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

Using the Stefan Boltzman Law you can calculate the temperature of the plate when it reaches equilibrium (400 W/m

^{2}) = 2 σ Teq^{4 }where σ is the Stefan Boltzmann constant 5.67 x 10^{-8 }W/(m^{2}K^{4}), factor of 2 for a two sided plate per m^{2}. 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/m

^{2}
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/m

^{2}and the energy going out is σT1^{4}+ σT2^{4}Since these will be equal at equilibrium
400 W/m

^{2}^{ }= σ T1^{4}+ σ T2^{4}
And there also has to be an equilibrium for the energy going in and out of the green plate

σ T1

^{4}= 2 σ T2^{4}
The bunnies can rearrange the second equation to get

σ T2

^{4}= 1/2 σ T1^{4}
and substitute for σ T2

^{4 }back into the first equation
400 W/m

^{2}^{ }= σ T1^{4}+ 1/2 σ T1^{4}
or

400 W/m

^{2}^{ }= 3/2 σ T1^{4}
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

I like this 'for dummies' explanation better than any other I've seen to date. Good work.

ReplyDeleteWhy thank you Kevin, Eli appreciates that.

ReplyDeleteTalking about plates in space, a colleague of mine (who sadly died before this was announced) was given a NASA award for his work on passive cooling of infrared space telescopes. This basically uses sun shields and radiators to cool a space telescope without needing cryogenic coolants.

ReplyDeleteAm I crazy or shouldn't that 2 in your 2nd to last equation line be a 1?

ReplyDelete2x + 1/2x = 5/2x but you want x + 1/2x = 3/2x, don't you?

Ah, another version of https://en.wikipedia.org/wiki/Idealized_greenhouse_model or http://scienceblogs.com/stoat/2014/02/16/the-idealised-greenhouse-effect-model-and-its-enemies/ but with nicer pictures. Surely *this time* people will understand :-)

ReplyDeleteMy entry on Spencers blog I think proves with a physical object that coll objects can make a warm object warmer:

ReplyDeleteSeptember 25, 2017 at 11:10 AM

Using whatever version of physics you believe in please explain how a room temperature thermal imaging camera can take thermographs of objects down to -50C

There are no “cold rays” so how does focusing an object at -40C onto a microbolometer at 25C change its temperature compared to focussing an object at -30C onto microbolometer at 25C?

https://3.bp.blogspot.com/-GT_Ar-9WWfQ/UNkU2Fb2nBI/AAAAAAAAA1M/NLxj8Rt7yRI/s1600/sky+high+low+cloud.jpg

This shows a thermal image of sky and clouds (in winter). The camera body was approx. 15 to 20C and the sensor was uncooled and therefore at a similar temperature perhaps hotter.

If you are correct and cold objects cannot cause warmer objects to warm then this picture should effectively be an even black (there is no object above the microbolometer temperature and in your view lower temperatures cannot warm hotter objects and remember there are no such things as cold rays). All objects in the field are less warm than the sensor and in your physics cannot change the temperature of the sensor.

However you can see temperatures from -2C to -35C. This is because the cold clouds are providing energy to the sensor changing its temperature.

If the object in the field of view were at abs zero then no heat would be transferred to the sensor and the sensor would be in equilibrium with its camera environment. Above absolute zero heat energy is transferred to the sensor and its temperature increases above the camera background.

At -50C (approx.) the heat from the object adds sufficient energy to the sensor for detection and the sensor warms until the energy OUT to the camera environment equals the energy IN from the camera environment plus the energy in through the lens.

This additional energy changes in microbolometer temperature and hence its resistance.

You should also be aware that the germanium lens used on thermal imaging cameras acts as a bandpass filter. So most radiation from CO2 and water vapour is not passed to the sensor. Hence no back radiation is seen from water vapour or CO2. Clouds of course are not water vapour but a cloud is an aerosol comprising a visible mass of minute liquid droplets (wiki) and hence thermal radiation is more like a black body allowing the camera to see the cloud.

In general a thermal imaging camera MUST be insensitive to GHG radiation otherwise hot air would fog the image.

http://www.flirmedia.com/MMC/THG/Brochures/T559243/T559243_EN.pdf

from flir explains all(?)

The alternative physics experts also need to explain how a co2 laser (10.6um==273K BB peak temp)can cut wood etc.

Prof Eli's diagrams are elegant, and I echo WC's comment - surely this cannot be argued against by the Denialati.

ReplyDeleteSurely.

And then I realised that I and WC are both being naïve - there will always be a dunce ready to step to the

plate...The question is who might bite first.

And is it sad that I can't get Millikan's oil drop experiment out of my mind?

Yep, copied it over from previous and forgot to kill the 2. The figures have it right tho. . . .(shame emoji)

ReplyDeleteWm, the point is to have something in big print and simple figures that can be pointed to in a tweet. Understanding from those with an agenda is not hoped for. Stopping them from confusing the crowd might be

ReplyDeleteIt's pretty terrible that a lowly psych stats person has to correct your math. Hopefully your colleagues never find out! :-)

ReplyDeleteEli was never into proofreading.

ReplyDeleteEli writes: "Eli was never into proofreading."

ReplyDeleteYou were one of the co-authors on the Harde response, correct? Apparently no in that group is much into proofreading. Though I suspect English as 2nd (or third) language is to blame.

No excuses though. There were native English speakers on the authorial team.

Nice Eli- I think Feynman would smile.

ReplyDeleteIt may interest andthentheresphysics to know that the same emissivity games could work to the advantage of ground based IR telescopes - you can tone down the thermal infrared background noise by slathering your instrument and the insdide of the dome with paint spiked with boron nitride, which though isoelectronic with graphite is a wide bandgap semiconductor whose strong bonds and light atoms make it an almost perfect IR reflector in the 10-13 micron range, which tears a hole in its emissivity in that range

When someone tries the 'cold atmosphere can't heat the warmer earth' ploy, I ask them to consider a thought experiment.

ReplyDeleteImagine two naked guys outside in the snow at -30 deg C. In front of them is a down sleeping bag which is also at -30 deg C. One of the naked guys gets into the sleeping bag. What happens?

@oldmoc...

ReplyDeleteI tried pretty much that same exact analogy on Victor-the-troll on realclimate when he said "the absence of cooling cannot produce warming". Didn't work. Denial is a river that runs very deep.

Bernard J

ReplyDeleteProf Eli's diagrams are elegant, and I echo WC's comment - surely this cannot be argued against by the Denialati.

Surely.

And then I realised that I and WC are both being naïve

I think the Stoat was being sarcastic...

ReplyDeletesurely this cannot be argued against by the Denialati.I linked here from my regular climate dust-up blog, and this was the rebuttal....

The blue sheet radiates 400 w/m2 on all sides, not half that amount on either side.

Apparently one does not add up the thermal emission from each side to get the total. In the same way, I suppose, one does not add up 400 w/m2 from each side of the sun to arrive at 800 w/m2.

Not being a physics graduate, I don't know how to respond to this. Anyone help?

(If you wade past the ramblings in his post where he enjoys G&T and roast Rabbett, you can see his own words on the matter: http://www.drroyspencer.com/2017/10/uah-global-temperature-update-for-september-2017-0-54-deg-c/#comment-267204)

ReplyDeleteIn the same way, I suppose, one does not add up 400 w/m2 from each side of the sun to arrive at 800 w/m2.With my limited abilities, I suspect I'm about to be told that the spherical sun does indeed emit more than 400 w/m2 (as received by the inside surface of a sphere surrounding the sun), and that this is the amount received from the surface of the sun facing the plate.

I'm not confident, though, so any help would be appreciated.

Eli set up a non rotating plate with the sun on one side so the sun only shines on one side. That's the model.

ReplyDeleteThe condition was chosen to simplify the discussion and avoid the need to discuss the effect of rotation of the Earth and the fact that the Earth is spherical

Should a bunny look at a more complex discussion such as the Weasel pointed to in Wikipedia or Arthur Smith published on arXiv he or she would find that the solar Flux is ~1400 W/m^2 on a surface perpendicular to the distance btw Earth and Sun. Because the sun shines only on one side at a time and the fact that the Earth is a sphere the average intensity integrated over the sphere is 1/4 of that or about 350 W/m^2.

Thanks. Please accept a handful of carrots.

ReplyDeleteI did a bit of mind experimenting to figure out that the sun is emitting more than 400 w/m2 in total, and that lone plate A should radiate 200 w/m2 each side, not 400 w/m2 either side.

A light bulb warms your hand. If we could manipulate reality to unfold the bulb to a flat plane, so that all its thermality was being emitted in one direction, we should get more heat.

There is a very simple way to do this IRL. Get a second bulb.

My co-interlocutor should thus be arguing that 2 bulbs provide no more heat to your hand than one.

"

ReplyDeleteI think the Stoat was being sarcastic..."I thought so too, and hence I was too...

And as Barry demonstrates the dunces didn't disappoint, even if they've thus far avoided this particular thread.

Pigs and mud-wrestling come to mind, as does talking to rocks. :-(

This may be an uninformed question, but has an analogous physical experiment been done to demonstrate this. It seems that one should easily be able to set up a radiative source with 2 plates and measure the temperatures. Would that be even more convincing for the 'skeptics'?

ReplyDeleteKen, this was well known even before 1900 so there is little publlshed work as such. However, it is inherent in the design of vacuum furnaces and cryostats. Cryostats are the inverse. You have a cooling unit with heat shields around it that keep the heat from the outside leaking in. Eli hisself operated a vacuum oven as a grad student with a several heat shield around the heating element and if they were not there the temperature of the oven was much lower than what was needed.

ReplyDeleteEli, you've mistaken me for my opponent at the blog I linked to. A contrarian there replied that the blue plate should emit 400 w/m2 on both sides. Knowing little of physics I overthought what he said and wasn't confident he was wrong, so came here to ask for help. The 2-bulb thing was what I came up with on my own to uncomplicate my thinking and see his error. Seems I haven't expressed myself clearly.

ReplyDeleteI've saved this page for future reference. It's the tidiest rebuttal to the simplistic 2nd Law mantra I've ever seen.

Fair enough. Apologies

ReplyDeleteSo the Bunny went over and looked at the food fight about this post that has broken out at Roy's place, and indeed Barry is right, he is doing good work.

ReplyDeleteEli has taken down a couple of the comments above which he was mistaken in attributing the ungodly position to Barry, but repeats them here for content

Most importantly the post was to show the placing a colder body near a warmer body can make the warmer body hotter when it is being heated by another source.

The relevance of this proof is that the Earth is heated by light from the Sun and layers of the atmosphere containing greenhouse gasses act as the colder third body.

What has been proved is that colder bodies can warm a hotter body through back radiation. Where there is an external or internal heat source in the system, the equilibrium temperature of the warm body will be higher than if the cold body were not present.

Yes as Gerlach & Tsheuscner point out the solar Flux at the Sun is much higher than the IR Flux from the Earth at the Earth but the solar flux at the Earth's surface matches the Earth's IR flux because of things like the 1/r2 falloff with distance from the Sun, the fact that only a half of the Earth is illuminated at a time by the Sun and the fact that the Earth is roughly spherical so that most of the surface is tilted to the impinging light from the Sun.

But anybunny claiming that the Sun has to heat both sides of the blue plate at the same time is claiming a binary star illuminating both sides of the Earth. There is this thing called night.

As to light plates, well Eli might ask folks to look at their cell phones. Most of them have flashlight apps.

The neverending at Roy's place

ReplyDeletehttp://www.drroyspencer.com/2017/10/uah-global-temperature-update-for-september-2017-0-54-deg-c/#comment-267204

"The neverending at Roy's place"

ReplyDeleteIs an improvement on the Reverending at Heartland's place on the Vegas strip

barry/Eli...

ReplyDeleteGood Lord there are some truly heavy duty deniers over there! :-o

A couple of points.

ReplyDeleteFirstly, this scenario claims that the green plate at a colder temperature is transferring heat to the blue plate as the only possible mechanism of causing the blue plate to increase in temperature.

The green plate has no other source of energy.

Initially the blue plate is at ~244 K and the green plate is never at a higher temperature than ~220 K.

This is indeed against the principles of the 2nd law of thermodynamics.

"In the absence of external work done on a body, heat can only move from warm to cold."

Further it was demonstrated more than 2 centuries ago by experiment that the radiation from a cold object cannot possibly cause a warmer object to increase in temperature.

Surely a well documented experiment in radiative transfer trumps any thought bubble ?

Ross, not only are you mis-stating Clausius' version of the second law, you are not understanding the proof.

ReplyDeleteFor the Clausius version of the second law, Eli refers you to Bindidon at Roy's place. Bindidon has it in both the original German and English translation

http://www.drroyspencer.com/2017/10/uah-global-temperature-update-for-september-2017-0-54-deg-c/#comment-267232

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

Moreover as Bindidon points out Clausius was well aware of what we call back radiation:

"What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one."

Both conditions are met in the Green Plate effect.

You appear to have missed the point that in this proof the sun (or at least another body) is shining on the blue plate transferring heat to it continuously.

If you just have a warm body sitting in space, it will cool by radiation.

If you have a warm body and a colder one near it, it will cool a little bit slower because of interchange of radiative energy btw the two The net interchange of heat will be from the warmer to the cooler. But in both cases the body(s) will cool down to the background temperature of the universe, like a few C

However, and here is what folk miss, if you have a heat source, like the sun, heating the warm body at a constant rate while it cools by radiation, the warm body will become hotter if there is a colder body near it because of the interchange of radiative energy between them.

Hey, this post has got a mention in the comments at Batshit Central (this is why Ross McLeod has turned up):

ReplyDeletehttps://climateofsophistry.com/2017/10/06/slayers-vindicated-by-additional-independent-researchers/#comment-31005

Joseph Postma claims that because the blue plate example always emits less than 400W back at the energy source (no matter how many green plates are added), this example refutes the RGHE. Of course this is idiotic because the blue plate does emit more than 400W if you consider both sides of it.

Topquark

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

ReplyDeletethermoDYNAMICS is not a dynamic process? LOL

If the blue plate starts at 0 kelvin and receives 400 W/m2, and MUST emit 200+200 W/m2 ... it never heats up!

Betty you neatly capture the difference between thermodynamics and dynamics in your example.

DeleteThe blue plate emits less than 400 w/m2 until it reaches equilibrium at 244 K and then it stops warming. Strictly speaking thermodynamics is only valid at equilibrium. Dynamics is how it gets there

Why would it go to 244K? I am assuming your plate is one molecule thick, so 400=sigma*T^4, T=~290K.

DeleteRoss:

ReplyDeleteAs thefordprefect notes above, how does a standard, easily purchased thermal imaging camera (e.g., http://www.flirmedia.com/MMC/THG/Brochures/IND_025/IND_025_EN.pdf) take pictures down to -40C with a room temperature sensor without violating your imaginary laws?

How does it take a picture of anything at all cooler than the sensor, say even at 18C?

You are dropped off in Antarctica. How do you feel cold?

DeleteBecause YOU are warming the environment.

The sensor detects thermal energy flowing from it to the colder object.

Despite what has been said my comment -

ReplyDelete"Further it was demonstrated more than 2 centuries ago by experiment that the radiation from a cold object cannot possibly cause a warmer object to increase in temperature." -

remains indisputably true.

There is modern experimental evidence which verifies the 2 centuries old experiment demonstrating that the radiation from a cold object does not induce warming in a warmer object !

There is another version of the second law and I think the scenario violates this one as well.

ReplyDelete"No process is possible whose sole result is the transfer of heat from a cooler to a hotter body."

I hope that it is OK to present an alternate interpretation to points made civilly.

ReplyDeleteI think Betty is right !

A temperature sensing device has a flux incident upon it.

It has an electrical current and a programmed "computer" that converts the incident flux into a temperature value for display.

Surely any value can be sensed ? As Betty said the flow of heat is from the sensor to the colder object !

It does not have to induce warming - it is the calculations using radiation physics that determines the temperature.

In no way does this involve the transfer of heat from a cold object !

Hence thefordprefect's argument is incorrect.

Someone once claimed a microwave oven proved that a cold object could heat a warmer object but that is just silly.

Google Pictet's reflection of cold to prove for yourself that the radiation from a cold object does not increase the temperature of a warmer object !

Similarly there are thousands of real world examples where the input of continuous energy does not necessarily involve increases in temperature.

The best know is water - at STP no matter how much energy you "throw" at water it will never increase in temperature above 100°C.

At even lower temperatures such as ocean surfaces water evaporates without the mass of water necessarily increasing in temperature.

The hottest land surface temperature recorded is 70.7°C whilst the natural hottest ocean temperature is ~36°C.

Similarly for melting of all objects that melt.

Viscosity of liquids decreases as they are "heated" and the temperature changes are NOT in accordance with the specific heats - I did experiments such as these in Chemistry classes at University 46 years ago.

The Stefan-Boltzmann, Planck's, and Wein's laws arose from the cavity radiation experiments and involved the emission of continuous spectra. No-one knows for certain that these laws apply to circumstances other than the emission of continuous spectra.

I hope you can accept alternate viewpoints to open discussion.

No one ever said the green plate was warming the blue plate. The sun is doing the warming. Greenhouse deniers consistently miss this very simple point. All the green plate is doing is affecting the radiative cooling rate of the blue plate as Eli has been at pains to point out.

ReplyDeleteWith just one plate, the sun warms it to 290K.

DeleteWith two plates, the sun warms both to 290K.

Delayed cooling is not heating.

Ross McLeod writes: "I hope you can accept alternate viewpoints to open discussion."

ReplyDeleteThis is math and physics. There really isn't an alternate view. "Views on shape of the earth differs" doesn't work here.

Eli has provided a very simple system diagram complete with the necessary numbers. It's either correct or it's incorrect. In your comments you never once show where it is incorrect. If you could, I'm sure you would have.

The question then becomes: what do *you* do in the face of evidence that proves your position incorrect? You can *deny* that evidence or you can accept it and modify your view accordingly.

I know where I'm placing my money :)

The plate in my kitchen is at room temperature. It radiates 20C worth of energy in all directions.

DeleteEli comes along and claims that that it actually emits 10C up and 10C down, and foolish people believe him. LOL

I found some time to create an Excel Spreadsheet using the EXACT algebra used in the post.

ReplyDeletePlanck's equation is the only equation that completely describes any blackbody radiant emission and the corresponding temperature relationship. The Stefan-Boltzmann equation describes the total power emitted at any temperature.

We can all agree on this surely ?

The area under the curve plotted using Planck's equation multiplied by pi is equal to the Stefan-Boltzmann calculated total power for the corresponding temperature.

Integrating Planck's equation times pi gives P = π∫L(λ)dλ= (2π^5 x k^4 )T^4/(15 h^3 c^2 ) = σT^4.

We can all agree on this surely ?

The rules of calculus state that the integral of a constant times f(x) is equal to c times the integral of f(x) => ∫c.f(x) dx = c∫f(x) dx.

The rules of calculus also state that the integral of (f(x) +- g(x)) is equal to the integral of f(x) +- the integral of g(x) => ∫(f(x) +- g(x)) dx = ∫f(x) dx +- ∫g(x) dx.

We can all agree on this surely ?

Thus I use the algebra shown in the post to see if it matches the results obtained by plotting Planck curves.

The claims made in the post are :-

1. σ T1^4 = 2 σ T2^4; and,

2. σ T2^4 = 1/2 σ T1^4; and,

3. 400 W/m2 = 3/2 σ T1^4; and,

4. T1 = ~262 K.

So I only used one 400 W/m2 in the Spreadsheet and let it calculate ALL of the temperatures based on the equations cited in the post.

Now if the algebra in the post is valid then a sum of the values given by Planck's equations for the appropriate temperatures should give the results of the algebra in the post.

Thus T1 = (2 x 400/3σ)^0.25 = ~262 K etc - EXACTLY what is written in the post.

I plotted curves for the various equations in the post and compared the results to the algebra.

None of the sums, fractions or multiples produce the results the simple algebraic manipulations claim !

It is also possible to extract a relationship for temperature versus the values calculated by Planck's law.

If the algebra in the post is correct then the results should be a value equal to the initial temperature.

Only the curves plotted for a single temperature display this necessity - the curves for sums, fractions and multiples are NOT Planck curves and therefore such algebraic manipulations are invalid.

I also verified the areas under the curves and the results are what they should be. For example the area under the curve for 262 K by pi is 266.666... W/m2.

I am prepared to forward this spreadsheet to the website for examination and discussion - I just need an email address.

The point I make is that if the algebra in the post is right one should obtain the same results using the Planck curves !

There is no doubt about this mathematically - none at all.

Ross writes:

ReplyDelete"No process is possible whose sole result is the transfer of heat from a cooler to a hotter body."

You did notice the heat source off to the left in all those diagrams? Maybe not, but it is hard discussing things with the conveniently oblivious

Betty:

ReplyDeleteYou do realize that you are specifically saying that whether you put a black body plate in front of your face or not, you still feel the exact same radiation from the sun, right? In other words, you are stating that said black body plate acts perfectly transparent to radiation.

Are you entirely sure this is what you really want to say? Black bodies were never assumed to be invisible to all incident radiation in my books.

That's not what is being said. The plate gains solar energy from 0K (assuming) to 290K. Once it reaches equilibrium, it absorbs and emits equally - not a "transparency".

DeleteWhat is being said explicitly at 9:18 is:

ReplyDeleteSUN -->> plate 1 (temp X) -->> plate 2 (temp X).

The only way for this to happen is for plate 1 to be transparent or effectively so by emitting the same 400W/m^2 it receives on the exposed side on the nonexposed side as well. That is you are saying that for plate 2 the above situation and:

SUN -->> transparent plate -->> plate 2 (temp X)

are completely equivalent. They are not. This is because plate 1 does NOT emit 400 W/m^2 from both sides equally.

At equilibrium, plate 1 (and 2) will be emitting 400 W/m2 from all sides. Equilibrium is not transparency.

DeleteThen since only 400 W/m2 are coming in you deny conservation of energy. Blatant idiocy in pursuit of perpetual denial

ReplyDelete400 in, 400 out on all sides. Both plates heat up to 290K - the max of 400 W/m2.

DeleteYou're projecting. You deny conservation of energy when you get 244K and 262K, when 400 W/m2 results in 290K. Where did 46K and 28K worth of energy disappear to?

DeleteBetty...

ReplyDeleteWell about all I can say to 1:08pm is you obviously don't know enough to come back into the shade!

Does your body emit 98.6F worth of radiation from all sides or does it emit 49.3F front and 49.3F back?

DeleteIs the sun in my body?

ReplyDeleteFalse analogies do not clarify anything.

Applies to every object. The sun is emitting 400 W/m2 at that distance in infinite directions. Does that mean the sun is emitting infinite W/m2 in accordance to your add-'em-up theory?

DeleteIf 400 W/m2 come into the system then 400 W/m2 go out from the ENTIRE system which is what this example. To avoid a valid accusation of politically convenient denial of conservation if energy please tell Eli how much thermal energy is coming out of the sun side of the blue plate, the green plate facing side of the blue plate, the blue plate facing side of the green plate and the space facing side of the green plate.

ReplyDeleteThe answer may surprise you but it is the same process which explains how the surface of Venus is so hot

You deny conservation of energy when you get 244K and 262K, when 400 W/m2 results in 290K. Where did 46K and 28K worth of energy disappear to?

DeleteEnergy is conserved. Temperature is not a conserved quantity. Besides which where did 290k come from? You need to get off those funny little pills

Delete400=sigma*T^4, T=290K

Delete244K and 262K have radiation equivalents of 200 W/m2 and 267 W/m2. You violated conservation of energy. Where did 200 and 133 W/m2 disappear to?

"The Green Plate Effect

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

Well THAT was singularly unsuccessful in the face of complete ignorance and denial!!!

Yeah! Eli refutes himself with his ignorance and denial. It's really funny to watch him create a fake math problem devoid of any physics.

DeleteThe algebra in the post is totally incorrect !

ReplyDeleteThis is provable mathematically using Planck's equation and the relationship between the Stefan-Boltzmann, Planck's and Wein's law!

The ONLY empirical evidence that exists as a result of the cavity oven experiments in the 19th century is that a black body emits radiation in proportion to its temperature.

No empirical evidence exists for any other algebraic manipulations of the Stefan-Boltzmann law !

If such algebraic manipulations of the type performed in this post were VALID then they would be verifiable using Planck's law AND THEY DO NOT SATISFY this necessary requirement.

Simple algebraic sums involving the Stefan-Boltzmann law are not valid so all of the huffing and puffing combined with insulting one another is totally pointless.

I find Betty's points totally supportable !

The standard argument for the greenhouse effect is that the "atmospheric layer" emits the same radiation on both sides in accordance with its temperature because it is a layer.

This post turns that argument on its head - can't you see that ?

Besides it is not valid to sum radiative flux using the Stefan-Boltzmann equation and calculate the final temperature.

This is not saying that there is any problem with IR thermometers as they do not use these algebraic manipulations.

I'll say it again - EVERY time you see some form of algebraic manipulation of the Stefan-Boltzmann equation such as performed here check it out using Planck's law.

If for example the equality as stated in the post -

σ T(1)4 = 2 σ T(2)4 AND

T(1) = ~262 K - which means T(2) = ~220.

Then plotting a Planck curve for T(1) - 220 K - and then multiplying every value by 2 MUST produce a curve for T(2) - 262 K - THE AUTHOR HAS SAID THIS IS TRUE !!!

BUT IT DOESN'T !!

The 2 x T(1) curve does indeed have an area under the curve of 266.66 satisfying the numerical requirement of the Stefan-Boltzmann equation but IT is not a true Planck curve and therefore this algebraic manipulation cannot be correct !

This is the crucial point.

As temperature increases the wavelength of the peak emission shifts to shorter wavelengths and there is no way to account for this FACT using the Stefan-Boltzmann equation ! Further, hot objects emit radiation at shorter wavelengths where the emission from a cooler object is near or zero.

This in no way implies there is anything wrong with the Stefan-Boltzmann equation or its useful applications - it is, however, stating categorically that there is a major problem with the way people use it using simple algebra !

Have a look at either the SB or Planck's equations - why would you expect there to be simple algebraic relationships ?

Try it for yourself using a spread sheet and plot some curves - you will see this is right.

This gives all the guidance needed to confirm what I say:-

http://www.spectralcalc.com/blackbody/CalculatingBlackbodyRadianceV2.pdf

From Betty Pound:

ReplyDelete"With just one plate, the sun warms it to 290K.

With two plates, the sun warms both to 290K."

If you had just the blue plate, and the other side of it was perfectly insulated, then yes, 400 W/m2 would warm it to 290K, and it would radiate 400 W/m2 back. If you then removed the insulation, the other side would also start radiating 400 W/m2. Because the plate is now radiating 800 W/m2, and only absorbing 400 W/m2, it can't stay in equilibrium and will then start cooling. It will reach equilibrium at 244K (absorbing 400 W/m2, radiating 200 W/m2 from both sides).

With Ross McLeod, I can't even tell what position he's defending. Is it:

a) With just the blue plate, it reaches 244K. When you add the green plate, both reach a temperature of 244K

b) With just the blue plate, it reaches 290K. When you add the green plate, both reach a temperature of 290K

c) Something else?

Over at Climate of Sophistry, I think they've decided on option b. Or, at least, the 2nd half of option b; I'm not sure what they think happens with just the blue plate.

The plate is one molecule thick. What other side?

DeleteWhat gives you permission to add radiative fluxes?

Tell us about this magical perfect insulator.

If the plate was an isocahedron would it be emitting 20 W/m^2?

Does your body emit 49.3F worth of radiation (292K) from your frontside and backside, each?

* 20 W/m2 from each face

DeleteEli Rabrtt said...

ReplyDelete"Energy is conserved. Temperature is not a conserved quantity. Besides which where did 290k come from? You need to get off those funny little pills"

YOU are conserving FLUX - W/m2 and that is most definitely NOT a conserved quantity !

If you look at the "net" form of the SB equation for the exchange of energy between 2 objects at T(1) and T(2) -

P(net) = σ (T(1)^4 - T(2)^4) -

the quantity in the brackets is heat - the energy exchanged between them due to the temperature difference.

If P(net) is positive T(1) is decreasing and presumably T(2) may increase.

If P(net) is negative then T(1) is increasing as T(2) decreases.

If they have the same temperature there is no exchange of heat and we call that thermal equilibrium.

Flux is not a conserved quantity.

"where did 290k come from?"

400/sigma = T^4 T = 290 K !

Nothing funny about that at all - the presence or absence of pills or otherwise !!

"The answer may surprise you but it is the same process which explains how the surface of Venus is so hot"

Absolute rubbish !

Venus is so hot because it has a huge number of active volcanoes continuously belching CO2, SO2, CH4 etc into it's already dense atmosphere such that it has a surface pressure of ~92 bar.

Sunlight doesn't even reach Venus's surfaces - the high altitude H2SO4 clouds reflect a significant portion of incident solar radiation and absorb a significant portion of the rest.

Venus albedo is at least 0.8 !

CO2 absorbs IR strongly at ~2.7 and ~4.3 micron and the Solar radiation has plenty of power at those wavelengths at Venus' orbit.

If nothing makes it to the ground it can't be a "greenhouse effect" can it ?

Harry Dale Huffman wrote this years ago - I've summarized it here:-

The Magellan spacecraft that took measurements of Venus' atmosphere as it descended to the surface before being crushed by the huge pressure ! It's readings show that the temperature high in Venus' atmosphere where the pressure is equivalent to 1 atmosphere on Earth - ~1000 millibar - is roughly ~339 K compared to ~288 K for Earth at the same pressure.

Venus is closer to the Sun - obviously. The average orbital radius of Venus is 108.21 million kilometres and for Earth it is 150 million kilometres - NASA Fact Sheets.

Using the inverse square relationships Venus receives roughly (150/108.21) squared = ~1.92 times the power per unit area that Earth receives.

Using the SB law Earth's ~288 K at one atmosphere - 1000 millibar - is raised by the 4th root of ~1.92 = 1.177 times ~288 = ~339 K which is the temperature the Magellan spacecraft recorded at equivalent pressures !

Coincidence ?

My position is simple Quokka -

ReplyDeleteAlgebraic manipulations using the SB equation to sum fluxes and calculate temperature do not agree with black body mathematics and Planck's equation.

If the algebra is right it must achieve the same result when applied to Planck's equation.

It doesn't and the result is obvious to see.

I've explained the mathematical relationship as set out in ALL the texts so there is no argument possible over this.

Try it yourself or remain closed minded - I don't care really.

Everyone should ALWAYS remember that these though bubbles have no bearing in reality.

ReplyDeleteThere is NO simple relationship between a so called energy balance and temperature in reality.

Different things heat up to different temperatures depending on their properties.

Surely the difference between the solid land and liquid oceans shows just how wrong this type of algebra is ?

It is nonsense !

The same radiation insolation that causes a 70.1°C surface temperature can NEVER raise the ocean temperature above ~36°C !

Yet the energy of evaporation is released without EVER achieving similar temperature increases !

At STP water will never exceed 100°C. Move higher in the atmosphere and the boiling point decreases markedly.

Seriously the algebra in the post does not work and the whole proposition has no basis in reality.

ReplyDeleteFirst Betty answers: "where did 290k come from?"

400/sigma = T^4 T = 290 K !

The problem with this is that the area radiating from the blue plate is twice as large as the area being illuminated. Only one side of the blue plate is being irradiated but when both sides are at some temperature T both sides will radiate energy. This would be the solution to a problem where the blue plate had only a single side.

Thus the proper equation

400 W/m2 x A = 2 A Sigma T^4

The area of the blue plate on both sides of the equation cancel.

If you want to change the problem, you have a different problem

Everything that followed from the wrong statement of the problem is chaff.

The plate is one molecule thick. There is no other side.

DeleteWasn't Stefan-Boltzmann Law derived empirically? Why didn't they divide by two? Surely the equipment had more than one side.

Your calculation is incorrect. The plate receives 400wm2 and radiates same in all directions.

Ross says:

ReplyDelete"There is modern experimental evidence which verifies the 2 centuries old experiment demonstrating that the radiation from a cold object does not induce warming in a warmer object !"

Please desist with mis-stating what was discussed in the post (Do it again and you are a simple liar). What was shown is that when a surface is heated at a constant rate, introduction of a second surface which captures some of the thermal radiation from the first surface and re-radiates a portion to the first will raise the equilibrium temperature of the heated surface.

In spite of your thrashing around, this does not violate the second law of thermodynamics and can be used to illuminate the role of greenhouse gases in maintaining the Earth's surface temperature.

Betty says:

ReplyDelete'You are dropped off in Antarctica. How do you feel cold?

Because YOU are warming the environment.

The sensor detects thermal energy flowing from it to the colder object."

On NET as Clausius said:

"What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one."

Go argue with him

Old language. Energy can flow both ways, while heat only one way. Use modern physics.

DeleteRoss says:

ReplyDelete'There is another version of the second law and I think the scenario violates this one as well.

No process is possible whose sole result is the transfer of heat from a cooler to a hotter body."

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

You left out the fact that one side of the blue plate is being heated by a power source. And as Clausius said (see above)

"What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one."

Go argue with him

Betty says:

ReplyDelete"The plate in my kitchen is at room temperature. It radiates 20C worth of energy in all directions.

Eli comes along and claims that that it actually emits 10C up and 10C down, and foolish people believe him. LOL"

If you heat the plate (use a metal one) on a gas stove, the top and the bottom will glow.

I won't sink to insulting or misrepresenting what anyone writes.

ReplyDeleteI say I have a spreadsheet which proves that all of the algebraic manipulations on this post are incorrect. None of them produce valid Planck curves for the sums etc of differing radiation values - NONE of them !

I'll supply this to anyone who wants it - it is ~2.2 MB and opens in Excel or OpenOffice or LibreOffice. (I guess it is up to the moderators to allow an exchange of documents etc.)

Prove my claim is wrong - ought to be simple !

EliRabett said..

ReplyDelete"Ross says:

"There is modern experimental evidence which verifies the 2 centuries old experiment demonstrating that the radiation from a cold object does not induce warming in a warmer object !"

Please desist with mis-stating what was discussed in the post (Do it again and you are a simple liar)."

I cannot believe you can be so angry that you call someone a liar for stating a simple TRUE FACT !

I said - "There is modern experimental evidence which verifies the 2 centuries old experiment demonstrating that the radiation from a cold object does not induce warming in a warmer object !"

This is FACT and it is TRUE. Pictet proved this more than 200 years ago !

You ARE claiming it is the back radiation from the green plate that makes the blue plate hotter no matter how you obfuscate your claims so this is relevant !!!

I simply made a factual statement about a 200 year old experiment that was replicated in February 1984 and the historical context of the science of the late 1790's early 1800's that lead to the original experiment was published after being accepted for re-publication in the American Journal of Physics in August 1985 !

I still say that experiment trumps thought bubble any day !

This document is available here :-

http://www2.ups.edu/physics/faculty/evans/Pictet%27s%20experiment.pdf

I am decidedly and provably NOT a liar and I do not "drop my bundle" when challenged !

I still say disprove my claims about the spreadsheet proving the algebra used in this post is wrong !

No need for any of the continual claim and counter claim.

I say your algebra fails mathematically and I can prove it !

I offer the proof to anyone so they can prove me wrong !

Ross darlin, the sum of two Planck curves is not a Planck curve, just like exp(ax) + exp(bx) is not an exponential. Learn some math.

ReplyDeleteBetty, Eli's plates are perfectly black and absorb all radiation that falls on one side. They don't have to be one molecule thick just their area A >> t^2 so edge effects don't matter and have high thermal conductivity so the temperature of one side is equal to the temperature of the other.

ReplyDeleteI dumbed it down for you.

Delete

ReplyDeleteBetty, the Stefan Boltzman relationship was derived empirically by Stefan and theoretically by Boltzmann from thermodynamics https://physics.stackexchange.com/questions/319861/boltzmann-s-original-derivation-of-the-stefan-boltzmann-law

Did they divide by 2?

DeleteWhat does SB Law describe?

The motion of molecules at a given level of radiation. A solid plate will conduct. All the molecules will dance in tune with the incoming frequencies. The max (equilibrium) temp is 400=sigma*T^4 , T = 290K

Betty: "The plate is one molecule thick. There is no other side."

ReplyDeleteConsider graphene

Anyhow that was not the statement of the problem which you are trying to distort

Responding to Betty at 5:43 PM:

ReplyDelete"The plate is one molecule thick. What other side?"

A one-molecule thick wall would not behave as a black body, so no, that's not the assumption. Light would sail through it mostly unimpeded, it would be transparent. Assume plate is thick enough to behave as a black body (the thickness of a sheet of alfoil would be enough) and thin enough that any temperature difference between the two sides is negligible. Such a plate

doeshave two sides."What gives you permission to add radiative fluxes?"

That would be the principle that says that blackbodies absorb all radiation falling upon them. Read the definition of a blackbody.

"Tell us about this magical perfect insulator."

Of course there's no perfect insulator. If it will make you happier, assume a good insulator (e.g. aerogel) thick enough that conductive loss can be ignored, to a first approximation. This sort of pointless nitpicking is just a waste of time.

"If the plate was an isocahedron would it be emitting 20 W/m^2 from each face?"

Why on earth would anyone think that? Obviously, if the icosahedron is at 244K, each bit of its surface will radiate at 200 W/m2, so multiply by the surface area of the icosahedron if you want the total rate of heat loss. For our thin wall, that also radiates at 200 W/m2 for the entire surface at 244 K. Since 1 m^2 of wall has 2 m^2 of surface area, it radiates 2 * 200 W/m2, in equilibrium with the incoming energy (based on the Stefan-Boltzmann equation).

How you would calculate results for such scenarios by plotting Planck curves, I can't imagine. All I'm relying upon here is the 1st law of thermo (conservation of energy), and the definition of a blackbody.

From Rosco: "Simple algebraic sums involving the Stefan-Boltzmann law are not valid".

No idea what you're talking about. Please give an example of one of these supposed sums. All I'm using S-B for is to give the power output of a surface at a given temperature.

Once again: is it possible to get a straight answer as to what you think the solution is, both for the single blue place and the blue and green plates?

It received 400. why is it emitting 200 from each face?

DeleteOh, here you don't multiply 200 by 12? Funny that you treat it differently from a flat plate.

You are wrong.

The plate you describe will still emit 400 in all directions at equilibrium (290K). And you can't add 400 to 400 to create a strawman 800 value to falsely disqualify it - otherwise you admitted an icosahedron will emit 20*200=4000 W/m2 (but receive 400 W/m2 ... cooling)

*12 = 20

Delete"That would be the principle that says that blackbodies absorb all radiation falling upon them. Read the definition of a blackbody."

DeleteSo you can add up the radiation from two bodies and get a higher temperatures? What is the temperature of two ice cubes vs one?

Ross: Picet did NOT consider the case where one of the bodies was receiving a constant amount of thermal energy from another source as Eli described in this post. As a matter of fact, Picet's case was considered in detail by Clausius in his work on the Mechanical Theory of Heat Chapter XII, pp 335 where he points out:

"What further regards heat radiation as happening in the usual manner, it is known that not only the warm body radiates heat to the cold one but that the cold body radiates to the warm one as well, however the total result of this simultaneous double heat exchange is, as can be viewed as evidence based on experience, that the cold body always experiences an increase in heat at the expense of the warmer one."

ReplyDeletehttps://ia600309.us.archive.org/0/items/diemechanischewr00clau/diemechanischewr00clau.pdf

Stop running your squirrel through Eli's patch.

Pictet's experiment proves that heat radiates from hot to cold.

ReplyDeleteWhen Pictet previously showed that a heated object at the focus of the mirror caused an increase in temperature registered by the thermometer he proved that heat radiates from hot to cold.

I note he was not the first to perform such experiments with a heated object but he was the first to perform the "reflection of cold" experiment.

When the flask of ice was situated at the focus of the second mirror the thermometer became the source of heat and it decreased in temperature.

This is entirely relevant !

This post is claiming the "back" radiation from the green plate heats the blue plate despite the green plate being at a lower temperature.

This is exactly what is being claimed !!!

I have not insulted any one and I am not making irrelevant comments. I am not "running your squirrel through Eli's patch".

I am offering realistic arguments that conflict with the claims of the post which I still say I can prove are wrong using mathematics involving Planck curves!

It doesn't matter whether or not there is a continuous amount of thermal energy input the whole claim of the post is that the back radiation from the green plate causes the blue plate to heat up !

It is deceptive to claim it isn't - the initial starting point had the input of 400 W/m2 causing the maximum effect it can on its own ! It was explicitly stated !

Quokka aid :-

"From Rosco: "Simple algebraic sums involving the Stefan-Boltzmann law are not valid".

No idea what you're talking about. Please give an example of one of these supposed sums. All I'm using S-B for is to give the power output of a surface at a given temperature."

If you can't see the inherent sums, multiples and fractions involving the SB equation in the the algebraic manipulations written as equations in the post then you must be blind !!!

The algebraic manipulations in this post CAN be shown to be mathematically incorrect using Planck curves !

There are only 2 possibilities :-

1. Algebraic sums DO NOT provide the correct answer when using the SB equation.

σT(1)^4 does not equal 2σT(2)^4.

Using the claims of the post T(1) equals the temperature of 400 W/m2 = 3/2 σT(1)^4 = ~262 K and therefore T(1) equals about 220 K.

Plot these 2 curves and then multiply every radiant emission value calculated for every

increment of wavelength by 2 and plot this curve.

The curve for 2σT(2)^4 has an area under the curve equal to the area under the curve for T(1) - 266.66... W/m2 BUT it is not the same curve for the T(1) curve = 262 K !

Peak emission is not shifted to shorter wavelengths by this simple algebraic multiplication AND the curve for 2 x T(2) is NOT contained completely within the curve for T(1) as EVERY other legitimate Planck curve IS !

OR -

2. The explicit proven mathematical relationships between the Stefan-Boltzmann, Planck's and Wein's laws are invalid.

My money is on 1.

Prove me wrong !

Typo alert

ReplyDelete"Using the claims of the post T(1) equals the temperature of 400 W/m2 = 3/2 σT(1)^4 = ~262 K and therefore T(2) equals about 220 K."

Again the maximum initial temperature the input is capable of on its own is the 4th root of 200/σ which equals 244 K as explicitly stated in the post !

Picot's experiment does not consider the case where the hotter body is receiving a constant input of thermal energy. Clausius analyzed it in detail. You can use Google translate.

ReplyDeleteIt's squirrels all the way down with you.

Quokka did a better job than Eli

ReplyDeleteBetty asks: "So you can add up the radiation from two bodies and get a higher temperatures? What is the temperature of two ice cubes vs one?"

ReplyDeleteWell, let's see. Ice is at 273K, so if the entire sky (as viewed from a surface) was ice, it would supply 315W. So consider a surface at 0K, facing a sky at 0K. Introduce an ice cube, occupying 1/1000th of the field of view. It will therefore supply 0.315W (on average; more at zenith, less if low on the horizon), which equates to a temperature of 48.5K (way higher than the cosmic background radiation!). A second ice cube will double the power to 0.63W, equating to 57.7K. See, not hard.

10,000 ice cubes and you get 3150W or 485K or 212C

DeleteYou can boil water with 10,000 ice cubes!

You just solved the world energy crisis. You must patent right away.

I demand a 10

I demand 10% royaltee fee for inspiration.

DeleteWhen will you be showing us this device?

Betts..

ReplyDeleteYou've already invented a very special black body that can take in 400W/m^2 on one side and then emit a total of 800W/m^2 from 2 sides which could then be harvested by suitable collectors stationed above and below said special black body. I should think you'd be rich enough already. Don't be greedy.

We've already established that fluxes can't be added, just like twmperatures can't be added. Try to keep up. 400+400=400 W/m^2

DeleteBetts...

ReplyDeleteFurther to your achievement, let's set up the following machinery:

CollectorA <<-- PlateA <<-->> PlateB -->> CollectorB

Now let's heat up PlateA and PlateB such that they really are emitting 800W/m^2, 400W/m^2 to the left and 400W/m^2 to the right. CollectorA and CollectorB harvest 400W/m^2 each, of course.

Now lets turn off the energy to the two plates. What happens under your theory?

Perpetual energy production!!! Plate A collects 400W/m^2 from PlateB, passes on 400W/m^2 to CollectorA while emitting 400W/m^2 back at PlateB which then does the same thing the other way!!!

Congratulations on your achievement!!! You will never need to collect royalties from anyone else ever again!!!

[Note: I'm ignoring some nasty positive feedbacks during the stage where both plates are being powered, but they also point out the absurdity.]

We've already established that fluxes can't be added, just like temperatures can't be added. Try to keep up. 400+400=400 W/m^2

DeleteBetty responds: "10,000 ice cubes and you get 3150W or 485K or 212C. You can boil water with 10,000 ice cubes! You just solved the world energy crisis. You must patent right away."

ReplyDeleteThe flaw in this reasoning is trivially obvious, but I'll spell it out for logic-challenged. The maximum amount of energy you can get from ice, 315 W/m2, is if the entire hemispherical field of view is filled with ice (i.e. 1000 cubes). Where were you planning to put the other 9000 cubes, unless you can find a hemispherical field of view of 20*pi steradians? (cf. the usual 2*pi steradians we get in our universe)

You 'forgot' to point out any error in my calculation with 1 and 2 ice cubes. That's because you can't, of course. But please do show us your calculation of how much energy an ice cube would supply, I'm sure we'd all enjoy a laugh.

From Betty: "Did they divide by 2? What does SB Law describe?"

ReplyDeleteIt describes how much energy is emitted from a surface, depending on its temperature. Whether that surface is on a thin plate, a thick plate, or a solid object such as a sphere is irrelevant (because they do not affect the two inputs to the SB law, temperature and surface area). But if that surface is on a plate, then we can also apply the SB law to the surface on the other side of the plate. If the whole plate is at 290K, then it tells us that both surfaces independently emit 400 W/m^2. If there is only 400 W/m^2 being absorbed, then the plate is losing 400 W/m^2 energy and therefore must cool down (due to an obscure principle, which you may even have heard of, called the Law of Conservation of Energy).

The idea that an icosahedron with 20 faces at 133C would only emit 20 W/m^2 from each face is of course ridiculous. The S-B law doesn't depend on how many planar faces there are, it cares about temperature and surface area. If we took our plate and put slight bends in it to form a concertina shape with many planar surfaces, the amount of energy it radiates would be unchanged because neither the temperature nor the surface area have changed.

Do you still claim each face emits 200? How much is each face emitting assuming the icosahedron is radiated on one face (the rest masked by "perfect" insulator). Answer the question, otherwise I will presume your old answer 20*200=4000 W/m^2.

Delete

ReplyDeleteThe 400 W/m2 is the input to the blue plate. It could be illumination, it could be an internal electrical heater or it could be a hotter body off somewhere to the left of the diagrams. The only important thing about it is that is a constant source of heat. It is the ONLY input of energy to the system, whether the system is the blue plate alone or the blue and green plates together

The surroundings are everything EXCEPT the blue plate alone or the blue and green plates when the green plate is present

That means that at equilibrium the system has to be rejecting 400 W/m2 to the surroundings to conserve energy. The amount of energy emitted from the system per second can never exceed 400 W/m2 but it can be less while any part of the system is heating up.

Pretty much everybunny who tries to falsify the example misses or misuses this

Quokka says:

ReplyDelete"Introduce an ice cube, occupying 1/1000th of the field of view...."Betty makes an interesting (but misguided) reply:

"10,000 ice cubes and you get 3150W or 485K or 212C"Betty, 1000 cubes that each cover 1/1000th of the field of view means the entire field of view is already covered by the first 1,000 blocks of ice. You basically have an igloo with a surface on the floor of the igloo that is receiving the radiation. The floor will receive 315 W/m^2 and will warm/cool to 0C (assuming it is insulated on the bottom side). A new ice block -- no matter where you place it, cannot cover more that 100% of the field of view and cannot provide more that 315 W/m^2 of radiation to the floor. Any blocks beyond the first 1000 will either be

* on the OUTSIDE of the igloo, and provide no radiation to the floor.

* on the INSIDE of the igloo, in which case they they do radiate to the floor, but simultaneously block and equal amount of radiation from the ice behind them.

This geometry seems to escape many people, and it leads people to incorrect conclusions. 10,000 blocks of ice in this situation would make the igloo's walls *thicker* but this does not provide more radiation.

What a bunch of assholes on this forum. They all become pedantic to a specific experiment setup.

ReplyDeleteQuokka arbitrarily chose 1/1000th field of view. Well, I can arbitrarily choose 1/10000th, and claim he invented an igloo oven.

Rather then call him out for his crackpot physics of adding fluxes, assholes actually defend him.

In his "physics" the size of the ice cubes, doesn't matter either. Did you notice that? So one can still fit 10,000 ice cubes 1/10th the original size.

Quokka, when are you going to patent your igloo oven?

One might more reasonably ask when are you going to patent your perpetual energy generator!

ReplyDeleteI just proved fluxes can't be added. Try to keep up. Your false setup doesn't produce perpetual energy. You are just deluding yourself with antiphysics.

DeleteFlux times area can be add Betty. Do try and keep up. If the areas of the two plates are the same, you can even add the fluxes as in F1*A + F2*A = (F1 + F2)*A and cancel the areas. Goodnight

ReplyDeleteI should have said flux densities. (W/m^2)

DeleteFlux is Watts.

You just added Watt*m^2.

Physically irrelevant. What are you doing?

If the plate is 1 m^2, then

Delete(400+400 W/m^2)/2 m^2 = 400

This comment has been removed by the author.

ReplyDelete"Well, I can arbitrarily choose 1/10000th"

ReplyDeleteYes, Betty, you could and then each would only provide 1/10 as much flux as before and the whole shell would still provide the same total flux of thermal IR. And you could still not exceed 315 W/m^2 total.

It is NOT about one specific thought experiment -- it is about general principles. Whether we divide the shell in two parts or 10 parts or 1000 parts or 10000 parts, the total shell sill provides the same total power.

"We've already established that fluxes can't be added"I didn't see anyone establish this as a general principle. It is more subtle.

* Fluxes from DIFFERENT directions can and do add. If we focus two different spotlights from two different directions down onto the ground, it will be brighter than one spotlight and will warm the ground more than a single spotlight. If we focus the flux from a thousand mirrors onto a tower (like a concentrated solar thermal plant), then the fluxes add and the tower gets much hotter than the flux just of the sun.

* Fluxes from the SAME direction cannot and do not add. If one spot light was placed directly behind the first, then the spot on the ground does not get brighter (because the 2nd spotlight is merely lighting up the back of the first spotlight!).

With the ice example, the first 1000 ice cubes which each cover 1/1000 of the hemisphere can add if they are carefully placed. We can get up to 0.315 X 1000 = 315 W/m^2 of radiation onto the ground. This is the best we can do with ice. We have thermal radiation already coming from every direction. Any more such ice cubes is the equivalent of putting one spotlight behind the other.

So what you're saying is that adding the flux densities emitted from each cube ...

Delete315+315+315+...+315 = 315 W/m2 max

That's what I was saying. You can't add flux densities.

You're too attached to the arbitrary setup of Quokka.

DeleteIf you place one-molecule thick graphene between two ice cubes, then according to Quokka et al this will receive 315+315=630 W/m2, and heat to 51C.

In my reality 400 + 400 = 800.

ReplyDeleteYours obviously varies.

800 over twice then area = 400 from each side.

Delete* 800 over twice the area = 400 W/m^2

DeleteIf 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.

ReplyDeleteIn your plate experiment, you have 200 leaving each side. The surface area of the plate is 2 m^2. Your flux is thus (200+200)/2 = 200. That's not the 400 necessary. You failed, big time, buster bunny.

"800 over twice then area = 400 from each side."

ReplyDeleteThat's precisely what leads to perpetual energy!!!

800 over 2 m^2 = 400 W/m^2

DeleteTry to keep up, dumb dumb.

"Dumb dumb"!!!

ReplyDeleteQuokka, for the ice cubes you divided by 1000 (then multiply by a thousand) to get 315 ... but for the icosahedron you don't divide by 20 (then multiply by 20) to get 400. How convenient.

ReplyDeleteEliRabett said...

ReplyDelete"Picot's experiment does not consider the case where the hotter body is receiving a constant input of thermal energy. Clausius analyzed it in detail. You can use Google translate.

It's squirrels all the way down with you."

Firstly it is Pictet.

Secondly your claim is nonsense.

Pictet allowed the air thermometer to come into thermal equilibrium with the air of the laboratory and that air temperature did not change to any significant extent. This was evidenced by the thermometer warming back to its initial temperature when moved away from the focus of the mirror.

Of course the thermometer is receiving a constant input of thermal from the air of the room via contact !!

And if your scenario is right when the flask of ice is placed at the second focus the thermometer is receiving more energy reflected by the two parabolic mirrors because the ice is emitting energy that was not present before it was placed there.

Thus the thermometer did indeed have a constant supply of thermal energy and placing the flask of ice at the focus supplied extra energy that was not there before.

Yet the ONLY reaction was a decrease in the temperature of the thermometer despite the extra thermal energy emitted by the ice to the thermometer.

About to break my policy !

If I'm squirrels you're nuts !

EliRabett said...

ReplyDelete"Ross darlin, the sum of two Planck curves is not a Planck curve, just like exp(ax) + exp(bx) is not an exponential. Learn some math."

Firstly, if the only argument you have is condescension then you've damaged your position before you've even begun.

Secondly I never made any stupid arguments "like exp(ax) + exp(bx) is not an exponential"

That is a typical tactic of someone who cannot argue against a plainly stated case - misdirection to ridicule when their opponent NEVER said any such thing !

I explicitly said there is a relationship between Planck's, the Stefan-Boltzmann and Wein's equations !

The Stefan-Boltzmann equation was derived empirically BUT it is also derived from Planck's by integration - THIS IS FACT !!!

Wein's equation was derived empirically BUT it is also derived from Planck's by differentiating Planck's equation and setting the expression to zero to establish peak emissions - THIS IS FACT !!!

These FACTS tend to strengthen the validity of each equation !

Bearing these FACTS - and they are the ONLY FACTS I submit as evidence - in mind let's analyse the math quoted here.

At T(1) = ~262 K the radiant emission is 266.66... W/m2.

At T(2) the radiant emission of 133.33...W/m2 is the emission from a temperature of ~220 K.

"The bunnies can rearrange the second equation to get

σ T(2)4 = 1/2 σ T(1)4"

So plot a curve for ~220 K - or better yet program the spreadsheet to do ALL calculations for you using its maximum precision and just input the accepted values for the SB constant, Planck's and Boltzmann's constants and the SB and Planck's equations etc.

Thus the only input is 400 and the various ratios as stated in the post - 2 times, 1/2, 3/2 etc.

Thus there is no denying that at ~220 K the green plate is emitting ~133.33 W/m2 and this numeric value is indeed half of what the blue plate is emitting at ~262 K which is indeed calculated to be ~266.66 W/m2 - AND I NEVER HAVE !!!

BUT there is another consideration in the algebraic "equality" "σ T(2)4 = 1/2 σ T(1)4" that the Stefan-Boltzmann equation CANNOT illustrate - only Planck's equation can !

Plot the 2 curves - ~262 K and ~220 K.

The area under each curve is σ T(1)4 and σ T(2)4 respectively !

This is indisputable - read the reference at SpectralCalc - some people who do know math !

The laws of calculus state that the integral of 1/2f(x) is the same as 1/2 times the integral of f(x).

So if I multiply every "y" axis value (Spectral Radiance) of the T(1) curve for the corresponding "x" axis value (wavelength, wavenumber, frequency or whatever is chosen) by 1/2 then this is EXACTLY equivalent numerically to 1/2 σ T(1)4 !

This is indisputable ! If you do dispute this then you don't know math !

And the curve for this mathematical transformation should be EXACTLY equivalent to "σ T(2)4" - Eli says it is so !

But it isn't !!

And that is the point Eli darlin !!

What I say is valid mathematics and it proves your assertions about adding up flux and calculating the temperature from the algebraic sums using the SB equation are wrong !

Don't say there are no sums or subtractions in the post as that is just a ridiculous assertion primary school children can refute !

There is nothing wrong with my math !

Eli Rabbet, you violated conservation of energy.

ReplyDeleteYou have 400 W/m^2 going in on 1 m^2 side, while 400 W / (2 m^2) leaving. You lost 200 W/m^2. Your "equilibrium" is not an equilibrium.

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

Delete

ReplyDeleteNo, Ross, the air is a constant TEMPERATURE heat bath which is exposed to the thermometer. The net amount of heat flowing into the thermometer from the air depends on the temperature difference between the air and the thermometer.

Betty, I have a really hard time following your points. To get to the core of these issues everyone needs to be more careful.

ReplyDeleteClarifying between

* the total power (Watts of IR)

* the flux density (W/m^2 of IR)

* (it doesn't help that both of these ideas are sometimes referred to as "flux" in different situations!)

Clarifying between

* incoming power and/or flux density

* outgoing power and/or flux density.

Clarifying

* exactly which object

* exactly which surface of which object.

* the orientations & positions & separations of various surfaces.

The original scenario provides clarity and is properly done. The various later ice scenarios are less clear and hence much tougher to analyze and to comment intelligently on.

As a start:

* outgoing flux densities from a flat bit of surface depend only on the temperature and emissivity. It makes no sense to try to add outgoing flux densities. (for example, the outward flux density from ice @ 0 C is ~ 315 W/m

* incoming flux densities to a flat bit of surface do add. You have to integrate the incoming flux densities over all the bits of solid angle.

"It makes no sense to try to add outgoing flux densities."

DeleteEli does it. 200+200=400, remember?

"Quokka arbitrarily chose 1/1000th field of view. Well, I can arbitrarily choose 1/10000th, and claim he invented an igloo oven."

ReplyDeleteYes, of course you could choose 1/10000th, either by using smaller ice cubes, or putting the ice cubes further away so that it now takes 10,000 ice cubes to cover the sky, rather than 1000. The point, which others have already made so I won't belabor it unduly, is that the maximum amount of energy that can be supplied by ice is 315 W/m^2, which happens when ice fills the entire field of view. Whether it is filled by one big ice cube close up, or trillions of them miles away, doesn't matter.

"Rather then call him out for his crackpot physics of adding fluxes, assholes actually defend him."

OK, show us your non-crackpot calculation of how much radiation an ice cube occupying 1/1000th of the sky provides. And then use that formula to calculate the radiation from two ice cubes near each other. I think you'll find that the amount of radiation doubles. Or are you really claiming that if we put a second sun in the sky, it wouldn't get any hotter?

"I just proved fluxes can't be added."

Huh, must have missed that. Which message was it exactly?

You answered differently for isocahedron.

DeleteYou set up the 1/1000th field of view example, I did not. My initial question had in mind material between two ice cubes:a sandwich, NOT 1 or 2 cubes really far away from a POINT.

Ross says:

ReplyDelete"And the curve for this mathematical transformation should be EXACTLY equivalent to "σ T(2)4" - Eli says it is so !"As near as i can tell, Eli only ever claimed that the integral for one curve is twice the integral of the other. Such a claim in no way implies that one integral is exactly twice the other at every point.

"What I say is valid mathematics and it proves your assertions about adding up flux and calculating the temperature from the algebraic sums using the SB equation are wrong !"

Why would this invalidate anything? If an object is radiating some amount of power like 267 W/m^2 (ie T=264 K), it will have a specific Planck curve related T=264K. It doesn't matter where it got the power. It might be from 5800 K sunlight. It might be an electric heater. It might be partly from 290 K radiation and partly from 220 K radiation. The outgoing Planck curve does not need to be the sum of the incoming Planck curves at each and every wavelength.

Eli, Your math says that having moving towards having infinite plates would lead to an equilibrium of T=290K for the first plate. What is the difference between a thick object and lots of plates? For example,

ReplyDeleteWhat is the difference between 100-molecule thick plate and 100 plates? Is not each column of molecules a plate?

Eli, you violate conservation of energy. Your plate is emitting 200 W/m2 toward the sun at the same time it's receiving 400 W/m2 - so it's really only receiving 200 W/m2.

ReplyDeleteThat's brilliant. By your logic the SB law must always have a 1/2 term. The equipment measuring solar energy must always show HALF of what they actually show, by your logic, because said equipment obviously has another side.

Betty asks: "What is the difference between a thick object and lots of plates?"

ReplyDeleteSpacing!

With 100 thin plates (for example 0.5 mm thick aluminum spaced 1 mm apart, painted black on each side to make the emissivity close to 1), there is empty space between each, preventing conduction. With a single plate with the same amount of material (50 mm thick with black paint on the outside to match the previous example), conduction will keep both sides of the plate nearly the same temperature.

Space is a heating mechanism! Wow!

DeleteLook, my argument was that the plate(s) reach an equilibrium of 290K. Conduction aides distributing the 400 W/m2 (290K) to the rest of the molecules of the plate - as you seem to agree.

Eli's argument was that the front plate reaches 244K alone, and 262K with another plate. As you add more plates you approach 290K.

So Eli and I agree for many plates. But it seems like you are arguing that SPACE is a warming agent. This looks very stupid.

"conduction will keep both sides of the plate nearly the same temperature."

DeleteThis seems like you agree with me that plate will receive 400 W/m2 on one side and emit (nearly) 400 W/m2 on the other.

This comment has been removed by the author.

ReplyDeleteNo, 400-200=200.

DeleteHe's saying the plate receives 200, and passes 200 through.

You are halfway there. Yes, it's receiving 400 and emitting 200 towards the Sun. The other 200 is emitted away from the Sun. This makes energy in = energy out thereby conserving energy as required by law.

ReplyDeleteIf you want 400 to be emitted away from the Sun, construct your plate out of perfectly transparent material. If you want 400 to be emitted towards the Sun, construct your plate out of perfectly reflecting material. If you want 200 to be towards the Sun and 200 to be emitted away from the Sun, construct your plate out of black body material. It's really that simple.

Any instrument measuring flux would of course need to be calibrated such that its own internal physics were not a biasing factor. That's hardly an issue. Or rather, it's an everyday issue with every measuring instrument.

Conduction and convection are not relevant to a discussion of radiation except in that at equilibrium, internally a black body itself is all at the same temp by definition.

As soon as the plate receives 400 from the sun, Eli claims it emits 200 toward the sun. So it only got 200.

DeleteThis is why two objects of the same temperature don't heat each other. Your claiming it should because, one receives radiation from the other. Well no, only the differential matters.

Since every instrument has two sides, why not include 1/2 term into SB Law?

DeleteWhen you say "pass through" you are not talking about black body radiation any more. You are talking either nonsense or conduction. It's not clear which.

ReplyDeleteYou simply don't understand the underlying definition of a black body, it appears.

pass through = blackbody absorbs and re-emits.

Deleteor absorbs, conducts, and re-emits. It just refers to emission on the other side after going through the material. You're not arguing that energy doesn't pass through, are you? Does it teleport to the other side?

Delete1. Instruments have many configurations. Your notion here is just plain silly.

ReplyDelete2. "As soon as the plate receives 400 from the sun, Eli claims it emits 200 toward the sun. So it only got 200." This is where you clearly show you have no Earthly idea what a black body is.

What anyone who understands the definition would say is that at equilibrium a black body absorbs every bit of incident radiation regardless of direction of incidence and emits an equal amount of radiation equally in all directions. In the case of a plate, only 2 directions are available, of course.

Wrong. Radiation is electromagnetic waves. Once the radiation vibrates the receiving electrons it will jiggle the molecules around. The jiggling of the molecules and electrons doos not automatically lose half its radiation to the very place it got it from. The jiggling would have to slow down for no reason other than your priest, Eli, wants it to.

DeleteThe smallest blackbody is a single molecule.

The internal physics of the black body are irrelevant. What is relevant is the radiation in and out.

ReplyDeleteMolecules radiate to molecules.

DeleteImagine the plate is one molecule. Eli says it gets hotter with a row of molecules with lots of spacing than just one molecule. Space is a heater! LOL

A black body is a theoretical construct and has no internal physics only a defined external physics. That said, it is an interesting physics which posits that individual molecules generally emit radiation in some preferred directions and not others! They're smart little buggers!

ReplyDeleteEmissions generally occur in all directions equally. When the emissions come from a vertical plate, half will go to the left and half will go to the right. Even when emissions come from a sphere, down to and including a sphere containing only a single molecule, half will go into one hemisphere and half into the opposite hemisphere.

Why would half the incoming radiation decide to go back?

DeleteRadiation is just conduction at a distance. A heated metal bar will not recycle half the conduction back to where it came from.

"Radiation is just conduction at a distance."

ReplyDeleteFrom this fundamental definitional error follows all of your incredibly wrong-headed thought processes and completely wrong conclusions. Trying to see any sense behind the nonsense you've said, I intuited, as I said above, you seem to be confusing radiation with conduction. Now you've said it. And again, that is the root of your problem.

Conduction requires a physical, material connection. In this case there is no such connection to allow conduction. Nor is there any gaseous or liquid medium to allow convection to proceed. Radiation equations neglect both conduction and convection. Correctly so. Consider the plates to be existing in a vacuum. There simply is neither conduction nor convection across a vacuum gap.

This is also why radiative energy goes both ways. The molecules simply emit radiation in all directions--half to one hemisphere and half to the opposite hemisphere--regardless of which way the energy came in. There is no way a molecule in the present situation can even "know" which way energy came in in the first place. And even if it did it cannot emit directionally or anti-directionally as a consequence of any such "knowledge".

I punch your stupid face. Which way do the molecules go?

Delete"

ReplyDeleteWhy would half the incoming radiation decide to go back?"John Garland's already pulled the wings off this fly, but the gobsmacking ignorance of Betty Pound's question deserves a moment of contemplative head-shaking...

When two objects are equilibrium they are both sending each other equal amounts of radiation.

DeleteFor example: two ice cubes. Each cube is sending the other 315 W/m2. They cancel out and there is no net heating.

Eli claims the sun sends 400 to the plate, the plate sends 200 to the sun, and yet they are somehow in equilibrium?

That is a contradiction.

BJ: I do sometimes wonder why such deniers bother to stand in the shade when it's really hot out. By Ross/Betty's reasoning it makes no difference.

ReplyDeleteNor does the number of blankets matter on a cold night. 1? 100? It's all the same.

Can any non-pedantic, non-asshole who actually understands physics answer my questions?

ReplyDeleteBetty Pound, I am afraid even the kindest of kindest physicist would be unable to answer in a way that would not insult you. After all, it seems that anything that corrects your mistaken views is considered an insult...

ReplyDeleteBetty, Betty, Betty...

ReplyDeleteThat would be conduction, again. You seem fixated on conduction.

Try shining a nice warm heat lamp on anyone's stupid face. Note the molecules don't go anywhere at all. They would only glow a little more brightly in all directions.

As for your ice cubes, where are they? If they are in deep space they will cool to near 0K but will do so a little more slowly than if they had not been near each other. Eli actually kindly answered this waayyy up thread but I guess the kindness didn't penetrate.

And blankets block convection, to keep you warm. You seem fixated on convection.

DeleteWait, you don't know that when sun shines on the ground, there is conduction to a few meters? That's sad.

If the ice cubes are surrounded by 0K, then obviously they are not at equilibrium. However, they will cool at the same rate, keeping equilibrium between each other.

Do you even know what radiation is? It's not teleportation. It's an electromagnetic wave. Imagine a rope tied to the object you want to heat. You wave the rope up and down. You claim it's possible for your heated object to dull your hand motion by 50%.

Betty, you do actually have a point about the word "equilibrium". Eli should have used the word "steady-state" in the top post.

ReplyDeleteAs stated at Wikipedia "In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either within a system or between systems." The net flows of energy that exist in the top post violate this definition. Strictly speaking, the system is NOT in equilibrium.

But only someone being pedantic would make an issue of this. It is clear what was meant -- the system has stopped changing. Furthermore, it is common for people colloquially to say "equilibrium" when the more correct term would be "steady-state".

fair enough!

DeleteBetty says

ReplyDelete" Eli claims the sun sends 400 to the plate, the plate sends 200 to the sun."The sun would be a very small bit of the sky, and nearly all the energy would be going somewhere OTHER than the sun. The 200 W/m^2 would be going to the 3 K background of deep space.

Now that is a very good point, I had not considered. Thank you, Timothy.

DeleteBecause the sun is a point source, the 200 would negligibly return to the sun - so we can throw it out.

That means the blue plate will reach 244K, not 290K as I previously stated.

Here's the problem, though:

The plates are not a point source - so backradiation will cancel out between the plates.

Blue to green: 200

Green to blue: 100

Means the green only gets 100, impossible.

I think that all plates will reach "equilibrium" at 244K. Space between plates is not a heater.

"I think that all plates will reach "equilibrium" at 244K"

ReplyDeleteHow do you think the 200 W/m^2 gets from the blue plate to the green plate in this case? The green plate is radiating 200 W/m^2 to the right, so it must be supplied with 200 W/m^2 to be at steadystate. But if the two plates are the same temperature (as you imagine), then no heat can flow from blue to green. And there is no other source around.

For heat to get from the blue plate to the green plate, the green plate MUST be cooler than the blue plate.

Right, the green plate gets heated to 244K, and then it comes to equilibrium with the blue plate. I was talking about the post equilibrium scenario, not pre.

DeleteThe discussion is about the green plate causing the blue to warm up further, which doesn't happen. QED.

I was talking post-steady-state too!

ReplyDeleteYou claim that the two plates are the same temperature, yet 200 W/m^2 of heat flows from the blue to the green plate (so that 200 W/m^2 can flow from the green plate to space). This cannot work.

Once achieving steadystate with the point source (sun), the plates can work on achieving Equilibrium between each other. When they do, Each plate (blue and green) radiates 200 W/m2 left and right. The 200 from blue to green, and 200 from green to blue cancel each other out, just like two ice cubes have 315 W/m2 going both ways.

DeletePretty straightforward stuff.

Betty, you are still missing a simple point!

ReplyDeleteYou are correct to say "The 200 from blue to green, and 200 from green to blue cancel each other out". Thus -- by your own words -- there is no net transfer from blue to green. There is also no net transfer from the sun to green -- no sunlight hits green. There is no net transfer to green at all!

And yet ... there is a 200 net transfer from green to space! The green plate is generating 200 W/m^2 out of thin air -- a clear violation of conservation of energy.

"There is no net transfer to green at all"

DeleteThat's what equlibrium means.

"And yet ... there is a 200 net transfer from green to space!"

No. The moment green loses energy to space, the blue transfers energy to green, to maintain equilibrium.

Minor correction:

Delete"There is no net transfer to green at all"

That's what equlibrium means.

"The green plate is generating 200 W/m^2 out of thin air -- a clear violation of conservation of energy."

No. The moment green loses energy to space, the blue transfers energy to green, to maintain equilibrium.

There is still two way transfer of radiation between Blue and Green. But at equilibrium, this implies no heat transfer.

You can't have it both ways!

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

"There is no net transfer to green at all"

EVERY SINGLE MOMENT green is transferring energy to space. Every single movement, blue is transferring energy to green. This is a *steady-state* situation which reaches a constant, fixed temperature difference with a fixed power flow; not an *equilibrium* situation which reaches a constant zero temperature difference with zero energy flow.

And the blue gets energy from the sun. Very good.

Delete"reaches a constant, fixed temperature difference"

And the problem stipulates that the plates are close by, so the gradient is negligible and therefore nonexistent. Obviously if the plates extended to infinity, it would be odd for them all to be at the first plate's temperature.

Have you worked out the numbers for many plates based on Eli's "physics"? It certainly doesn't progress in "fixed temperature differential" chunks.

You can't have it both ways.

Curiously, Tim, I can't tell what you're defending anymore. You still buy into Eli's green plate theory?

Ross says: "And the curve for this mathematical transformation should be EXACTLY equivalent to "σ T(2)4" - Eli says it is so !"

ReplyDeleteAs near as i can tell, Eli only ever claimed that the integral for one curve is twice the integral of the other. Such a claim in no way implies that one integral is exactly twice the other at every point."

You are wrong !

Just how do figure this can work given the explicit relationship betwwen Planck's, the Stefan-Boltzmann and Wein's equations ?

Seriously that is one of the most incompetent statements ever.

This is from the post

σ T(1)^4 = 2 σ T(2)^4 AND σ T(1)^4 is the integral for the Planck curve for T(1).

What is the curve for 2 σ T(2)^4 = σ T(2)^4 + σ T(2)^4 = σ T(1)^4 IF IT IS NOT A PLANCK CURVE ?

If it is NOT a Planck curve then the area under it is NOT σ T(1)^4 and all of the use of the SB equation is invalid!

This is fact !

This comment has been removed by the author.

ReplyDelete"I think that all plates will reach "equilibrium" at 244K. Space between plates is not a heater."

ReplyDeleteWrong. Yes space is not a heater. But what you don't seem to realize is it's a perfect conductive/convective insulator at the same time which with one plate being in the Sun and the other being in the shade prevents them from being at the same temp. Just like going into the shade on a hot sunny day prevents overheating. You are positing a conductive relationship like if the plates were in contact.

"No. The moment green loses energy to space, the blue transfers energy to green, to maintain equilibrium."

No. This is conduction rearing its wrong headed head again. As such, as Tim points out, you have posited a perpetual energy machine yet again.

I'm not going to talk with you anymore since 1/3rd of what you say is nonsense, 1/3rd is superficially true but irrelevant, and 1/3rd mistating my position.

DeleteFurther to "space is not a heater"...

ReplyDeleteI finally may understand your fundamental problem: If the plates are in perfect thermal contact while being illuminated with 400 W/m^2 they exhibit one temp--244K. Now separate them by a slight vacuum space and things change. The green plate cools to 205K as being in the shade it only receives 200W/m^2. The blue plate warms to 262K because of the back radiation of 100W/m^2 from the now colder green plate.

Space is not a "heater" but heat can most definitely can be a redistributed around through it with various ways including half mirrors (which is what a disk black body effectively is). Perhaps if the original article stated the problem in terms of no mirror behind the blue plate, a half mirror (i.e., black body) behind it, and a perfect mirror behind it (leading to a 290K blue plate) you would have understood the situation better?

You can't add 100 from the green to the blue any more than you can add an ice cubes's radiation to the room's radiation.

DeleteYou failed to learn a single thing from this lengthy conversation.

Well that won't necessarily keep me from pointing out your errors. See if you understand what I just posted whether you hold your breath and pout or not. Bothers me not at all!

ReplyDeleteYou claim separating the plates leads to 262K and 205K. What prevents blue from heating green (and cooling itself)? You violated equilibrium.

DeleteI should have made clear: "perfect thermal contact" = conduction.

ReplyDeleteI did make one error: Solving for the temp of the green plate at equilibrium you get 220K, not 205K--I solved for the starting point not the equilibrium point, stupidly.

ReplyDeleteSorry, but the one who is failing to learn is you. Even in space, it matters a lot whether you are surrounded by, but not in conductive contact with, rock at 400C, rock at 200C, rock at 37C, rock at -100C, rock at -200C, or nothing at all. The fact that you don't believe in back radiation or radiation coming from objects colder than other objects means nothing. It's true whether or not you believe it.

It's simply a fact you will radiate away less total heat per unit time when surrounded by 0C rock than when surrounded by near 0K space. You can say it's not true, but you'd be wrong.

Yes, you are correct about equilibrium. 205 is the start point at 200W/m^2.

ReplyDeleteNow that you see this, apply it all the way through and you'll get it.

On closer reading, you got part but not all of it. 262K is the end point for the blue plate. It doesn't cool from there. It is the point where energy in = energy out.

ReplyDeleteFollowing then on through, 220K is the point where energy in = energy out for the green plate.

The point is your criticism is very much on point and will lead you to the right answer if you just follow through the math remembering that energy in always must equal energy out. If there is more going out than coming in, as you consistently have been proposing, you're in the realm of perpetual motion.

No, it is not the end point. Heat flows spontaneously from hot to cold.

DeleteHow did get 220K not 205K?

John, you ADDED radiation from a colder source to a warmer source. Delayed cooling is not heating. You are deluding yourself with math devoid of PHYSICS.

ReplyDeleteSeveral points:

ReplyDelete1. You are thinking in the physics of conduction not radiation. Wrong PHYSICS. Every object in the universe above 0K radiates to EVERY other object. Radiation does NOT "flow from hot to cold". It is everywhere going to and from all objects in radiative sight of each other. That's just how radiation works. You continually deny this, but it really is a simple, observable fact. The math merely describes the fact.

2. Yes, I added the incoming radiation from being surrounded by but not touching, say, 30C rock in space to your outgoing radiation at 37C (assuming a transparent spacesuit). Correctly so. And I added the lesser incoming radiation from being surrounded by -200C rock and noted the second situation results in you radiating your heat away faster. The backradiation from the 30C vs -200C rock defines the half the radiative situation in the first place. Both are cooler than 37C, but "delay" has nothing to do with it.

3. 220K results from substituting 262K (i.e., T sub 1) back into the original simultaneous equations and solving for T sub 2.

3) OK, but 262K still heats 220K. Not equilibrium.

Delete'Radiation does NOT "flow from hot to cold'

Never said it did. You fabricate quotes.

I have just recently come across the topic of GHE and was never aware of how hard it seems to be to grasp even the most trivial points correctly.

ReplyDeleteI can only see two possibilities:

1) Complete conceptual confusion, which even cannot be solved by a proper physical education.

2) Trying to bend the facts intentively.

I'm not able to tell, which of the two is right in the case of some commenters here, especially Betty Pound.

But it is so simple to grasp the main point, that I must assume possibility 2) to be much more probable.

-----

The discussion NEVER was about, if a colder object is able to heat a warmer one by itself.

The question always and only was, if a constant energy source is able to bring an object A to a higher temperature, as long as another object B is present in its proximity, compared to the different case, where object B is NOT present and object A is completely alone with the energy source. Question: Will the steady-state temperature of A differ from the first case to the second, assuming the energy source is completely constant in any case.

If we assume that radiative interaction between A and B is not prohibited, then in the first case ( energy source heating object A, object B in proximity ) object A MUST have a higher steady-state temperature than in the second case ( energy source heating object A in the nothingness ).

This is not violating any law of physics, on the contrary, it is the only way to satisfy the so far known laws of thermodynamic and radiative energy transfer.

And of course in the first case B is NOT heating A in physical terms ( although you might say so colloquially ).

B is slowing the rate of the cooling of A, therefore the constantly incoming energy from the energy source can rise the internal energy of A to a higher level than it would without B in the proximity of A. This results in a higher temperature of A.

Physically spoken, the energy source heats A. A cools radiatively to nothingness. If B is brought in proximity to A, A cools to nothingness AND to B, B heats up in return. Therefore, at steady-state, B will have a certain temperature and thus radiate energy towards A. In return A has then a slower total loss of energy ( since now energy is coming in, that was not coming in before ). This leads finally to a higher temperature, for the constantly energy providing source is unaffected by the exchange between A and B.

So, summing up: The energy source heats A, A cools towards B, B is thus heated by A, as soon as B has a temperature > 0 it will radiate energy. Of this energy some reaches A, thus slowing the radiative energy loss of A, thus the ENERGY SOURCE (!!!) heats up A to a higher steady-state.

So it is the energy source heating everything, not B heating A. Nevertheless A will end up with a higher temperature when B is there compared to the absence of B.

It needs much words to describe this, so a picture is better. But as the comments show, even the most simple picture cannot do its job if the confusion is too big or the intention is to strong.

-----

The only way out for those who seem to deny those effect, is as far as I can see to say that the radiation from B cannot be absorbed by A. Or to break it down to a discussion about semantics, for when it would be right to use the term "heat".

Well, then they have invented new physics and can go have a shot for the Nobel Prize.

You violated the laws of physics.

DeleteObjects can't warm themselves by their own reflected sun-warmed radiation.

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

In Eli's case you are left with plates at 262K and 220K. You then need to come up with BS rationale for why the 262 can not then warm the 220.

You need to understand the difference between rhetoric and physics.

Betty: The second law of thermodynamics is merely a summary of the consequences of the detailed, physical, causal mechanisms of the universe. Eli's example and the multiple responses to you by commenters, are explanations of those mechanisms. Correctly understanding and interpreting the summary (the second law) requires understanding the multiple of those detailed, physical, causal, mechanistic explanations, because there are multiple, different, situations needing explanation.

ReplyDeleteTwo of those situations are (1) a system of objects in physical contact, which includes conduction as a mechanism of energy transfer between objects, and (2) a system of objects

notin physical contact--perfectly insulated from each other's physical contact by vacuum. Situation 2 includes radiation as the only mechanism of energy transfer. Eli's example is of situation 2, not situation 1.The behaviors of the systems in both situation 1 and situation 2 legitimately can be summarized by the second law. But the mechanisms within those systems differ.

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. Those distortions/violations should clue you that your understandings of the detailed physical causal mechanisms are incorrect.

It's not nearly as hard as you are making it. Just accept that there are two situations with two mechanisms. Deal with each of them on their own terms. You will then see that both can be summarized by the second law.

My comments on conduction are separate from those of radiation. You obviously did not read them, and got hoaxed by John Garland, the lying accuser.

DeleteBetty,

ReplyDeletePlease. Just give it up. You're going to hurt yourself trying to wrap your little brain around this stuff!

"You need to understand the difference between rhetoric and physics."

ReplyDelete... says someone who only talks about the ideas, but never does any calculations to support her position.

Betty...

ReplyDelete1. You have repeatedly said an ice cube or other colder object does not radiate heat energy to anything warmer. That is neither a hoax nor a misquote nor a misrepresentation of what you've said. But a large ice cube next to you really does radiate more heat your way than, say, a block of dry ice the same size whether you like it or not. Deny it all you want. You are wrong.

2. The very simple reason that the 262K plate cannot warm a nearby plate on the shaded side to more than 220K is that the radiation emitted by said 262K plate on each side is not 400W/m^2 but rather (by a quickie bunch of keypresses on a calculator that I'm open to having corrected) 267W/m^2. This leads to having the green plate emit 133W/m^2 on each side. This means 400 W/m^2 enters the system from the sun and 267 + 133 = 400 W/m^2 exit out the extreme left and extreme right sides of the system. Energy in= energy out. Everyone but you--well and Ross too--is happy.

3. This lower radiation emission rather than your preferred 400W/m^2 emission by each side of the blue plate has been from the beginning and remains till now the essential point you deny. It gets just plain silly for you to continue to deny this as anyone with an honest understanding of even first year physics knows you're wrong. If you were right you would have invented perpetual energy production.

4. The math involved--2 equations with 2 unknowns and taking 4th roots--is quite literally junior high school or early senior high school level depending on your track at the time. It's just not esoteric either.

Betty Pound

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

NOT receiveD! Constantly recieveS! It makes all the difference.

NOT "boost itself". Nobody ever claimed that anything happens "by itself" here, but by the constant energy supply of a energy source.

If your above citet statement was true, explain one thing: How does clothing warm the human body in winter? The clothing is always colder than the body. The body has its own constant energy supply.

If you were right and slowed cooling of a body with a constant energy supply can in no way lead to a higher temperatur of the body, clothing is meaningless and humans can as well be naked during winter.

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

Or, also alternatively, you are simply wrong.