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.
