Trenberth's Energy Budget

[IanC]

No offense gslack, but I am talking about the skin touching the blanket. Because the blanket is in open air and able to bleed off heat into the atmosphere, it will never reach equilibrium with the skin...
....When you look at the void between you and the bed (the tented area) you can see the radiation spread out and the temperature will eventually get relatively close, but never in equilibrium with your body. Even after hours in the bed, even with two bodies, the air space never gets to body temperature no matter how many blankets you have.

so you did your own experiment and found that objects around a heated object never reach equilibrium at the same temperature, even with conduction and convection added to the arguably weakest form of energy transfer, radiation.

yet you mock me for saying the same thing.

explain in your own words how both the heated and unheated bars in Spencer's experiment will end up at the same temperature. and while you're at it explain how all the surfaces will radiate away the same amount of energy even though some are radiating out to the cold and some to the warm.

 

[wirebender]

No offense gslack, but I am talking about the skin touching the blanket. Because the blanket is in open air and able to bleed off heat into the atmosphere, it will never reach equilibrium with the skin...
....When you look at the void between you and the bed (the tented area) you can see the radiation spread out and the temperature will eventually get relatively close, but never in equilibrium with your body. Even after hours in the bed, even with two bodies, the air space never gets to body temperature no matter how many blankets you have.

so you did your own experiment and found that objects around a heated object never reach equilibrium at the same temperature, even with conduction and convection added to the arguably weakest form of energy transfer, radiation.

yet you mock me for saying the same thing.

explain in your own words how both the heated and unheated bars in Spencer's experiment will end up at the same temperature. and while you're at it explain how all the surfaces will radiate away the same amount of energy even though some are radiating out to the cold and some to the warm.

My experiment happened in the open atmosphere. Spencer's experiment happened in a cooled vacuum and only involved what you clearly stated was the weakest form of energy transfer. Sorry that you are unable to grasp the difference; and terribly sorry that you beleive the two to be the "same thing".

 

[IanC]

No offense gslack, but I am talking about the skin touching the blanket. Because the blanket is in open air and able to bleed off heat into the atmosphere, it will never reach equilibrium with the skin...
....When you look at the void between you and the bed (the tented area) you can see the radiation spread out and the temperature will eventually get relatively close, but never in equilibrium with your body. Even after hours in the bed, even with two bodies, the air space never gets to body temperature no matter how many blankets you have.

so you did your own experiment and found that objects around a heated object never reach equilibrium at the same temperature, even with conduction and convection added to the arguably weakest form of energy transfer, radiation.

yet you mock me for saying the same thing.

explain in your own words how both the heated and unheated bars in Spencer's experiment will end up at the same temperature. and while you're at it explain how all the surfaces will radiate away the same amount of energy even though some are radiating out to the cold and some to the warm.

My experiment happened in the open atmosphere. Spencer's experiment happened in a cooled vacuum and only involved what you clearly stated was the weakest form of energy transfer. Sorry that you are unable to grasp the difference; and terribly sorry that you beleive the two to be the "same thing".

OK, explain what difference the open atmosphere vs vacuum is making regarding the equilibrium.

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change. in the original the bars were side by side with no specified distance between them. lets rotate the bars 90 degrees and keep them the same distance apart. now it is just two of the smaller faces available to transfer radiation into the unheated bar. from your original 'math' this will make no difference even though there is half the area to transfer heat and lead to equilibrium. not to mention an extra 10% of the surface area radiating directly into the cold outside. Am I correct in my assumption that you are sticking with your original 'math'? the bars will come to equilibrium at an equal temperature?

hahahaha, ludicrous

 

[gslack]

so you did your own experiment and found that objects around a heated object never reach equilibrium at the same temperature, even with conduction and convection added to the arguably weakest form of energy transfer, radiation.

yet you mock me for saying the same thing.

explain in your own words how both the heated and unheated bars in Spencer's experiment will end up at the same temperature. and while you're at it explain how all the surfaces will radiate away the same amount of energy even though some are radiating out to the cold and some to the warm.

My experiment happened in the open atmosphere. Spencer's experiment happened in a cooled vacuum and only involved what you clearly stated was the weakest form of energy transfer. Sorry that you are unable to grasp the difference; and terribly sorry that you beleive the two to be the "same thing".

OK, explain what difference the open atmosphere vs vacuum is making regarding the equilibrium.

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change. in the original the bars were side by side with no specified distance between them. lets rotate the bars 90 degrees and keep them the same distance apart. now it is just two of the smaller faces available to transfer radiation into the unheated bar. from your original 'math' this will make no difference even though there is half the area to transfer heat and lead to equilibrium. not to mention an extra 10% of the surface area radiating directly into the cold outside. Am I correct in my assumption that you are sticking with your original 'math'? the bars will come to equilibrium at an equal temperature?

hahahaha, ludicrous

Example #1 Changing the experiment or parameters to try and make it work.

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change.

Changing the experiment again? Its what you have done from the start and do so every time you cannot accept what it tells you..

Example #2 Doing the same thing over and again while expecting a different outcome, and pretending to get one regardless.

in the original the bars were side by side with no specified distance between them. lets rotate the bars 90 degrees and keep them the same distance apart. now it is just two of the smaller faces available to transfer radiation into the unheated bar.

Ian, try and get your head around this... The size of the face close to the heat source changes nothing in this experiment. Shrinking the radiator and receptor surface area size will also shrink the radiating surface area and thus shrinking the amount of radiating heat from that vector to the receptor. Ian...seriously...

you keep changing the parameters to make it work and it still falls flat. I explained it, wire explained it, hell wire showed you the math, I explained how spencer took an existing experiment to show zeroth law in action and removed a parameter to make an unclear and deceptive claim about energy transfer in the atmosphere, and you still try and pretend we are both wrong... Why is that Ian? Can't accept you are wrong? yeah...

Ian you BS is tiring, you make less and less sense and your arguments are getting ignorant now...

 

[wirebender]

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change.

Hard to pin down? What the hell are you talking about. I gave you the expanation in the most explicit language possible; mathematics. Sorry you don't speak that language. That being said, it has been explained to you in verbage as well and you didn't get that either. The bottom line, Ian, is that the topic is apparently so far over your head that it can't be explained to you.

 

[IanC]

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change.

Hard to pin down? What the hell are you talking about. I gave you the expanation in the most explicit language possible; mathematics. Sorry you don't speak that language. That being said, it has been explained to you in verbage as well and you didn't get that either. The bottom line, Ian, is that the topic is apparently so far over your head that it can't be explained to you.

your math is highly suspect. if shrinking the area available for heat flow and expanding the area available for heat loss doesnt effect the final answer then your modelling of the problem is wrong. its as simple as that.

 

[gslack]

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change.

Hard to pin down? What the hell are you talking about. I gave you the expanation in the most explicit language possible; mathematics. Sorry you don't speak that language. That being said, it has been explained to you in verbage as well and you didn't get that either. The bottom line, Ian, is that the topic is apparently so far over your head that it can't be explained to you.

your math is highly suspect. if shrinking the area available for heat flow and expanding the area available for heat loss doesnt effect the final answer then your modelling of the problem is wrong. its as simple as that.

Nice try with the slight of hand there Ian but you said to turn BOTH objects to the smaller areas face one another at the same distance... In effect you shrank BOTH surface areas that transfer the heat... Increasing the distance from the heat will change the entire experiment again.. The original experiment was not to see if altering the heated and receiving areas geometry relative to position and vector would alter the result, and attempting to do so is a deliberate distraction from the original...

Stop trying to confound it Ian....

 

[IanC]

My experiment happened in the open atmosphere. Spencer's experiment happened in a cooled vacuum and only involved what you clearly stated was the weakest form of energy transfer. Sorry that you are unable to grasp the difference; and terribly sorry that you beleive the two to be the "same thing".

OK, explain what difference the open atmosphere vs vacuum is making regarding the equilibrium.

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change. in the original the bars were side by side with no specified distance between them. lets rotate the bars 90 degrees and keep them the same distance apart. now it is just two of the smaller faces available to transfer radiation into the unheated bar. from your original 'math' this will make no difference even though there is half the area to transfer heat and lead to equilibrium. not to mention an extra 10% of the surface area radiating directly into the cold outside. Am I correct in my assumption that you are sticking with your original 'math'? the bars will come to equilibrium at an equal temperature?

hahahaha, ludicrous

Example #1 Changing the experiment or parameters to try and make it work.

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change.

Changing the experiment again? Its what you have done from the start and do so every time you cannot accept what it tells you..

Example #2 Doing the same thing over and again while expecting a different outcome, and pretending to get one regardless.

in the original the bars were side by side with no specified distance between them. lets rotate the bars 90 degrees and keep them the same distance apart. now it is just two of the smaller faces available to transfer radiation into the unheated bar.

Ian, try and get your head around this... The size of the face close to the heat source changes nothing in this experiment. Shrinking the radiator and receptor surface area size will also shrink the radiating surface area and thus shrinking the amount of radiating heat from that vector to the receptor. Ian...seriously...

you keep changing the parameters to make it work and it still falls flat. I explained it, wire explained it, hell wire showed you the math, I explained how spencer took an existing experiment to show zeroth law in action and removed a parameter to make an unclear and deceptive claim about energy transfer in the atmosphere, and you still try and pretend we are both wrong... Why is that Ian? Can't accept you are wrong? yeah...

Ian you BS is tiring, you make less and less sense and your arguments are getting ignorant now...

Ian, try and get your head around this... The size of the face close to the heat source changes nothing in this experiment. Shrinking the radiator and receptor surface area size will also shrink the radiating surface area and thus shrinking the amount of radiating heat from that vector to the receptor. Ian...seriously...

hahahaha. what a bunch of nonsense. worthy of Cliff Clavin! explain how less available heat absorbance and more available heat shedding adds up to the same result. add up the warming surface area in both cases, the cooling surface area in both cases. 50% less warming, 10% more cooling. yet you two guys say there is no change. what a couple of retards

 

[gslack]

OK, explain what difference the open atmosphere vs vacuum is making regarding the equilibrium.

because you are so hard to pin down on stating your explanation of the heat flows in Spencer's experiment lets make a small change. in the original the bars were side by side with no specified distance between them. lets rotate the bars 90 degrees and keep them the same distance apart. now it is just two of the smaller faces available to transfer radiation into the unheated bar. from your original 'math' this will make no difference even though there is half the area to transfer heat and lead to equilibrium. not to mention an extra 10% of the surface area radiating directly into the cold outside. Am I correct in my assumption that you are sticking with your original 'math'? the bars will come to equilibrium at an equal temperature?

hahahaha, ludicrous

Example #1 Changing the experiment or parameters to try and make it work.



Changing the experiment again? Its what you have done from the start and do so every time you cannot accept what it tells you..

Example #2 Doing the same thing over and again while expecting a different outcome, and pretending to get one regardless.



Ian, try and get your head around this... The size of the face close to the heat source changes nothing in this experiment. Shrinking the radiator and receptor surface area size will also shrink the radiating surface area and thus shrinking the amount of radiating heat from that vector to the receptor. Ian...seriously...

you keep changing the parameters to make it work and it still falls flat. I explained it, wire explained it, hell wire showed you the math, I explained how spencer took an existing experiment to show zeroth law in action and removed a parameter to make an unclear and deceptive claim about energy transfer in the atmosphere, and you still try and pretend we are both wrong... Why is that Ian? Can't accept you are wrong? yeah...

Ian you BS is tiring, you make less and less sense and your arguments are getting ignorant now...

Ian, try and get your head around this... The size of the face close to the heat source changes nothing in this experiment. Shrinking the radiator and receptor surface area size will also shrink the radiating surface area and thus shrinking the amount of radiating heat from that vector to the receptor. Ian...seriously...

hahahaha. what a bunch of nonsense. worthy of Cliff Clavin! explain how less available heat absorbance and more available heat shedding adds up to the same result. add up the warming surface area in both cases, the cooling surface area in both cases. 50% less warming, 10% more cooling. yet you two guys say there is no change. what a couple of retards

Please show where I said "no change"... yeah, i said changes nothing in the experiment.. The experiment was not and is not an attempt to show changes in vector or geometry relative to heat surface area and rate of heat loss...

you tried again to alter the topic to cover your ass Ian.. Now you can stop lying anytime...

 

[IanC]

wirebender's 'math' gives the same answer regardless of orientation. the orientation obviously makes a difference in heat flow. therefore wirebender's math is incorrect.

can I make it any simpler?

 

[wirebender]

your math is highly suspect. if shrinking the area available for heat flow and expanding the area available for heat loss doesnt effect the final answer then your modelling of the problem is wrong. its as simple as that.

You lose again Ian. I already told you that the only way to increase the temperature of the heated bar was to either decrease its area or increase the power going to it. Clearly adding the heat sink added to the radiating area and the final result was both radiating at a considerably lower temperature than the heated bar alone.

If the math is wrong, then prove the math is wrong. Or you might man up and simply admit that you don't know whether the math is wrong.

By the way Ian, it isn't "my" math and it isn't gord's math either. It is Jozef Stefan's math and it has stood for a very long time and my bet is that you are not going to tear it down either.

 

[wirebender]

hahahaha. what a bunch of nonsense. worthy of Cliff Clavin! explain how less available heat absorbance and more available heat shedding adds up to the same result. add up the warming surface area in both cases, the cooling surface area in both cases. 50% less warming, 10% more cooling. yet you two guys say there is no change. what a couple of retards

Prove the Stefan-Boltzman equation wrong Ian or prove that it isn't applicable in problems of radiation.

 

[wirebender]

wirebender's 'math' gives the same answer regardless of orientation. the orientation obviously makes a difference in heat flow. therefore wirebender's math is incorrect.

can I make it any simpler?

Wirebender doesn't have any math. Neither does gord. It is a straight forward application of the Stefan-Bolzman law to a problem of radiation. The answer is what it is Ian. Either prove the Stefan-Boltzman law wrong or prove that it isn't applicable to radiating bodies and we have something to talk about.

Going on and on about the math being wrong, or far fetched isn't getting the job done Ian. Either you can prove it wrong, in the language of mathematics, or you can't. My bet is that you can't because it isn't wrong. Since you can't prove it wrong, then you might consider admitting that you are working from a position of faith, or instinct, or wishfullness or something other than a position of hard, unblinking science and a willingness to accept the laws of physics no matter what the little guy behind your eyes is screaming to the contrary.

Here is the mathematical sequence again Ian. You tell me where it goes off the tracks and show me the math to prove it. By the way, this sequence addresses spencer's thought experiment. He didn't specifiy the emissivity of the plates so it was assumed to be 1 and he didn't specify the surface area of the plates so they were assumed to be one square meter. Those numbers were chosen for convenience. You can change the emissivity if you like and you can change the surface area if you like. The final answer will be different in accordance with the changes you make but the bottom line will remain the same. The heated plate and the cooled plate situated in a cooled vaccum will attain an equilibrium temperature considerably lower than the heated plate in the chamber alone.

Last time, you got off the tracks by assuming that because the heated plate was cooled by the addition of the other plate that perhaps more energy was flowing through the heated plate via the power cord attached to the heated plate. If more energy is coming in, then spencer is still wrong because the cooled plate didn't warm the heated plate, the additional energy input did. Remember the proof that to raise the temperature of the heated plate the wattage going in would have to increase from 744.95 to 795 watts? The energy input remains the same. Spencer is trying (and failing) to show that backradiation is possible in the atmosphere. Letting more power flow into the bar because it now has essentially more surface area makes it behave in a manner other than as a blackbody and the whole experiment is geared as an attempt to prove something about blackbody radiation.

Re: Vacuum Chamber with plates.

First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

P = e*BC*A*T^4

Where P = net radiated power (Watts), e = emissivity, BC = Stefan’s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts

(***That’s ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)

The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2
——–
If another identical “non-heated” and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:

The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.

The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.

The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

And the equilibrium temperature for both plates will be 284.69 K or 53 deg

You can either prove the math wrong or you can't Ian. I am betting that you can't in which case, the discussion has been over for some time and you merely didn't realize that you lost. The orientation of the plates doesn't matter because there is no such thing as backradiation. The failure of the experiment was due to the fact that spencer ignored the 2nd law of thermodynamics and the law of conservation of energy and assumed that backradiation is possible. Tell me Ian, do you also believe in back conduction, or back convection? how might they work? If you believe backradiation is possible, you must also believe back convection and back conduction are possible. how me the math that supports such nonsense in the physics literature.

 

[IanC]

you really dont think these things through, do you wirebender. I will ignore your strawman defense of accusing me of trying to prove thermodynamics wrong. in fact they are what I use to imagine the energy flow in these type of problems.

lets try again to make you 'see the light' (bad pun)

both bars are at 150C and are in separate vacuum containers. what happens with respect to time and equilibrium? the heated bar continues to receive 750W therefore continues to radiate 750W @ 150C with an area of 1m2. assuming a regular rectangular shape and even heat distribution that means four faces radiating 150 W each and 2 faces at 75w. total =750w= 150+150+150+150+75+75
no change with respect to time for the heated bar.

the unheated bar starts at 150C and also radiates 750W with six faces adding up to 1m2, same as the heated bar. because the unheated bar is losing heat but not receiving any replacement heat the temperature will drop, exponentially until it is essentially the same as the cooled container.

next variation- the unheated block somehow magically recieves the 150w of radiation from one of the large sides of the heated block. this heat is evenly distributed so the six sides of the now heated block would radiate at 30+30+30+30+15+15=150w

next variation- the unheated block somehow magically receives the 75w of radiation from on e of the small sides of the heated block, evenly distributed. 15+15+15+15+7.5+7.5=75w

your 'math' of what is happening in the thought experiment is already smashed to bits but lets keep going.

next variation-if the heated bar and the unheated bar were placed in the cooled vacuum container close enough to totally exchange the radiation from the two inside faces but not touching so there is no conduction what would happen? the heated face would radiate 150w to the unheated block but get 30w back. because there is thermal impedance the heat would build up at both faces but I will ignore that as an unnecessary complication. the 30w back radiation is not heat flowing into the warmer block, it is just part of the calculation. net Power= emitted Power- absorbed Power. the 30w back radiation needs to be equally divided between all the faces of the heated block.....

sorry, to be continued when I have time.

 

[gslack]

Ian if you can prove his math wrong as it is, why do you keep making all these variations?Why all the complication if it really is so simple to show it wrong just do it already...

Jesus man, you are like a bullshit fountain... Show his math wrong already! Damn man..

EDIT: Ian his post was not a strawman, you ARE trying to claim thermodynamics laws wrong in your continued pretense by altering the original experiment enough you can somehow make the laws different. Strawmen have been your entire MO in all of this. Everytime you bring a variation or a different experiment or change the parameters like you just did, its a strawman.. Wow man you going to wake up today sometime?

 

[wirebender]

you really dont think these things through, do you wirebender. I will ignore your strawman defense of accusing me of trying to prove thermodynamics wrong. in fact they are what I use to imagine the energy flow in these type of problems.

There is your first problem Ian. You are imagining rather than doing the math. I have done the math and like it or not, it trumps whatever imaginings you care to do. The math trumped spencer on his own blog because he made no effort to address it. He simply sat quitely hoping that it would go away. It didn't.

You have the math concerning radiation and your "sides" fantasy has no place in it. Of course if you like, feel free to show me the Stefan-Boltzman derivitive that shows an increase of energy as the result of adding sides. Do be sure to square that increase in energy with the law of conservation of energy.

 

[wirebender]

Jesus man, you are like a bullshit fountain... Show his math wrong already! Damn man..

I am afraid that isn't in the cards. Imagine, proving the Stefan-Boltzman law wrong by simply adding sides. Who would have thought? What sort of thinking leads one to actually believe such gibberish.

 

[IanC]

you really dont think these things through, do you wirebender. I will ignore your strawman defense of accusing me of trying to prove thermodynamics wrong. in fact they are what I use to imagine the energy flow in these type of problems.

lets try again to make you 'see the light' (bad pun)

both bars are at 150C and are in separate vacuum containers. what happens with respect to time and equilibrium? the heated bar continues to receive 750W therefore continues to radiate 750W @ 150C with an area of 1m2. assuming a regular rectangular shape and even heat distribution that means four faces radiating 150 W each and 2 faces at 75w. total =750w= 150+150+150+150+75+75
no change with respect to time for the heated bar.

the unheated bar starts at 150C and also radiates 750W with six faces adding up to 1m2, same as the heated bar. because the unheated bar is losing heat but not receiving any replacement heat the temperature will drop, exponentially until it is essentially the same as the cooled container.

next variation- the unheated block somehow magically recieves the 150w of radiation from one of the large sides of the heated block. this heat is evenly distributed so the six sides of the now heated block would radiate at 30+30+30+30+15+15=150w

next variation- the unheated block somehow magically receives the 75w of radiation from on e of the small sides of the heated block, evenly distributed. 15+15+15+15+7.5+7.5=75w

your 'math' of what is happening in the thought experiment is already smashed to bits but lets keep going.

next variation-if the heated bar and the unheated bar were placed in the cooled vacuum container close enough to totally exchange the radiation from the two inside faces but not touching so there is no conduction what would happen? the heated face would radiate 150w to the unheated block but get 30w back. because there is thermal impedance the heat would build up at both faces but I will ignore that as an unnecessary complication. the 30w back radiation is not heat flowing into the warmer block, it is just part of the calculation. net Power= emitted Power- absorbed Power. the 30w back radiation needs to be equally divided between all the faces of the heated block.....

sorry, to be continued when I have time.

continued...

so in the variation with the large (0.2m2) heated face radiating 150w into the large (0.2m2) unheated we have a choice of bookkeeping methods. I prefer to imagine 150w out of the heated bar and 30w back from the second bar, for a net exchange of 120w. wirebender is vehemently against any mention of back radiation so we can instead imagine only 120w out of the heated face and 0w out of the unheated face. either way works for calculating, I just find it easier to visualize with back radiation.

what are the energy balances now that the two blocks are together? the heater is still inputting 750w into the heated bar but now one of the large faces is only outputting 120w instead of 150w, meaning 30w is not radiated away but is used to heat up the bar until the faces are radiating 6w hotter for the large faces and 3w hotter for the smaller faces. 156+156+156+78+78+[(156-30)the heated face radiating into the second bar]= 750w. the faces of the heated bar are emitting more power which means they are warmer than before. the temperature of the heated bar has increased!!!!

what about the unheated bar? it has received 120w through the inside face and also nullified an extra 30w of power. I prefer to imagine it as absorbing 150w and reradiating 30w back at the source but it is a matter of preference. so the remaining 0.8m2 is radiating 120w. 30+30+30+15+15+0(for the inside face)=120w. of course this is vastly oversimplified. the heat flow is coming in through the side instead of from the centre so the inside face would be much warmer than the face opposite it. and the extra 6w from the heated bar after it equilibrates for the nullified 30w would slightly heat the unheated bar which would cause more back radiation, oops I mean power nullification and warm up the heated bar by a smidge, etc.

it is easy to see that the heated bar will always be warmer than the unheated bar. it is also easy to see that any thing warmer than the temperature of the cooled container will impede the loss of heat from the heated bar which will force the heated bar to warm up until the increased flow from the unimpeded sides balances the power in=power out.

I have asked wirebender to explain how the two bars will be exactly the same temperature at equilibrium--no answer. his 'math' states that they are, therefore it is incorrect.

perhaps he can explain how a heat source can bleed away the same amount of energy when it has become more insulated without becoming warmer itself. but I doubt it.

personally I think the small amount of temperature increase of the earth's surface caused by CO2 is mitigated by increased water evaporation which causes rising air and clouds that release latent heat are much more efficient at transferring energy than radiation. just look at the tropics and the afternoon thunderstorms that cool the surface in minutes. but I dont make up the inputs for the computer models, or decide the various values of Trenberth's energy pathway graph (which has changed considerably over the years, I wonder if this one is right?).

anyways wirebender. two simple questions for you. you've ducked in the past and I'm sure you will duck now. explain how the bars come to equilibrium at the same temperature when one is heated and the other is not. explain how a heated insulated object can come to equilibrium without warming up.

 

[gslack]

Ian why are you confounding this even more? Why not stick with the original thought experiment and prove it right, instead of continually altering it? That ones already established here, its been written out, and yet you continue to claim its right but refuse to use it to show its right...

If wire wants to continue playing your silly deflection its on him, but your "physics" persona is done for.. You know about as much physics as my 13 year old...

BTW Ian, remember the chamber, and the vacuum.. And thats all the help im giving you..

 

[wirebender]

I have asked wirebender to explain how the two bars will be exactly the same temperature at equilibrium--no answer. his 'math' states that they are, therefore it is incorrect.

Incorrect based on what Ian? Your intuition? The math is correct but do feel free to prove the Stefan-Boltzman Law wrong.

I answered your question in the most explicit language possible. Sorry you don't speak it. That being the case, you really are in no position to question the answer.

You have lost the point, you have lost the discussion and have lost any respect that I may have had for you. Why continue?