You assume equilibrium as the final state of the system, thus the plates finally has to become equal in temperature, but in no way this can be a higher temperature than that of the formerly warmest plate.

- Correction: The system never is in equilibrium, it is reaching steady-state. That is, because there is a constant exchange of energy with the envoirenment. In steady-state it is not against any law of physics, that one plate will be heated more effectively by the energy source, as soon as a second plate is in proximity to the first one.

Christian Anders said...

Betty,

We could have an argument about semantics, because "equilibrium" and "steady-state" are mixed up. That's true and it's not good.
But again, that is not the only problem, you produce two more by your very own misinterpretation of things.

You seem to be thinking the whole thing in a closed system, where energys can only balance "themselves" following the laws of thermodynamic. But it is not a closed system. It is an open one, with constant supply and constant loss of energy in its steady state. In here, the correct application of well known physical laws will lead to the result presented in the article.

If one slows cooling of a body while not slowing the energy supply of the very same body, this body will end up with higher steady-state temperature than before.

If you don't agree with that and even with a constant energy supply it shall be impossible for a warmer body ( supplied with the energy ) to heat further in presence of a colder body, please answer a very basic question: Why are we using clothing in winter?

Quokka said...

Betty discusses an ice cube as if the entire radiative output of the cube was being radiated through a single face:

"If an ice cube is wrapped in paper, then by your logic of adding fluxes 315*6 (sides of cube) = 1890 W/m2 or 427K - the paper burns."

How, exactly, do you think you would place one ice cube, let alone an entire hemisphere of ice cubes, so that all of them have all of their faces radiating at the same bit of surface? Because that's what would be needed to get your 1890 W/m2 of radiation. I know I probably shouldn't say that was a stupid comment. But - damn, that was a stupid comment.

You can add fluxes that are radiating *at the same surface*. That's why, in my earlier example, two ice cubes in the sky had twice the warming effect of one. This is really really obvious, so I shall assume, possibly optimistically, that you don't dispute that point.

Elsewhere you ask why the SB Law doesn't have a factor of two to accommodate plates radiating on both sides: "Since every instrument has two sides, why not include 1/2 term into SB Law?"

Well, that's because it would be relying on an invalid assumption. There may *not* be an other side, if the surface is part of a solid sphere, say. Or the other side may have different emissivity, different surface area, or different temperature. So the SB law tells you how much a single surface is radiating. If there is an 'other' side, and you want to know how much it radiates, then you apply SB independently again to that other side. In the special case of a thin homogenous plate, then yes, they will be almost exactly the same.

Where on earth do you go to learn this stuff? Because I would definitely be asking for your money back.

John Garland said...

This retarded lying piece of filth and general craphead--and many other such undesirable creatures--have repeatedly noted to you that a machine that takes 400W/m^2 in and emits 800W/m^2 out with no additional inputs is a machine capable of perpetual energy generation. I demonstrated the setup waayyy up thread.

Machines which release more energy than they take in, you see, can use all that excess energy to generate ever more energy themselves even if disconnected from the original power source. And that, dear friend, is what retarded lying pieces of filth and general crapheads call perpetual energy generators.

Such is the brilliance and clarity of mind you possess that you have developed a source of infinite energy that can save mankind a lot of trouble on into the future!!! I suggest you consult a patent attorney very soon as your method is out now and you'll need to act fast to protect it and to establish your priority.

Other crapheads have pointed out other ideas of yours which could also be turned into perpetual energy generators. But none are so elegant as the 400 in, 800 out design. That one is truly brilliant! Pure, um, "inventive genius"!

Bob Loblaw said...

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

Now, anybunny with a hare brain can see that Eli's post has drawn out a plethora of ignorant "fools with an agenda".

When this bunny was a prof, one of the things that he tried to emphasize in student assignments was that units matter. If you multiply a number in W/m^2 by a number in m^2, you get W. Conversely, if you want to balance W, and you've got numbers in W/m^2 with different surface areas, you'd better include some numbers in m^2 before you start hopping off at the mouth.

The ignoranti here won't even realize I'm talking about them, but the big issue here is not that the Law of Conservation of Energy is being violated, but rather that the Law of Conservation of Units is being violated.

Christian Anders said...

Bob Loblaw,

"Conversely, if you want to balance W, and you've got numbers in W/m^2 with different surface areas, you'd better include some numbers in m^2 before you start hopping off at the mouth."

And where in this whole discussion is that of any importance?

The two starting assumptions, namely that the plate nearest to the energy source receives the energy flux on its entire area facing the energy source, and that the two plates are of identical size, are enough to dismiss your critic.

Christian Anders said...

Betty,

"My argument was that 2 plates is no different from 1 fat plate (assuming no vertical heat loss)."

You can easily modify the thought experiment from the article in that way. Let us assume, one plate is heated to steady-state by the energy source and then it is separated into two plates, thus matching the setup of the article.

1) In a single plate conduction will evenly distribute the full incoming energy of 400 W/m² until the plate has constant temperature throughout its entire volume. Then it emits 400 W/m², 200 W/m² from each surface area. That is the steady-state.
Do you agree?

2) If you now separate this plate into two, and arrange them parallely, you will have the same setup as the article describes.
Do you agree?

3) According to you, nothing is then going to happen anymore: Both plates have the same temperature, since they have been brought to steady-state before separating them. So we should be in the state, that is also sloppily called "equilibrium".
Do you agree?

I am asking for your agreement with every single point, because it is interesting for me to find the very point, where the disagreement sets in.

-----

If you agree with all four points, we are left with a serious problem. Because it leads to:

a) The second plate, which is not directly supplied by the energy source with 400 W/m² radiation, has to emit 200 W/m² into the nothingness facing away from the first plate. This has to be because it was beforehand heated up to a temperature to do exactly so ( by heating to steady-state before the separation of the one "fat" plate into two plates ).
Do you agree?

b) Its internal energy is evenly distributed and therefore it has to emit the same amount of energy from both of its surface areas. So 200 W/m² into nothingness and 200 W/m² directed to the first plate. Adds up to a total radiation of 400 W/m².
Do you agree?

c) The first plate emits in the same manner: It receives 400 W/m² from the energy source and emits 200 W/m² from each of its two surface areas.
Do yo agree?

d) Only one surface area of the first plate faces the second plate, thus the second plate receives 200 W/m² from the first plate.
Do you agree?

e) The first plate receives 200 W/m² from one surface area of the second plate.
Do you agree?

If you agree with all that, you have created one body ( plate one )receiving 600 W/m², emitting 400 W/m² and having constant temperature all the same.
You have created another body ( plate 2 ), receiving 200 W/m², emitting 400 W/m² and having constant temperature all the same.

Despite this physical impossibility you are speaking against the one logical solution to that problem, namely let temperature changes happen in both of the plates to come to a new steady state, where they differ in temperature and thus have a clean budget of energy absorbance/emission.

Or, alternatevily, you does not agree with one ore more of all the points above and can explain, in which way you have been misunderstood.

Bob Loblaw said...

Christian Anders: "The two starting assumptions, namely that the plate nearest to the energy source receives the energy flux on its entire area facing the energy source, and that the two plates are of identical size, are enough to dismiss your critic."

I think you fail to understand who I am criticizing.

Hint: just as there are two sides to this argument, there are two sides to the plates. The two are not acting the same.

Betty Pound said...

No, idiot, it's like to ice cubes next to each other. They radiate 315 W/m2 in all directions in space, including to each other.

You create an unphysical strawman that you can add 315 to 315 and get 630.

The green NEVER heats the blue. The blue heats the green, and its energy is replaced by the sun.

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

Steady state heating applies to material, not EM radiation. Look it up.

Everything else you say is tautological BS devoid of physics.

Betty Pound said...

The paper is thin, and sheer conduction within the paper obviously exists.

Spheres have another side, as does any 3D object.

Betty Pound said...

Craphead refuses to learn. You can't add flux densities eminating from object. You look like an idiot when you do it.

An ice cube releases 315*6=1890 W/m2 from its 6 sides. That's what you think. You think an ice cube is therefore a perpetual motion machine because of your own stupid logic of adding outgoing fluxes.

Betty Pound said...

An cube heated by X W/m2 will, at equilibrium, emit X W/m2 in all 6 directions.

It will NOT emit X/6 W/m2 in each direction.

It will NOT emit a total of 6*X W/m2 in all 6 directions.

It will emit X W/m2 in each direction AND X W/m2 total.

Betty Pound said...

The plate is one molecule thick. There is no two sides. Retards need two sides to peddle their pseudophysics of energy splitting.

Betty Pound said...

* flux density splitting

Betty Pound said...

Retards believe that a cold body will prevent a hot body from cooling (to it)!!!

It's like saying a rock in free fall will not hit the ground because the ground blocks gravitation energy.

Any sensible person with integrity can see what they are doing: tautological math devoid of physics.

Betty Pound said...

Retards now claim "steady state heating", something that doesn't apply to EM radiation, or the setup of the problem: thin plates very close to each other.

John Garland said...

Well I guess that little storm explains the technical and scientific issues fully!

One hint to guide your further thinking: An ice cube which is not receiving a constant energy input from some source--like if it is placed out in deep space far from any sun--may start emitting at 315W/m^2 but then as it continues to radiate it will radiate less and less energy until it emits none at all at near 0K. That you think it would continue to radiate at 315W/m^2 in perpetuity somehow is just another example of your invoking perpetual energy generation.

You seem very interested in perpetual energy generation.

Betty Pound said...

"If one slows cooling of a body while not slowing the energy supply of the very same body, this body will end up with higher steady-state temperature than before. "

You can't slow the cooling of a body with a colder body. Heat flows spontaneously from hot to cold. The warmer body does not warm itself because of the presence of a cooler body. Forget Eli's BS, and go learn physics.

Betty Pound said...

Since you like to put words in other people's mouth, I will shit in your mouth the next time it opens.

Betty Pound said...

As it cools, you still believe you can multiply its flux density by 6 times.

Timothy Folkerts said...

Betty says: "The green NEVER heats the blue. The blue heats the green, and its energy is replaced by the sun."

That much is true enough. And for blue to heat green, blue must be warmer than green. If both are the same temperature, then there can be no heat flow from one to the other.

So thanks, Betty, for recognizing that blue is warmer then green.

Betty Pound said...

The heat flow goes to zero to create equilibrium.

If blue is warmer than green, the it's still heating green and not at equilibrium.

Betty Pound said...

Retards believe that warm can't heat cold all the way to equilibrium. Instead they believe that warm can heat cold half way and then itself half more.

Amazing retards!

Timothy Folkerts said...

OK Betty, here's a chance to show off your scientific skills in a couple simple situations.

Consider a 1m x 1m red plate -- much like the blue plate but heated instead by electricity, not sunlight. The electric heater within the red plate is designed to always send 400 W of electrical power to the red plate. Assume the red plate is a black body (emissivity = 1). Show that the temperature of the red plate will be
a) 244 K if the red plate is in deep space, far from any star, planet, etc.
b) 268 K if the red plate is enclosed by a sphere @ 200 K
c) 309 K if the red plate is enclosed by a sphere of ice @ 0 C.
d) 282 K if the red plate is half enclosed by a sphere of ice @ 0 C, with the other half exposed to deep space.

(It took me longer to type this than to calculate the results.)

Betty Pound said...

Grand Retard Eli draws a phony diagram of Blue sending 200 to Green, and Green sending 100 to Blue.

There is no fucking room for the 100, because every bit of space is occupied by 200 B->G. EVERY SINGLE BIT OF SPACE.

EVERY SINGLE BIT OF SPACE is occupied by warm heating cold until equilibrium, then maintaining that equilibrium.

Grand Hoaxer Eli has fooled a bunch of retards with a phony diagram, an illusion!

He drew a waterfall with half the water going up at half speed!

Congratulations, Eli, for finding a bunch of inbred low IQ cretins to defend your hoax.

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

Oh dear, Betts...you seem a tad agitated.

re. "There is no fucking room for the 100, because every bit of space is occupied by 200 B->G. EVERY SINGLE BIT OF SPACE."

Did you know that radiation going one way has no effect on radiation going the other way? As well, just because a surface is emitting radiation at one level doesn't at all mean it cannot receive radiation at another level. If you shine two light beams of the same or different energies at one another, they really do just ignore each other and go along their merry ways to affect whatever surfaces they should encounter!

I see, however, that you're tired of being shown to be a complete fool and are now trying to get Eli to "repress" your "free speech" by posting obscenities and calling people names. This might work if people cared about your opinions.

Christian Anders said...

Betty Pound,

at least you found one thing, where you certainly have a profound expertise:

Ignoring simply everything related to the topic, but simultaneously demanding all others should stick to it.

You have not answered even one of the very simple and totally topic related questions adressed at you.

Of course you cannot do so, for you are showing lack of knowledge of fundamental physical principles.

Example: "Grand Retard Eli draws a phony diagram of Blue sending 200 to Green, and Green sending 100 to Blue.There is no fucking room for the 100, because every bit of space is occupied by 200 B->G. EVERY SINGLE BIT OF SPACE.
EVERY SINGLE BIT OF SPACE is occupied by warm heating cold until equilibrium, then maintaining that equilibrium."

That is so plainly wrong that it's amazing you can state such nonsense. What you state is against the laws of radiative energy transfer, which have proofed to be right since they have been discovered.

Opposing radiations do not cancel out. They simply don't do, and it can be measured ( and was measured countless times ), that they don't do.

Thus you are argueing not only against the likes of Planck, but also against experimental proof.

Therefore I'm convinced, that you do not have any clue about physics at all and would recommend you to study the subject, before stating intellectual retardation of others who have far more expertise on the field than you.

Of course you could proof me to be wrong by answering the basic questions adressed at you, ore by giving a conclusive calculation of your alternative theory.

Betty Pound said...

Shining a light on a light does not make that light emit more light. It doesn't make it brighter.

But according to you, it should.
It doesn't and you have debunked yourself.

Ever heard of destructive interference?

Also, you are cheating by using two raw energy sources.

Radiation is electromagnetic waves. They are physically real. they occupy space.

If the plates are 100 molecules tall, then all 100 molecules have spinning electrons vibrating electromagnetic waves propagating in the horizontal direction to the next plate with 100 molecules. Each bucket is filled, so to speak.

There is no room for 100 to travel. None. All molecules are busy with the 200.

You live in a fantasy world where two things can occupy the same space. Fantasy is not reality.

Betty Pound said...

All surface molecules are busy transfering 200. Where is there room for 100?

Why are you mentally retarded?

Your mother did not breast feed you? Poor baby.

You need psychological help.

Betty Pound said...

* assuming light of same intensity.

If they are not same intensity, the brighter will make the darker match it. Because the differential gets transfered.

Christian Anders said...

Then it should be also impossible to radiate 300 into one direction, if all molecules are "busy" with 200 and for additional 100 there is no space. So rising the temperature will not lead to more radiation...

Just joking. Of course your most fundamental error has now become crystal clear. "Destructive interference" "space occupied by EM waves" "not two at the same space possible" and so on:

You have completely missed the development of quantum mechanics, which claims all those things to be possible which you claim to be impossible.

Yes, EM waves of opposite direction are allowed to cover the same space. Yes, molecules can absorb and emit radiation at the same time.

The claims of QM are backed up by tons of empirical evidence. Your claim is falsified by those empirical evidence.

Timothy Folkerts said...

This quote comes to mind ...

"I often say that when you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind."
Lord Kelvin

John Garland said...

Wow! "Destructive interference"!!! So, your 2 surfaces which are radiating at different energies are both emitting coherent radiation at the exact same wavelength and exact same amplitudes and the plates are placed at exactly a multiple of a half wavelength apart???

Could you show us exactly how you would construct such a setup with a random sun and two random plates?

That's a really good one!

EliRabett said...

Betty: "You can't slow the cooling of a body with a colder body."

Styrofoam coffee cup.

Thank you for playing

Betty Pound said...

It's not slowing the cooling. The coffee heats the cup.

Tell us what is slowing the transfer from the coffee to the cup.

Betty Pound said...

All that happens is B-200->G.

Show us how you can have 200 and 100 flowing to and fro with the same molecules at the same time.

Betty Pound said...

^ waves magic wand.

no, waves two magic wands in the same space in two opposite directions.

Betty Pound said...

According to Grand Retard Eli, a coffee cup will slow the coffee from cooling, and thereby warm up the coffee some more.

Quokka said...

Betty has repeatedly claimed that Eli's model has the green plate sending heat to the blue plate and is therefore unphysical:

"The green NEVER heats the blue. The blue heats the green, and its energy is replaced by the sun."

Just to summarize again, Eli has the blue plate at 262K, radiating 267 W/m2 to each side, and the green plate at 220K, radiating 133 to each side. So between the blue and green plates, B->G is 267 W/m2 and G->B is 133 W/m2, so the amount of heat being transferred (taking the definition of heat as net energy flow, not, as is often implied in colloquial language, as thermal energy) is 133 W/m2 from blue to green. So the blue *is* heating the green, exactly what Betty claims should happen. And similarly, the green is heating the empty space on the other side of it, exactly as Betty says should happen. And the sun is heating the blue plate, exactly as Betty says should happen (400 W/m2 S->B, 267 W/m2 B->S). So it's all good!

And, to forestall the inevitable next stupid claim, no, the space between the blue and green plates does not fill up with photons (Got a supporting reference for that, Betty?). The very definition of a blackbody makes it clear that they both absorb and emit radiation. At the same time, even.

John Garland said...

"The very definition of a blackbody makes it clear that they both absorb and emit radiation. At the same time, even."

The very definition twas pointed out to Betts many scores of posts ago. Did it help?

Nope.

Timothy Folkerts said...

Could we all agree to stop using the word "equilibrium"?

Nothing in the top post is in equilibrium, and using this word seems to be causing endless confusion in certain commenters when they try to force "equilibrium" physics on non-equilibrium situations.

This is a steady-state situation. Energy is continuously flowing, which by definition makes it non-equilibrium. ("n thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either within a system or between systems." -- Wikipedia)

Betty Pound said...

"So it's all good!"

No, it's not. Because the blue is arbitrarily restricted from heating green to equilibrium.

It's like saying the coffee cup restricts the coffee and cup from reaching room temperature.

"absorb and emit radiation. At the same time, even"

You have revealed your complete scientific incompetence.

An electron can not go into an excited and ground state at the same time.

It also can't go to different excited states at the same time.

And, being at 267 excited level and being MAGICALLY given 133 level would not boost it to 400 anyway.

You are a fucking dumbass. You believe you can add cold to hot.

Betty Pound said...

Tell us how an electron can be going up to excitation level AND down to ground state at the same time.

Betty Pound said...

You're coming up with a word that rationalizes your pseuduphysics.

There is tons with conduction. But not one with EM radiation.

You will not show us, because you know you are full of shit.

Betty Pound said...

"Energy is continuously flowing, which by definition makes it non-equilibrium."

Energy is continuously flowing through everything, including objects in equilibrium.

"... no net macroscopic ..."

Key word is NET. There is equal energy flowing in and out.

In reality, both blue and green will come to 244K, because there is nothing preventing blue from heating green. Nothing. The green does not force blue not to heat it to EQUILIBRIUM with blue. Nothing.

Quokka said...

Betty: "An electron can not go into an excited and ground state at the same time. It also can't go to different excited states at the same time."

Psst, here's a secret: blackbodies have *lots* of molecules in them. Some of them can be emitting and some of them can be absorbing at the same time. And your claim is demonstrably false. Put a piece of black metal in full sunlight. We know it's radiating (because all objects >0K radiate). And we know it's absorbing, because it gets warmer. At the same time, even.

Betty's comment raises another question. I thought/assumed the Planck curve spectrum of radiation from a blackbody was related to the distribution of kinetic energy of the molecules in the blackbody, i.e. absorbed thermal energy is stored as molecular kinetic energy, not different electron excitation states. Can someone knowledgable (i.e. not Betty) confirm/deny/clarify?

Betty Pound said...

Retard, all the molecules on the right side of the blue plate are busy sending 200 W/m2 (my scenario, 267 yours) to the green plate. All of them!

"Some of them can be emitting and some of them can be absorbing at the same time."

OK, so half the molecules are sending 267 and half are receiving 133?

Did you think you can add two different electron excitation levels of two different molecules?

They average! So you get 200 W/m2. Exactly as I said!

Thank you for accidently reconfirming what I've been saying.

You are quite a dumbass.

Betty Pound said...

molecular motion -> excited electrons -> EM radiation thru space --> excited electrons --> molecular motion.

Long distance harmonics.

Christian Anders said...

Where did you get a cube from? It was about one plate vs. two plates, each plate assumed to have no vertical heat loss at all.

Christian Anders said...

Betty Pound,

to cite you:

"According to Grand Retard Eli, a coffee cup will slow the coffee from cooling..."

- Which is perfectly right. Let's assume standard cups, where the mass ratio of coffee to cup is non-negligible. The cups are of completely identical shape and mass, but having different material properties and thus different heat conductivity. So the coffee will cool with different rates depending on the cup it is in, if poured in hot.
Do you agree?

"...and thereby warm up the coffee some more."

- Which is perfectly wrong. Of course a colder cup can never heat warmer coffee to an even higher temperature.
Do you agree?

For both of those two statements we should simply get a "Yes" from you, because all of this is really undergraduate school physics.

Thank god no one has ever claimed colder things could heat warmer things just by themselves. Thus we are all standing on solid ground of applied physics right now.

So now we can go a step further and add an energy source to the two coffees in their cups:

Two cups are of completely identical shape and mass, but having different material properties and thus different heat conductivity. The same amount of coffee is filled in each cup. The coffee has room temperature.

Two identical immersion heaters are now added to the cups, so there is one in each cup, and begin to heat up the coffee simultaneously with identical power.

Will the two coffees be heated up with the same rate and thus reach their final temperature at the same time? Yes or No?

Again I'm sure you will have the right answer, which of course is "No". Very good. We are still standing on the ground of basic physics and can therefore go another step ahaed.

Now we assume as an thought experiment the cups with their immersion heaters as an encloser ( imagine some kind of pressureproof cap if you like ), that is brought into the nothingness of vacuum. The difference in materials remains unchanged. The immersion heaters are constantly supplying energy.

Presuming that, I'd like you to answer two basic questions:

1) Has the amount of energy radiatet away by the outer surface of the cups to match the amount of energy transfered into the coffee by the immersion heaters, when steady-state is reached? Yes or No?

2) Is the amount of energy radiating away from the outer surface of the cups depending on the temperature of those very surfaces? Yes or No?

For the third time I'm sure you will not leave the ground of physics and give the right answers, which are simple to deduce by basic physics and of course are "Yes" in both cases.

Very well. We have come very far and only two questions remain:

1) Is the differential of temperature throughout the the material of the cups the same, even though they differ in heat conductivity? Yes or No?

2) If 1) has to be answered with "No", how does this affect the temperatures of the two coffees in their different cups? Do they differ, or do they not differ?

-----

Here we have the interesting point. If you have come so far just using logical deduction under the presumption of the laws of physics, you can only come to the conclusion that the temperatures of the two coffees have to differ in steady-state.

But that is clearly what you state all the time is not allowed to happen by the laws of physics.

So if your answers on the questions above are different from my predicition let us know. If they are according to my predictions, let us know how you can bring this together with all what you have stated earlier.

Betty Pound said...

First of all, never take someone else's sentence, chop it up into two quotes, and then comment on it. Especially when there is an "and".

"Thank god no one has ever claimed colder things could heat warmer things just by themselves."

Wow, you are one very stupid human being.

THAT is the central thesis of the whole fucking article!

Eli says the blue plate reaches 244K, and then he adds a green plate (initially 0K) which boosts (obviously all by itself, because its the only thing we added to the experiment) the blue plate to 262K.
---

If the immersion heaters are set to radiate 60C worth of energy, then that is what the coffee AND cup will be. The insulating material will simply delay heating to 60C. So unless this experiment has a time expiry, it doesn't matter.

I encourage you to stop being an idiot.

John Garland said...

re:
-----------------------
"Thank god no one has ever claimed colder things could heat warmer things just by themselves."

Wow, you are one very stupid human being.

THAT is the central thesis of the whole fucking article!
-----------------------

No it isn't.

The central thesis is how radiation from a source on to 2 plates one of which is fully illuminated and the second of which is totally shaded from the original source but receives all energy emitted from the nonillimmuminated side of the first plate acts to produce 2 different plate temps.

Re. "then he adds a green plate (initially 0K) which boosts (obviously all by itself, because its the only thing we added to the experiment) the blue plate to 262K."

What is "obvious" to you is your mistaken assumptions. Adding an element to a system that then interacts with other elements is very common. Any feedback situation, for example.

Finally, there is no such thing as so many degrees worth of energy. Energy is not defined that way. You appear to have a thermostat on your heater which turns on whenever the water temps go below, say, 60C. This whole setup as well as Eli's has a heater which is constantly on and examines the heat flows.

There is, for example, an energy setting for the heater which will lead to a 60C cup of coffee in one container. However that same setting will not heat another cup of coffee with greater or less conductivity to 60C. This is why insulation on your house saves money in the winter.

Christian Anders said...

Betty,

First I have to apologize for my chopping of your sentence. I assure you that I would not have commented the two parts of this sentence differently if I had left them together.

Now: Thank you for stating some things in a clear and unmistakable manner.

Because you did so, you showed the kind of your misconceptions beyond any doubt.

I qoute you ( and hope to do it correctly now ):

"THAT is the central thesis of the whole fucking article!
Eli says the blue plate reaches 244K, and then he adds a green plate (initially 0K) which boosts (obviously all by itself, because its the only thing we added to the experiment) the blue plate to 262K."

It is not the only thing we are adding. For the second plate has not an internal energy of zero as soon as its temeperature becomes >0K, we are adding additional energy to the experiment.

But I must admit that this is not the key to your misconception. This key is in your next statement, which I also try to quote correctly:

"If the immersion heaters are set to radiate 60C worth of energy, then that is what the coffee AND cup will be. The insulating material will simply delay heating to 60C. So unless this experiment has a time expiry, it doesn't matter."

That is plainly wrong. For you have not answered the questions in detail, I cannot find out where exactly you got it wrong, but you got it wrong.

By analyzing your statement it is eyecatching that you don't use the correct term for energy ( Joule ) or power ( Watts ), but a term that already includes your misconception ( "60C worth of energy" ). That's circular reasoning!

Energy is always an absolute quantity, measured by the unit of J. You made it to a relative quantity in the unit "worth of absolute temperature". Temperature is an ensemble attribute, related to energy by the specifications of the material in question and the time needed for energy to be transported inside the material ( for conduction is always of limited pace and can differ between different materials ).
That simply means, that for two different materials with different properties, "60C worth of energy" are different quantities in the unit of J. Thus, they are also different quantities for supplied power in W.

-----

Back to the experiment of cups of coffee. The immersion heaters are constantly supplying the same amount of energy in J/s each. Joule per second! Not "worth a temperature".

So I again would like to ask you a question:

If you assume this constant supply of energy into each cup with identical power in W ( that is what I have stated to be the experiment ), will you stay with your statement, that the cups as well as the coffee will end up at the same temperature ( regardless of how long it takes to get there )?

Think before answering! You have said yourself that the times for heating differ between the two cups, as I quote you again:

"The insulating material will simply delay heating to 60C."

What happens to the coffee when two different heating rates ( you stated those as well as I, see the quote above ) are combined with identical energy supply into identical masses of identically shaped cups of coffee?

John Garland said...

There is no need to apologise, Christian. If Betty is asserting that the statement (A and B) is true, she is also of course asserting that both A is true and that B is true independently.

Christian Anders said...

John,

It's about being polite, regardless of the behavior of the counterpart. And I want to prohibit that my arguments are left aside because there are other points to attack, may the attack be justified or not.

I find the cup experiment not as simple as the one in the article, but easier to imagine, since you have not to make any assumptions that are impossible in real life ( for example no vertical radiation from the edges, no fixing points in space ).

Since we will come to the same result, this does not really matter, it only reduces the need of abstraction.

What really amazes me is, that Betty does state a different heating rate ( which is of course correct and brings a step further ), and still comes to the same steady-state temperature in conclusion. Her way to do so was to simply skip the assumption of the constant energy supply to that of an inconstant energy supply.

Well, in that very case she can be right. But that was not the experiment. The experiment was about a constant supply and with this, the complete misconception becomes obvious and crushes down.

What can be the way out of this? I would bet on the "fact" that such an energy source cannot exist, because it would violate the 2nd law of TD.

Prepare to talk against this!

John Garland said...

You're a better man than I, Master Anders.

Re. heat source, yeah...it appears to be assumed to be connected to a thermostat or some equivalent thing if the 60C is going to be maintained regardless of heat flows.

Christian Anders said...

Which would be an energy source that is strictly forbidden by the initial conditions of the experiment.

Each source delivering energy with constant power, each source working with identical power. Done.

Just to clarify this beyond any doubt.

Betty Pound said...

The immersion heater is electric. So it heats by EM radiation. We apply SB Law.

So, 60C worth of energy supply means:

sigma*(273.15+60)^4

618 W/m^2

Since your stupid ass did not provide any numbers, I had to.

The various conductive materials merely delay the heating process: the amount of time needed to reach 60C.

Betty Pound said...

So Eli did not add a cold green plate to boost the blue plate's temperature? That was not the central thesis of the "the green plate theory"?

I have no time for pathological liars like you.

Betty Pound said...

Christian,

Don't fall for Folkert's steady state hoax.

Betty Pound said...
This comment has been removed by the author.
Betty Pound said...

What would restrict the immersion heater from emitting 618 W/m2 (60C)?

No, you assume some sort of thermostat. The cup material tells the immersion heater what to emit? You're crazy - and a pathological liar.

Betty Pound said...

Oops, 698 W/m2

Betty Pound said...

* 698 W/m2

Timothy Folkerts said...

Betty,

1) The very blog you quoted about steady-state vs equilibrium supports the blue plate/green plate systems as NOT is equilibrium.

To paraphrase, constantly heat one side of the blue plate -- the combination of the blue and green plates is our system. The end being heated (blue plate) will be warm, but the other end (green plate) will be cooler. There will be a temperature gradient between the blue plate and the green plate. We can definitely measure a fixed unchanging temperature at different locations, yet this is not in a state of equilibrium but only in a steady state.

2) Immersions heaters generally heat by CONDUCTION, not RADIATION.

3) Even if we did have a radiative immersion heater, the correct answer should be 698 W/m^2, not 618 W/m^2.

4) Even using 698 W/m^2, this is only the answer if the radiator is radiating directly to the cold depths of outerspace. In an ice-cold room, this 698 W/m^2 "60 C heater" would warm to 93 C. In a 20 C room, the same heater would be 101 C.

5) Even in the cold depths of outer space, if the heater was surrounded by any sort of insulation (for example some coffee or a styrofoam cup), then the "60 C heater" would also be warmer than 60 C.

Timothy Folkerts said...

"What would restrict the immersion heater from emitting 698 W/m2 (60C)?"

Two simple, related questions for you, Betty. Put your "60 C" radiative heater in the room that is 60 C already.
1) How many W/m^2 of electrical power would you need to supply to keep the heater at 60 C?
2) How warm would the heater get if you ACTUALLY supplied 698 W/m^2 of electrical power?

Christian Anders said...

Betty,

you gaave me a link to teach me something:

But what do I find? The author of the text provided by your(!) link states exactly what I and others tell you all the time. He also confirms the use of the term steady-state is perfectly matching the experimental conditions of the two plates or the two coffee cups.

I cite from the text that you personally have provided:

"Constantly heat one end only of a long metal rod – the metal rod is our system. The end being heated will be quite hot, but the other end of the rod will be cooler. There will be a temperature gradient running down the length of the rod. We can definitely measure a fixed unchanging temperature at different locations along the length of the rod, yet this is not in a state of equilibrium but only in a steady state. Why? Because if we were to isolate the system – the metal rod – from the surroundings, and to do this we place an adiabatic container around the rod, then we would notice a change taking place in the rod. The hot end would cool down as the cool end would heat up until the entire rod came to exist at a single fixed unchanging temperature. At this point our rod would now be in an equilibrium state, but it wasn’t before – it was only in a steady state."

The essence of that is simply, that a system where energy constantly enters and leaves is never in euilibrium, but in steady-state. For the plate experiment as well as the cup experiment this is exactly the setup, for they receive and emit constantly energy supplied by a source.

Also the text clearly states that in such a system a temperature gradient(!!!) will characterize the steady-state, for it would be impossible to transport heat without it.

That is the very point you stated to be physically impossible, still you personally have provided this point.

Are you simply kidding?

Betty Pound said...

1) You suffer from reading comprehension problems. Try reading to the end, imbecile.

"At this point our rod would now be in an equilibrium state, but it wasn’t before – it was only in a steady state."

2) No, retard, they heat by radiation to the neighboring molecules, which THEN sets off conduction.

3) you're a fucking piece of shit. I corrected my type-o.

4) You can't add temperatures, dumb fuck. 60C max.

5) No, it wouldn't.

You are mentally retarded. Please seek professional help.

John Garland said...

Betts...

TF provides the technical detail, but at a more general level all you need do is examine the units which are units of energy not absolute degrees K.

Betty Pound said...

1) 698 W/m2. The heater has no regulator. It outputs what you set it to.

2) 60C

Betty Pound said...

Are you fucking retarded? Yes.

"At this point our rod would now be in an equilibrium state, but it wasn’t before – it was only in a steady state."

It's steady state heating TO EQUILIBRIUM.

It gets to equilibrium, you dumb fuck.

Christian Anders said...

Betty,

an immersion heater does heat via conduction, not radiation.

Your calculated 618 W/m² are, what the outer surface of the cups emit at 60°C surface temperature.

Since we have learned from the link given by your very self, to bring the 60°C to the outer surface, we need a temperature gradient to the inner surface, for heat could otherwise not be conductet to the outer surface and lead there to the emission you have calculated. Thus the temperature of the inner surface has to be more than 60°C.

You have beaten yourself. Or you miscalculated. Choose!

Betty Pound said...

* by regulator I mean the external environment controlling the output.

John Garland said...

The definition of a steady state is that no change in the system is occurring even while energy is flowing through. The derivative w.r.t. time of state variables is 0.

What, then, is steady state heating TO any other state at all? Be specific. That is, how does a steady temperature state evolve over time "to EQUILIBRIUM" if the derivative of temperature W.R.T. time is 0?

Betty Pound said...

Shut the fuck up, you dumb piece of shit.

Christian Anders said...

Betty,

First a correction: I wrote 618 W/m² where you have calculated 698 W/m². It hast to be 698.

You seem to be unable to read. I cite again, but now highlight the vital parts so you won't miss them again:

"Constantly heat one end only of a long metal rod – the metal rod is our system. The end being heated will be quite hot, but the other end of the rod will be cooler. There will be a temperature gradient running down the length of the rod. We can definitely measure a fixed unchanging temperature at different Locations along the length of the rod, yet this is not in a state of equilibrium but only in a steady state. Why? Because if we were to isolate the system – the metal rod – from the surroundings, and to do this we place an adiabatic container around the rod, then we would notice a change taking place in the rod. The hot end would cool down as the cool end would heat up until the entire rod came to exist at a single fixed unchanging temperature. At this point our rod would now be in an equilibrium state, but it wasn’t before – it was only in a steady state."

Summing up: The system has to be isolated from the energy source ( the adiabatic container does this ) to come from steady-state to equilibrium state. That is what the text from your link states.

There is no such thing as "heating to equilibrium". As long as you heat, there will never be Equilibrium at all. That is what the text from your link states.

Remember: No system of the here discussed experiments was ever closed adiabatically and thus in the definition of the text delivered by you able to reach equilibrium in any way. But steady-state.

You can criticize Eli on having that point semantically wrong, I would agree.

Christian Anders said...
This comment has been removed by the author.
Betty Pound said...

"Remember: No system of the here discussed experiments was ever closed adiabatically "

What the hell do you think blue and green plates IN SPACE is?

You are one dumb sack of shit.

The text explicitly stated steady-state heating occurs BEFORE equilibrium. But it does reach equilibrium.

Betty Pound said...

The RATE of heating never changes. It still reaches equilibrium.

"how does a steady temperature state evolve over time "to EQUILIBRIUM"

That is the definition of equilibrium, you dumb fuck.

"if the derivative of temperature W.R.T. time is 0?"

How do you get from NY to LA if your speed never changes from 60MPH?

What the fuck is wrong with you?

Betty Pound said...

The people on this forum suffer from serious mental retardation.

https://i.stack.imgur.com/hmOTU.png

The steady state heat flow (left) will eventually come to equilibrium (right).

John Garland said...

Uh Betts...

1. When you travel from point A to point B the first derivative of distance w.r.t. time IS velocity.

2. If you take the derivative of velocity and find it's 0--as you will at constant velocity--then you have shown that there is no acceleration. That is velocity is not evolving. But the point of travel isn't whether velocity is evolving but rather whether the distance to the destination is evolving (decreasing).

This is about 2nd or 3rd week calculus and in generally taught at the high school level for those in a science track.

Do you just string words together? Or what?

Christian Anders said...

Betty,

to cite you:

"What the hell do you think blue and green plates IN SPACE is?"

When supplied with energy from an energy source, a system is never adiabatic, not in space and nowhere else. How do you get the absurd idea it was?

"The text explicitly stated steady-state heating occurs BEFORE equilibrium. But it does reach equilibrium."

The text actually states a totally different thing. It states that steady-state and equilibrium-state are different for the example of the rod, and that you need to cut off the energy supply at the hotter end of the rod to move the system towards equlibrium. That is, what "adiabatic container" means.

For in all here discussed experiments the energy source is never cut off and thus the system never is adiabatic, the equlibrium case never is fulfilled.

Do you really know the defintion of adiabatic? It does not seem so.

You cannot heat a system to equilibrium. That is physically impossible.

John Garland said...

Your picture shows how conduction works when a rod is heated at one end and then the heat is turned off. That is not the situation under discussion.

A steady state cannot evolve along that state. To quote your distance/time example, the velocity is steady-state and unless something acts to change it you will simply go on at 60mph forever and ignore your destination altogether.

Betty Pound said...

You fucking retard.
You just explained the obvious without realizing what it means.

Right, distance is key.
That is getting to your temperature destination. The velocity only determines how soon you there.

You agree that RATE of heating will not prevent reaching equilibrium.

Betty Pound said...

Retard, on earth there is convection. In space there is not. On earth you need an adiabatic isolator to emulate space. Do you understand equivalences?

"you need to cut off the energy supply at the hotter end of the rod to move the system towards equlibrium."

The text said no such thing. Nice try, filthy liar.

Text:
"Constantly heat one end only of a long metal rod ..."

Betty Pound said...

The mental gymnastics needed by these retards is getting quite high.

John Garland said...

Um Betty...

In a steady state system set up as described, the rate of change for temperature at any particular location is 0. No evolution to any other state is possible. By the very definition of steady state.

You just can't get to any other there from any particular here. By definition.

snarkrates said...

Betty,
I'm begging you. Just stop. It's clear you have neither the mental capacity nor the inclination to do the work needed to understand these examples.

Many of the people on this board have advanced degrees in engineering and science. We can all tell you are bullshitting. You are just embarrassing yourself.

Timothy Folkerts said...

"1) You suffer from reading comprehension problems. Try reading to the end, imbecile.

"At this point our rod would now be in an equilibrium state, but it wasn’t before – it was only in a steady state.""

I did read to the end, but clearly you didn't. You complained about someong taking one of your statements out of context, but yet here you do it. In context, the blog says:
Because if we were to isolate the system – the metal rod [or plates in our case] – from the surroundings, and to do this we place an adiabatic container around the rod [plates], then we would notice a change taking place in the rod [plates]. The hot end [blue] would cool down as the cool end [green] would heat up until the entire rod [both plates] came to exist at a single fixed unchanging temperature. At this point our rod [plates] would now be in an equilibrium state, but it wasn’t before – it was only in a steady state."

Only AFTER surrounding the plates with adiabatic walls (which includes cutting off the incoming heat from the 'sun' and cutting off outgoing heat to space), will it be equilibrium!

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

Betty,

so you do not know for sure what adiabatic means.

A system receiving energy from an energy producing source is NEVER adiabatic.

Thus it follows logically, that the rod has to be cut off from its energy source, because it is encapsulated in an adiabatic container. By the definition of "adiabatic" the container cannot have the energy source within. Otherwise it would not be an adiabatic container.

Only then it could be explained how you can think something could be heated first to steady-state, and then further to equilibrium.

"We can definitely measure a fixed unchanging temperature at different Locations along the length of the rod, yet this is not in a state of equilibrium but only in a steady state."

Why does this state of already fixed temperatures change further to get to equilibirium? What is the mechanism? Please explain, I want to learn.

Betty Pound said...

An equilibrium of 0K. Heat can not be shielded from the sun, nor can it be shielded from leaving UNLESS you put WORK into. Since you put WORK into it, you tarnished the experiment.

John Garland said...
This comment has been removed by the author.
Betty Pound said...

You fucking retard! you just described equilbrium.

"There will be a temperature gradient running down the length of the rod."

There will be a differential between locations.

There will be no differential AT a location.

"We can definitely measure a fixed unchanging temperature at different Locations along the length of the rod"

Yeah the temperature at 1cm will not be different from temp at 1.01cm at same time period.

the temperature at 2cm will not be different from temp at 2.01cm at same time period.

There will be a difference between 1 and 2 cm locations.

But this is all temporary. The rod will go to equilibrium, when convection is removed.

Betty Pound said...

The gradient slope is steady while heating, Not the temperature, dumb fuck. A steady temperature is equilibrium, by definition, dumb fuck.

John Garland said...

======================
"In a steady state system set up as described, the rate of change for temperature at any particular location is 0."

That is equilibrium.
======================

Nope. Equilibrium is a very special case of steady state.

As stated before, a steady state is when, well, er, a particular state variable stays, well, steady. There is no requirement that for a state variable to remain steady the system system has to be at some sort of static equilibrium than there is a requirement for an object travelling at 60 mph to stay in the same location.

You should remember your very own link said: "We can definitely measure a fixed unchanging temperature at different Locations along the length of the rod, yet this is not in a state of equilibrium but only in a steady state."

Betty Pound said...

The cranks on this forum will not stop with their abuse of language and physics. They've moved away from EM radiation to conduction, and yet they are still wrong.

https://www.forbesmarshall.com/fm_micro/images/8.Conduction.gif

Betty Pound said...

The gradient slope is steady while heating, Not the temperature, dumb fuck. A steady temperature throughout is equilibrium, by definition, dumb fuck.

"There will be a temperature gradient running down the length of the rod."

There will be a differential between locations.

There will be no differential AT a location.

"We can definitely measure a fixed unchanging temperature at different Locations along the length of the rod"

Yeah the temperature at 1cm will not be different from temp at 1.01cm at same time period.

the temperature at 2cm will not be different from temp at 2.01cm at same time period.

There will be a difference between 1 and 2 cm locations.

But this is all temporary. The rod will go to equilibrium.

John Garland said...
This comment has been removed by the author.
John Garland said...

Re. "The cranks on this forum will not stop with their abuse of language and physics. They've moved away from EM radiation to conduction, and yet they are still wrong."

I guess that means you now repudiate your statement of 15/10/17 3:04 AM where you said "Radiation is just conduction at a distance."???!!!

John Garland said...

I should add, you OUGHT to repudiate it. But not for the reasons you've brought up since.

Betty Pound said...

Repudiate a simile?

Radiation heat transfer takes place at the speed of light.

Convection heat transfer takes place at the speed of convective medium.

Conduction heat transfer takes place at the speed of thermal conductivity

And ALL three heat from hot to cold, SPONTANEOUSLY.

Betty Pound said...

When are you going to repudiate your BS?

https://www.forbesmarshall.com/fm_micro/images/8.Conduction.gif

Timothy Folkerts said...

"A steady temperature throughout is equilibrium, by definition"

really? Can you quote a single thermodynamics textbook that supports this definition? Can you find a single reputable webpage that supports this definition?

Or perhaps definitions in physics are determined solely by Betty.

... cue more foul language but no content ...

Betty Pound said...

Stop playing games, Tim.

People know what steady means - fixed, unchanging, same.

If the temperature is the same throughout, it's in equilibrium.

How could it not be. The temperature is the same through out.

What is wrong with you Tim? Please get professional help.

Betty Pound said...

I can't believe how stupid these people are.

Of course they will never address this:
https://www.forbesmarshall.com/fm_micro/images/8.Conduction.gif

John Garland said...

Re. "They can't distinguish the difference between a steady level (equilibrium) and a steady slope (steady-state heating)."

How, exactly, does a state variable with a nonzero slope define a steady state?! You are violating the definition of a steady state.

Betty Pound said...

The mental degenetates on this forum should watch this video:

https://youtu.be/dRWGwusNeI0

There are many like it.

Betty Pound said...

What state variable?
Crime, Inflation, Debt, etc.
Which one are you talking about?

Picture an isosceles triangle mountain. It's a steady state climb to the top.

Betty Pound said...

Another video for the sick freaks on this forum:

https://youtu.be/bGT2tEBu6f4

Betty Pound said...

This video gets cut off:

https://youtu.be/AiZWFrb76pM

Which of you lame brains wants to guess how it ends?

John Garland said...
This comment has been removed by the author.
John Garland said...

18/10/17 8:26 PM"

You simply don't know what you are talking about. A system that is climbing, even if at a linear, steady (i.e. constant) first derivative rate is simply not in a steady state. By definition. See any valid textbook in the world.

Christian Anders said...
This comment has been removed by the author.
Christian Anders said...

Betty,

You seem again to have misread one of my statements. You are commenting a quote of a statement of mine:

You fucking retard! you just described equilbrium.

Wtf? I explicitly stated the same as you ( first sentence of the quote, look at it, there is the word Gradient in it ), namely that steady state creates a temperature gradient. You replied to that, that I would have described equlibrium, because steady state instead of equilibrium creates a temperature gradient.
We said exactly the same here.
Look: I stated the gradient to be unchanging, not the temperature over the length of the rod. Temperature of course changes over length of rod in steady state.

To quote you further:

2) "There will be a differential between locations."

I agree.

3) "There will be no differential AT a location."

I agree

4) "Yeah the temperature at 1cm will not be different from temp at 1.01cm at same time period.
the temperature at 2cm will not be different from temp at 2.01cm at same time period."

Looking at this as an approximation: I agree

5) "There will be a difference between 1 and 2 cm locations."

I agree.

-----

We have five points now, and on all of them I agree with you, since in point 1 you just misread my statement. So: Agreement in five vital points about this very experiment with the rod.

How comes that our conclusions differ so gravely? This secret lies in Point 6:

6)"But this is all temporary. The rod will go to equilibrium, when convection is removed."

On this I disagree.

A hypothetical adiabatic container does ot remove convection inside. It simply prevents the contained system from transfering energy to the outside of the container. If not so, it would not be adiabatic by definition. But it necessarily has to be adiabatic, for that is one restriction of the experimental setup.

How can the container do this? Only in the way, that he is a perfect insulator and the inside of his walls are perfectly reflective for every kind of radiation. If it absorbed, it would heat up, which is forbidden by the defintion of adiabatic. And the rod inside is radiating, because its temperature is >0K.

And that is the very problem with your approach: As long as energy is delivered by a source inside the container, it will not be an adiabatic container, for the contained energy constantly rises. It is not allowed to get out, but gets pumped in by the source constantly anyway.
That violates the restrictions of the experiment and is therefore forbidden.

Thus the energy source has to be removed to ever come to equilibrium instead of steady state.

The boldly typed above is the conclusion you seem to be unable to reach, because if you did, your concept crushes down.

-----

Why is that of any interest for the experiment from the article? Because your failure to grasp this vital point is so great, that you even deliver videos which show nothing of interest for the question processed by all the experiments ( plates, cups ).

What do the videos show? We can see that a heating device can bring a surface to an evenly distributed temperature that remains fixed over time.

Nobody ever doubtet that!

But everybody besides you, Betty, stated that the other large surface of the material( container, iron, cup, whatever ), namely the one facing towards the heating device, must be warmer than the surface facing away from the heating device and being measured in the videos.

That is what all the fuzz is about:

Will the opposing surfaces of anything with infinitesimal thickness get to identical temperatures over time, when only one of those opposing surfaces is heated by any kind of power source.

Yes or No, Betty?

EliRabett said...

It's actually quite easy if you are not deep into denial to distinguish between a steady state situation and one with a steady slope.

In a steady state, state things are constant. For a steady slope things are CHANGING at a constant rate.

The problem under discussion here is in a steady state.

Betty has a choice here both in regard to foul mouthism and wasting everyotherbunny's time. Repeating nonsense is simply a waste of everybunny;s time. Calling them names is simply nonsense.

Behave yourself

EliRabett said...

Just to concentrate minds. The problem assumes infinitely thin, infinitely large, perfectly conducting, flat plates with two sides.

This is physics, not engineering and such idealizations are common, clarifying and useful for understanding.

Blathering about the shape of the plates, how many sides they have, their thickness, their composition, how close they are to each other, etc are attempts (successful Eli might add) at distraction from the basic physics that the example provides.

Christian Anders said...

Eli,

right you are.

I wanted to find out what exactly the root of Betty's misconception is and since I din't know if it was for lacking the ability to downgrade complexity by abstraction, I chose to set up a more engineer-like real world experiment compared to yours ( two coffee cups + heating unit, differing in material only ).

Thankfully Betty herself provided another setup with a one-side heated rod, where the very same conclusion is drawn than I have with any of my two coffe cups. With an additional rod of a more or less conductive material than the first the analogy to the two cups would have been completed.
Unfortunately she is so deep into some alternative physics, that she did not even realize what she delivered there.

But the root of misconception seems to be clear now.
Betty states that a real world material of non-zero thickness can be heated and reach equilibrium state after time anyway, keeping the heating device activatet.

Such is physically impossible. I tried to point this out, discussing the "what" and "why" in length. Unsuccessfully, I must admit.

Betty reamined to be the only being of the universe that knows adiabatic containers which contain a working heating device, for that is how she interpreted the text she linked. I have asked for exactly that.

My effort was simply aimed to show that you can never have heat transported over any real distance without a temperature gradient over this very distance.

Not from the inside to the outside of a coffee cup, not from one end of a rod to the other end, not from a radiating energy source to an infinitely thin, infinitely large and perfectly conducting plate, not between two of those plates, not with conduction, convection, or radiation. Simply never.

For that reason in your experiment both plates differ in temperature at steady state, for that reason a steadily heated coffe within a cup is warmer than the outside of the cup in steady state, and so on.

But since Betty demands for equilibrium and equilibrium is the very state without any temperature gradient that might enable heat to be transfered, she is unable to tell what happens with the heat from the heating device.

She demands it is not to be cut off. She also demands that the system will come to equilibrium after time anyway. These two points are physically impossible to unite, which is the very root of Betty's misconseption about thermodynamics.

Bernard J. said...

"Betty Pound" (who is most definitely male) or any other of the ilk, I am curious...

What is the mechanism whereby a radiatively-active gas can absorb photons and reradiate them in every direction but not the direction from whence the original radiation came?

And does the same thing happen with fluorescence? Think carefully here, because I might just posit for you an experiment where you can see the direction of the reradiation...

John Garland said...

I agree re. your gender assignment, Bernard.

snarkrates said...

So, why are we so concerned about D students (and, yes, I'm being charitable) who have neither the ability or inclination to understand anything more challenging than a blunt instrument? Isn't the appropriate thing to do simply to point and laugh?

Betty Pound said...

And you're still wrong Eli.
There is nothing preventing 262K Blue from heating 220K Green.

You are a scientific failure.

Betty Pound said...

With people like you, it's the arguments you can't respond to that reveal you're a fraud.

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...

I posted videos and diagrams proving you 100% wrong.

Acknowledge or STFU.

Betty Pound said...

To reach 244K equilibrium.

John Garland said...

Betty:

Nope. Eli right. Betty the Perpetual Energist wrong.

You have been "responded to" repeatedly. In detail. With textbook math.

Pointingly and laughingly yours.

snarkrates said...

Could Betty simply be a bot with Tourette Syndrome?

Betty Pound said...

No, he's 100% wrong. He lives in fantasy land, where he created his own funny physics. The math "works", but math is not physics. In math you can add outgoing flux densities or temperatures, in physics you can't.

snarkrates said...

Or is Betty a gifted comedian--carrying out a hilarious portrayal of a brain-damaged and delusional imbecile in a piece of performance art?

Timothy Folkerts said...

Betty says: "No, he's 100% wrong. "

Assuming you mean "Eli is 100% wrong in the top post", I would be interested to hear what specific ally you think is wrong. How about this -- tell us the first line that he writes that you specifically think is incorrect.

I'll even go first -- the first error is when Eli first says "equilibrium" ["Using the Stefan Boltzman Law you can calculate the temperature of the plate when it reaches equilibrium (400 W/m2) ... "]. To be technically correct, that should read "steady-state". And all subsequent uses of "equilibrium". should read "steady-state". OK -- fixed that.

Your turn. What do YOU think is wrong next???

Betty Pound said...

Stop playing dumb, Tim.
I am not going to repeat myself since you won't read it and absorb it anyway.

I know precisely which of my arguments remain unanswered or answered with gibberish.

Christian Anders said...

Timothy,

you really want to play it like this? I tried and I failed. In that game Betty is a true genius.

A main argument was, that the radiation from the green plate cannot be absorbed by the blue one. Either it is cancelled out by interference before reaching the plate, or cannot be absorbed, because the molecules of blue are all "busy" emitting and can therefore not absorb.

Do you have an idea, how to formulate a proper statement about such an argument? The gaps in logic are obvious, but stating the obvious simply does not do the trick.

But have a try...

Bob Loblaw said...

"Stop playing dumb, Tim."

Take that warning seriously, Tim. You are entering a battle where you have to play a role against someone to whom it comes naturally - someone who is a real expert; a natural talent the likes of which I don't think I have ever seen before. You'll just be bringing a knife to a nuclear war.

Betty Pound said...

"gaps in logic are obvious"

Projection. Your priest Eli thinks that Blue will not heat Green to equilibrium, but will instead bounce half its energy off of Green and warm up itself.

Nothing works like that in reality. Nothing.

There is no such law in physics that allows that.

Perhaps there is such a thing in post-modern physics: where you get to make up new rules along the way, and mere acceptance by others makes it "true".

That's all that's going on here: fantasy.

Christian Anders said...

Betty,

"Nothing works like that in reality. Nothing.
There is no such law in physics that allows that."

That's perfectly true.

So thank God that no one has ever claimed such a thing like:

"Your priest Eli thinks that blue will not heat green to equilibrium, but will instead bounce half its energy off of green and warm up itself."

To prevent the experiment from stating this, someone has put a constantly working power supply on the left side. Have you noticed it?

But wait. Wasn't that pointed out before? A few times maybe? Anyway: I like to bring up this point again, for you obviously have forgotten it, because for some reason you used the term "itself.

Luckily no one has ever claimed any of the plates heating anything by themselves.

Betty Pound said...

That is exactly what is claimed.

John Garland said...

Nope. The only thing doing the heating is the Sun. Everything else is merely how the Sun's heat radiates away. The laws of radiation are about as verified as anything in physics.

And, contrary to your perpetual energy assertions of 400 in and 800 out, if you put a plate in front of your face, it will radiate away heat to you and away from you such that you will always be exposed to less thermal radiation than standing directly in the Sun even if you stay out for a long time.

John Garland said...

Further, if a put a perfect thermal reflector behind the first plate, no one, well no one with a brain, would say the mirror is heating the plate as if you turn out the Sun, you will find the plate goes to 0K. Energetically, all Eli's black body is is a half mirror.

Christian Anders said...

Betty,

But you don't deny that a power supply actually is working in the experiment?

To ask that once more for perfect clarification beyond any doubt:

You say that: An adiabatically enclosed system in which heat is transferred will go into a state of perfectly even distributed temperature ( equilibrium ), while powered by an energy source that is eternally working.

True or False?

( That is, what I am able to extract from all your comments as the key principle you defend )

Betty Pound said...

And what prevents the blue from heating the green to equilibrium?

Where did you get 800? You can't add outgoing flux densities.

The plate will get to same temperature in the sun, regardless if you are behind it.

Betty Pound said...

Did you see the videos and gifs I posted?

( then you lack reading comprehension and critical thinking faculties )

Christian Anders said...

Betty,

For all of that I'm nearly sure that you would verify the request of my last comment as "true".

I just wanted clarification to be sure beyond any doubt.

If you could simply state "true" or "false" to this request of mine:

"You say that: An adiabatically enclosed system in which heat is transferred will go into a state of perfectly even distributed temperature ( equilibrium ), while powered by an energy source that is eternally working.

True or False?"

It is only one word for you to type and you can show me, how good or bad my skills in comprehensive reading and video watching are.

EliRabett said...

BP:And what prevents the blue from heating the green to equilibrium?

The fact that both the blue and green plates are radiating energy to space as well as each other limits the temperatures that they can reach.

Since the blue plate receives more energy than the green one, it must reach a steady state temperature higher than the green one does.

Your posts have indeed been a fine introduction to conservation of energy denial. Please let us know when you get a working model going.

Betty Pound said...

Retard, what space?

There is no space to the right of blue, only the green plate.

You deny conservation retard. You claim Blue can warm itself with half its own emited radiation.

Spontaneous self-warming is forbidden.

You confuse heat and energy.

Anyone with integrity can see that you're just using rhetoric and denying science.

Christian Anders said...

Betty,

you really forget quickly.

Only two comments ago you have been reminded that there is a heat source in the experiment that constantly produces "fresh heat".

So your term "spontaneous self-warming" is without substance.

( Give special attention to the boldly typed word! )

Do you have something like Alzheimer disease, as far as you can remember?

Betty Pound said...

As has been explained to you repeatedly, the sun alone heats the blue plate to 244K.

According to Eli's religion, adding a green plate causes the blue to heat some more, beyond what the sun could heat it to.

And the mechanism he uses is to say that the blue backheats itself 50%. That is what his diagrams say. There is no denying it. Eli is not claiming that the green plate tells the sun to heat the blue some more because he knows how ridiculous that sounds.

So in fact, Eli's diagrams claim that the green plate provides extra heating to the blue, beyond what the blue receives from the sun alone.

Of course it's self-heating after solar heating.

Nothing in nature warms itself by recycling its own received energy.

It doesn't matter what Eli says, his diagrams violate the laws of physics.

Betty Pound said...

Come on, people, how can you not see Eli's error?

Eli claims that blue will heat to 262K with green plate present. That means every molecule will emit 267 W/m2. He turned the sun's 400 W/m2 to 2*267=533 W/m2 using his own logic of adding flux densities.

Betty Pound said...

So, let's go back to what Eli said about conservation of energy:

"400 w/m2 x 1 m2 = 400 W = 200 W/m2 x 1 m2 + 200 W/m2 x 1 m2 = 400 W"

That's all well and good for one blue plate. But by adding green plate, the blue is now ...

Receiving:
400 w/m2 x 1 m2 = 400 W

Sending:
267 W/m2 x 1 m2 + 267 W/m2 x 1 m2 = 533 W

533 W > 400 W

Thus Eli is the one violating conservation of energy while projecting his error onto others.

Christian Anders said...

Betty,

according to your calculation there is zero of the energy that the green plate radiates absorbed by blue.

Is there no such radiation directed towards blue?

If there is, does it

- cancel out in between the plates?

- be reflected by blue?

- be absorbed by blue?

Betty Pound said...

It doesn't cancel out. It becomes averaged.

My scenario:
(200 b2g + 200 g2b)/2 = 200

Eli scenario:
(267 b2g + 133 g2b)/2 = 200

Equilibrium is reached at 200 W/m2 (244K) and Eli's scenario transforms into mine.

I already explained this averaging before. Where you paying attention?

Christian Anders said...

Ah, thank you for that explanation:

"Equilibrium is reached at 200 W/m^2 ( 244K) and Eli's scenario transforms into mine."

So you say that Eli's scenario is just an intermediate state, correct? Thus he is wrong by claiming that to be the final state?

I actually did miss that, sorry :( .

How is in your scenario the heat that is coming in from the source on the left transfered over the distance between blue and green, so green can constantly emit 200 W/m^2 towards the nothingness on the right? That I do yet not understand.

Another question about the mathematics you used: You do average two fluxes ( b2g and g2b, your notation ) but you claimed also that flux densities are not allowed to simply be added (that is how I interprete your reply to Eli from 21/10 4:43. maybe misinterpretation).

How is something averaged that is not to be added before?

On this I need assistance as well, sorry :(

Quokka said...

Averaging those values is meaningless. If you've got 267 b->g, and 133 g->b, you have to subtract one from the other to get the net energy flow (i.e. heat), which is 133 b->g. This is more obvious if you select other values, e.g. suppose you had
500 b->g, and 100 g->b. Then 500-100 represents something physically meaningful, the heat flow from from b->g. 500+100 doesn't represent anything useful, and neither does (500+100)/2.

Yes, you've said all this before. Problem is, it doesn't make any more sense now than it did then. You have 600 W/m2 radiated at the blue plate, but only 400 radiated away. You have 200 radiated at the green plate, but 400 radiated away. For any system in steady state, every part of the system must be radiating exactly as much as it is absorbing. So if both plates were at 200K, as you claim, then the blue plate would be heating up and the green plate would be cooling down (because of the law of conservation of energy).

barry said...

This by Tom Dayton...

"Your approach is backwards: You are treating the second law like a holy revealed truth, and the detailed, physical, causal, mechanistic explanations like incomplete, rough, approximate explanations. You understand Situation 1's mechanism. You do not understand Situation 2's mechanism, perhaps because you do not recognize the very existence of Situation 2. You are treating Situation 2's mechanism as if it is identical to Situation 1's mechanism. Ironically, you are distorting the meaning of the second law (and/or violating the first law) to accommodate your misunderstanding of Situation 2's mechanism."

... summarizes precisely the routine error that skeptics make.

They rebut by removing the heat source from the Situation. Situation 1 is a mantra whenever Situation 2 is described. Getting them to unfixate is near impossible.

barry said...

This by Tom Dayton...

"Your approach is backwards: You are treating the second law like a holy revealed truth, and the detailed, physical, causal, mechanistic explanations like incomplete, rough, approximate explanations. You understand Situation 1's mechanism. You do not understand Situation 2's mechanism, perhaps because you do not recognize the very existence of Situation 2. You are treating Situation 2's mechanism as if it is identical to Situation 1's mechanism. Ironically, you are distorting the meaning of the second law (and/or violating the first law) to accommodate your misunderstanding of Situation 2's mechanism."

... summarizes precisely the routine error made by skeptics.

They remove the heat source from the Situation, 'rebutting' Situation 2 with Situation 1. Getting them to unfixate is near impossible.

Betty Pound said...

"For any system in steady state, every part of the system must be radiating exactly as much as it is absorbing."

"you have to subtract one from the other to get the net energy flow (i.e. heat), which is 133 b->g."

Now you have a situation where blue emits 267 left and 133 right. However,

"for any system in steady state, every part of the system must be radiating exactly as much as it is absorbing."

You contradict yourself. An ice cube emits 315 W/m2 all around, not 215 one way and 415 another.

Conduction would ensure that the left side and right side come to equilibrium, as well.

Even more obvious when the plate is one molecule thick.

"So if both plates were at 200K, as you claim, then the blue plate would be heating up and the green plate would be cooling down "

244K not 200K

No, they wouldn't, because they are at equilibrium. There is now 0 heat transfer despite 200 flowing both ways.

"Averaging those values is meaningless. If you've got 267 b->g, and 133 g->b"

Wrong. Your molecules' electrons can not be in the same state at the same time. The electrons can not be both at excitation level 267 and 133 at the same time.
They get averaged.

Again, there is nothing stopping from blue heating green to equilibrium temperature. Blue does not get to bounce back 50% of its energy off of Green in order to warm itseld further. Which is what Eli does in his false reality diagram.

Christian Anders said...

Betty,

to quote you:

"There is now 0 heat transfer despite 200 flowing both ways"

What happens physically to the "200 flowing both ways" when arriving at a surface of either blue or green?
Averaging is a mathematical operation that gives no information about the physics going on at the plates.

How can you have "0 heat transfer" but at the same time green not cooling due to the 200 W/m^2 that leave towards the nothingness to the right side of green in your scenario?

Betty Pound said...

How do you explain things to an idiot?

The green is gaining and losing 200 W/m2 in my scenario: 0 heat transfer.
The green never sends heat to blue, only energy. Energy is not heat.

Your electrons can't be at two different states at the same time. So you need to average the states.

Christian Anders said...

Betty,

"The green is gaining and losing 200 W/m2 in my scenario: 0 heat transfer."

In your scenario you showed 200W/m^2 coming in from the blue plate left of green and 200W/m^2 leaving green on its left towards blue. This leads in total to a heat transfer of 0, for energy does not necessarily equals heat.

Understood.

What is the value of the flux leaving green to its right, directed towards nothingness?
You only told the value of the flux leaving green to its left.

"Your electrons can't be at two different states at the same time. So you need to average the states."

What is the physical representation of the mathematical operation of "to average the states"?

To my knowledge radiation arriving at a surface can 1) be absorbed, 2) be reflected, or 3) be partly absorbed while the other part is reflected.

Is there a fourth opportunity I do not know? If not, which of 1), 2) or 3) is representing the mathematical operation of "to average the states" physically?

Betty Pound said...

1)
200 W/m2 of heat goes right
200 W/m2 of energy goes left

2) why average?
the right side of blue plate can not send 267 and receive 133 at the same time.

an electron cannot be at 267 excitation level (emit) and told to be 133 at the same time (absorb).

I get tired of explaining common sense reality to idiots.

If you're going to repeat the same nonsense ad nauseum, it shows you are uneducable: a failure.

Christian Anders said...

"1)
200 W/m2 of heat goes right
200 W/m2 of energy goes left"

For blue this seems perfectly right to me, since blue is fed by 400 W/m^2 from the power source at the very left.

For green I have a problem adding up the quantities correctly: Green is fed by 200 W/m^2 from blue at the left. Green radiates 200 W/m^ towards blue and 200 W/m^2 to the nothingness right of green.

That you have confirmed.

How are the 1 * 200 W/m^2 that green receives from blue are converted into 2 * 200 W/m^2? That is the value of energy green radiates away according to the equilibrium temperature of 244K in your scenario.

I guess the key for understanding this is in the physical process that describes the interaction between radiation and the opposing surfaces of blue and green.

For this you gave me no answer:

"2) why average?
the right side of blue plate can not send 267 and receive 133 at the same time."

I did not ask for what is not happening in Eli's scenario. I asked for what is happening in your scenario.

Your answer so far was "it averages". That is a mathematical statement, no physical one. I asked for the physics. Since you state that absorbing 267 and emitting 133 at the same time is impossible, I wonder what exactly happens when 200 are emitted while simultaneously 200 arrive ( right side of blue <-> left side of green, your scenario ).
You did not explain the physics of that. You just stated it to be.

"If you're going to repeat the same nonsense ad nauseum, it shows you are uneducable: a failure."

Maybe I would repeat less frequently if you simply answer questions. I try to understand how your scenario works in detail. Not why Eli's scenario is not working. I'm ready to assume that as true.

Quokka said...

"You contradict yourself. An ice cube emits 315 W/m2 all around, not 215 one way and 415 another."

I know ice emits 315 W/m2 all around. This in no way addresses my statement that in steady state, each part of the system must be radiating and emitting the same amount of energy (or it would be changing temperature).

"Conduction would ensure that the left side and right side come to equilibrium, as well."

Irrelevant. I have been assuming all along that the plate is thin enough that both sides are essentially at the same temperature.

"No, they wouldn't, because [the plates] are at equilibrium. There is now 0 heat transfer despite 200 flowing both ways."

Yes, you have 200W flowing both ways between the plates. BUT as already explained, the plates can't be at equilibrium if they're radiating and emitting different amounts of energy (conservation of energy). You are assuming the plates are in equilibrium, when that's what you're supposedly trying to prove. To prove it, you need to show that they're emitting and absorbing the same amount of energy.

"Quokka: Averaging those values is meaningless. If you've got 267 b->g, and 133 g->b"

"Wrong. [...] The electrons can not be both at excitation level 267 and 133 at the same time."

I think you are making this up. Find a reference showing that thermal energy is stored in electron excitation levels. Bet you can't. Even if they were, we *know* that black bodies can absorb and radiate at the same time. It's implicit in the definition of a blackbody. And we know it from real-world experience, because a cold black body placed in the sun warms up and starts radiating. But it must also be absorbing the same amount of energy (once its temperature stabilizes) to replace the energy being lost. If you don't agree, what do you think it will do: absorb or emit?

Quokka said...

"Again, there is nothing stopping from blue heating green to equilibrium temperature. Blue does not get to bounce back 50% of its energy off of Green in order to warm itself further. Which is what Eli does in his false reality diagram."

Now this is a fascinating comment. It implies that the energy from the green plate can't be absorbed by the blue plate because it originally came from the blue plate. But of course this is nonsense, because photons don't have a memory of where they originally came from. They aren't even the same photons; the green plate absorbs photons from the blue plate, and then emits new photons.

Also, Betty appears to be implying that if the green plate was replaced by a 200W background radiation, then that energy *could* be absorbed by the blue plate, because the blue plate would be radiating into distant space, and the energy in the incoming photons *didn't* come from the blue plate. Is that what you actually think? Because it's nonsense. And if it isn't what you think, then your objection that "Blue does not get to bounce back 50% of its energy off of Green in order to warm itself further" is silly because your claim that Blue can't absorb 200W doesn't actually depend on the original source of the energy like you say it does.

"The green is gaining and losing 200 W/m2 in my scenario: 0 heat transfer.
The green never sends heat to blue, only energy. Energy is not heat."

Another fascinating comment. It implies that a surface at 244K sometimes sends 200W/m2 of energy, sometimes sends 200W/m2 of heat (and, presumably, sometimes a mixture). This is nonsense. There is no substance "heat" which is different from "energy". Heat is simply the net energy flow. And the idea of a net energy flow implies that two surfaces have to be able to send energy both directions to each othr, contra Betty's insistence that this can't happen. If you examine the radiation coming from 244K surfaces that you claim are emitting "heat" and "energy", they will both emit photons with identical radiation curves. *All* absorbed energy adds to the energy in a blackbody, *all* emitted energy subtracts from it. The net difference determines whether the blackbody warms up or cools down.

You still haven't addressed the fact that energy is not conserved in your scenario.

Bernard J. said...

"Now this is a fascinating comment. It implies that the energy from the green plate can't be absorbed by the blue plate because it originally came from the blue plate. But of course this is nonsense, because photons don't have a memory of where they originally came from."

I've been tempted to make exactly this same point over the last several days, to follow on from my last (unanswered) question to BP, but pigs and mud-wrestling...

Still, it would be delightful to get confirmation from 'Betty' Pound that this is what (s)he believes. Because if it is, the direct implication is that it would be possible to design a CO₂ laser that doesn't require any of that messy electron-excited nitrogen or cooled helium nonsense...

Betty Pound said...

Energy is not heat.
200 of heat green 2 space
0 heat green 2 blue, but 200 energy.

(200+200)/2=200

Betty Pound said...

"Irrelevant. I have been assuming all along that the plate is thin enough that both sides are essentially at the same temperature."

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.

Conduction will ensure that such a plate will reach thermal equilibrium: (267+133)/2=200.

"Find a reference showing that thermal energy is stored in electron excitation levels"

Radiation is not energy storage, but delivery.

http://www.astronomynotes.com/light/s8.htm

EliRabett said...

"How do you explain things to an idiot?"

An interesting question. Eli's answer is that he is simply going to start deleting anything Betty Pound says that derides any of the bunnies trying to educate Betty Pound.

Betty Pound said...

Eli claims that blue will heat to 262K with green plate present. That means every molecule will emit 267 W/m2. He turned the sun's 400 W/m2 to 2*267=533 W/m2 using his own logic of adding flux densities.

So, let's go back to what Eli said about conservation of energy:

"400 w/m2 x 1 m2 = 400 W = 200 W/m2 x 1 m2 + 200 W/m2 x 1 m2 = 400 W"

That's all well and good for one blue plate. But by adding green plate, the blue is now ...

Receiving:
400 w/m2 x 1 m2 = 400 W

Sending:
267 W/m2 x 1 m2 + 267 W/m2 x 1 m2 = 533 W

533 W > 400 W

Thus Eli is the one violating conservation of energy while projecting his error onto others.

EliRabett said...

Let Eli establish a few ground rules here.

A surface at temperature T emits electromagnetic energy whose spectrum and other properties are described by the Planck Law and the total of which is described by the Stefan Boltzmann law.

All of the radiation in the Green Plate Effect is thermal radiation.

Transfer of energy by thermal radiation is transfer of heat. That's just the way it is. The key to understanding why this is, not that Eli believes that Betty Pound cares about understanding is that thermal radiation from body can be characterized by the temperature of the emitting body.

Anybunny trying to confuse this issue from now on will simply have their post deleted.

Christian Anders said...

Betty,

"The green sends 200 heat to space

The green sends 200 energy to blue
The green received 200 energy from blue"

Do you really not see, that with these three simple lines yo invented physics yet thought to be impossible?

The 1st law does not distinguish between Joules that transfer heat and Joules that do not. One Joule is simply always one Joule, regardless of any heat transfer it might cause.

Show any textbook or paper claiming differently!

You really let green receive 200 Joule per second and m^2 but emit 400 Joule per second and m^2 ( 200 for each surface ). And no, it makes no difference at all if the Joules transfer heat or not. You create 200 Joules/m^2 energy out of thin air each second at the location of green.

The irony, that you yourself do exactly what you claim to be wrong in Eli's experiment, is unparalleled.

Of course the deeper reason for this colossal confusion is that you think the system to be adiabatic and therefore able to go to equilibrium.

It was stated often before, even by a text that you delivered yourself, that a system with a power source is not adiabatic and can therefore never reach equilibrium state. But you are even unable to comprehensively read your own choice of texts.

That is completely amazing from the point of view of cognition psychology.

And of course somewhere deep inside you seem to know the weak spots of your argument, for you constantly refuse to tell what happens to the radiation in between the plates when arriving at the surface of the opposing plate, regardless of how often requested from you.

You could chose between reflection, absorption, or a mixture of the two. You decided for "averaging", something nowhere to be found in physic's literature like you used it.

Simply: Amazing

Betty Pound said...

"*All* absorbed energy adds to the energy in a blackbody,"

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

That's why flux densities and temperatures can't be added. It is a Max() function, not an Add() function.

The blackbody will not even absorb lower energy photons.

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.

Betty Pound said...

"All of the radiation in the Green Plate Effect is thermal radiation.

Transfer of energy by thermal radiation is transfer of heat."

OK, you defined thermal = heat. Fair enough.

But Heat only exists when traveling from hot to cold.

Which is the case in G to B.

Therefore not all radiation from Green is thermal/heat.

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

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

Eli is a physics failure.

"going to start deleting anything Betty Pound says"

Already saved. You will be publicly shamed for the rest of your life, unless you correct yourself and tell the truth.

Sorry that lady justice is such a bitch.

EliRabett said...

BP: Eli claims that blue will heat to 262K with green plate present. That means every molecule will emit 267 W/m2."

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

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

The green plate will be at 220 K and emit 133 W/m2 from each side.

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

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 Blue Plate EMITS EXACTLY THE SAME NUMBER OF W/m2 AS IT IS ABSORBING

The green plate is emitting 133 W/m2 from each side for a total of 267 W/m2

That is EXACTLY THE SAME NUMBER OF W/m2 AS IT ABSORBS FROM THE BLUE PLATE

First law is well satisfied