We now insert our hollow sphere (somehow, suddenly and magically, it doesn't matter, this is assumed) which fully encloses the heat source. It is 7mm away from the heat source, and it is 7mm away from the wall. And we let things come back up to radiative steady-state.
And now, to do this properly, we must make a bit of a mental leap, which may be difficult for some people: the total heat transfer now from heat source to the chamber wall is equal to: (heat transfer from heat source to the inside of the enclosing plate) PLUS (heat transfer from the outside of the enclosing plate to the wall).
This is a place where "thermodynamic thinking" will mess you up. Some people will insist that the TOTAL heat transfer must take place between EACH object. But that is simply not true. This was CLUE #3: a quote from a heat transfer engineering textbook about how "thermodynamic thinking" will lead one astray. [Jane Q. Public, 2014-09-10]
As I said, in the unlikely event that you wrote down equations, they'd violate conservation of energy. Thermodynamic thinking like this leads one back to reality, not astray. Draw a boundary inside the inner surface of the enclosing shell at your steady-state values. Since nothing inside that boundary is changing, power in = power out. But that's completely impossible. Your solution violates conservation of energy, as predicted.
Jane's obligations include continuing to spread misinformation about ocean acidification even after I've repeatedly debunked him.
So I predict that Jane's answer won't include any equations that could be used to calculate the enclosed source temperature. Instead, he'll probably grace us with another lengthy, incoherent rant about "problems" in my analysis which are (as usual) too vague to be expressed in equations. In the extremely unlikely event that Jane musters up the courage and competence to actually write down an equation that could be used to calculate the enclosed source temperature, it will almost certainly violate conservation of energy.
... Slashdot has a time limit on these old threads. If you don't post the rest by tomorrow, they will likely close the thread and archive it. I don't know the exact time limit but I have given you plenty of time already, and overly indulged you, but that is ending now. You stated yourself, just above, that it is not difficult to do.
... If this thread is archived before you post the last bit of your supposed refutation (you still have plenty of time), I am going to declare you a fraud and a failure. ... [Jane Q. Public, 2014-09-09]
After this thread is closed, this conversation can continue here.
Just so we are absolutely clear on what your claim is: starting at the agreed-upon initial conditions, heat source at 150F, when a hollow sphere is suddenly inserted into the chamber, completely surrounding the heat source, of the specified dimensions, then when allowed to reach steady-state the actual temperature of the heat source is 234.1 degees F.
Did I summarize that accurately enough? I don't want to re-hash the initial conditions we agreed upon. I still agree with them.
Yes, that summary is accurate enough.
We will most certainly have to disagree on that, because it's wrong. That equation is for finding Q, the net heat transfer, which is not "equal" to power at all. It is energy in Joules. [Jane Q. Public, 2014-09-09]
No, Jane. I linked to Wikipedia's equation for radiative heat transfer, which is in Watts, not Joules. You can verify this by noticing the "dot" over the heat transfer "Q" on the left hand side of that equation. In physics-speak, a "dot" means a "time derivative" so that equation is in units of power (Watts). Or you could've checked the units on the right hand side, and verified that they're also in units of Watts, just like I said.
... So where is your final answer for the temperature of the heat source at stead-state? THAT was what you said you were calculating, so where is it?
... [Jane Q. Public, 2014-09-09]
The final answer for the enclosed source at steady-state is 385.4 K (234.1 F). Anyone with a calculator could have verified this based on my comment yesterday.
... How could I possibly be "wrong"? I'm not doing anything. This is YOUR claim, not mine.
... [Jane Q. Public, 2014-09-09]
Your claim that the source doesn't warm after the passive plate is added is wrong.
... Why should I do that? YOU said you were going to refute Latour. It wasn't my claim. You got partway through, now you refuse to finish, and you're trying to blame ME somehow? How do you figure?
... [Jane Q. Public, 2014-09-09]
The final answer for the enclosed source at steady-state is 385.4 K (234.1 F). Anyone with a calculator could have verified this based on my comment yesterday.
... Hahaha! I've been WAITING for you to show me how this is done. I've asked you about five times now to show me. What are you waiting for? I want you to show us how you did what you claimed you have already done -- refute Latour -- so I, and anyone else who reads this later, can check your work.
... Here's your incentive: if you can actually, successfully complete a refutation of Latour, and show us, and it checks out, I will be happy to declare to everyone that I was wrong and you were right about that issue. You have my word. I will shout it out loud. I'll admit it here on Slashdot and even open a Twitter account and post it there. ... I have said what I have to say, unless and until you decide to post the rest of YOUR refutation of Latour. if I have to finish your problem for you, using YOUR methods, I'm still going to declare you a failure, regardless of whether the answer turns out to be correct. Because YOU claimed you could do it. So show us. [Jane Q. Public, 2014-09-09]
The final answer for the enclosed source at steady-state is 385.4 K (234.1 F). Anyone with a calculator could have verified this based on my comment yesterday.
So I successfully completed a refutation of Dr. Latour, and showed you. Does it check out? If so, will you really be happy to declare to everyone that you were wrong? Or was that a lie?
As before, that net radiative power is described by Wikipedia’s equation which accounts for areas and view factors. [Dumb Scientist]
... You say you're using the equation for radiative power, when you're linking to the equation for heat transfer. We already know what the equation for radiative power is: (epsilon)(sigma)T^4. [Jane Q. Public, 2014-09-09]
And then:
Once again, we'll obviously have to agree to disagree about the net heat transfer between two gray surfaces. [Dumb Scientist]
What the HELL are you talking about? I understand the equation from Wikipedia. I just happened to mention that you called it a power equation rather than a heat transfer equation. THAT IS ALL. [Jane Q. Public, 2014-09-09]
We'll obviously have to agree to disagree that I explicitly used the equation for net radiative power, and linked to an equation described as: "The radiative heat transfer from one surface to another is equal to the radiation entering the first surface from the other, minus the radiation leaving the first surface."
We've agreed that net radiative power is power out minus power in through a boundary, but we'll obviously have to agree to disagree that Wikipedia's radiative heat transfer is "equal" to net radiative power.
... THEN I gave you an equation for radiant emittance: (epsilon)(sigma)T^4, and you called it a "heat transfer" equation having something to do with 0K black bodies, which is simply false.
... [Jane Q. Public, 2014-09-09]
Again, we'll obviously have to agree to disagree that I explicitly used the equation for net radiative power (or net heat transfer). If I hadn't, it might make sense for Jane to say "We already know what the equation for radiative power is: (epsilon)(sigma)T^4."
Again, we'll obviously have to agree to disagree that I explicitly used the equation for net radiative power (in Watts). If I hadn't, it might make sense for Jane to say we already know that equation is the equation for radiant emittance (in W/m^2).
Once again, if I had explicitly used the equation for net radiative power, Jane's equation would only be valid for net radiative power (or net heat transfer) to a 0K blackbody. But I've obviously failed to explain net radiative power (or net heat transfer) between two gray surfaces, so we'll have to agree to disagree once again.
... Now, when I simply pointed out these apparent MISTAKES in terminology to you, in order to try to keep things straight, you're throwing a fit. Well, don't try to blame this on me. I was just explaining why the things YOU have been saying lead to confusion. I will not apologize for simply trying to sort out basic misunderstandings.
... [Jane Q. Public, 2014-09-09]
If I was "throwing a fit" by saying we'll obviously have to agree to disagree, then what's this?
"... non-person... disingenuous and intended to mislead
"Jesus, you're a dumbshit.
"... what a despicable human being you are
GISS ADMITTED GODDARD WAS RIGHT. YOU DIDN’T KNOW. YOU’RE IGNORANT OF THE FACTS. LEARN THEM. MEANTIME, GO AWAY. [Lonny Eachus, 2014-08-30]
Try Google, dumbshit. Unless you don’t know how. It took me all of 20 seconds.
... Why? Why should I do this for you? Would you like me to wash your balls too? Answer: no. ... The fact I WON’T wash your balls for you is not evidence that they don’t exist. The fact that YOU won’t, IS. ... Correct. To all outside observers, so far, your balls don’t exist. Why don’t you prove that they do? show us. ... Should we just ASSUME it? Or, like you, should we require that you SHOW US? ... To make an even better analogy: there is a picture of them that has been posted online by your girlfriend. ... BUT we don’t believe you really have any. Should we ask you to prove they’re yours? Every time we discuss it? [Lonny Eachus, 2014-08-30]
Sorry, dude. You aren’t going to get me to wash your balls. The rest of us are looking at pictures of your girlfriend. wondering when you’re going to say “I won’t hang them out again just for you. Look it up.” [Lonny Eachus, 2014-08-30]
Are you REALLY that fucking stupid? [Lonny Eachus, 2014-08-30]
I’ve insulted you because you deserve it. Arguments were made. Your inability to absorb them is not evidence. [Lonny Eachus, 2014-08-30]
I guarantee something: that doesn’t make ME an asshole. I’ll leave it up to others what it does mean. [Lonny Eachus, 2014-08-30]
I am participating in only as a courtesy. I'm just following along. Either get on with it, or not. But if you refuse to do what you told me I was coming here to see you do (refute Latour), then you refuse. That has nothing to do with me, and you don't get blame it on me. Get the hell on with it, or not. Whichever you do, it's YOUR choice. I am very, very close to calling you full of shit and posting this where everyone can see it. [Jane Q. Public, 2014-09-09]
Feel free to post this where everyone can see it. But you never will. And you'll never solve for the enclosed source temperature. Instead you'll just keep saying things like this:
... I have repeatedly tried to engage you in a thorough analysis of this experiment. EVERY TIME, you have done (usually incorrectly) a partial analysis... NEVER daring to face the full problem in real detail.
... You have NEVER, ONCE, tackled the problem head-on. ... Always weaseling sideways... [Jane Q. Public, 2014-08-30]
No, you haven't even taken a single solitary step towards solving for the enclosed source temperature. But I've repeatedly tackled the full problem in real detail.
See? Same shit different day. You won't sit down and do the calculations start-to-finish... simply another illustration of the depths of hand-waving you will go to, rather than actually doing all the calculations on the actual experiment from start to finish.
... The only reasonable conclusion to be drawn here is that you won't do it because you know you're wrong. [Jane Q. Public, 2014-08-30]
It would only take you a few minutes to write down the equation and values that could be used to solve for the enclosed source temperature. Is the only reasonable conclusion to be drawn here that you won't do it because you know you're wrong?
... I thought we'd actually settle this scientifically, once and for all, but I see that you were never really interested in that anyway.
... [Jane Q. Public, 2014-08-30]
This might be the most ironic statement ever, coming from someone who refuses to take a single baby step towards solving for the enclosed source temperature.
My honest opinion is that YOU are the one who is trolling, and never intended to actually refute anybody at all. You simply wanted to waste more of my time. [Jane Q. Public, 2014-09-09]
In the time it's taken you to write all these incoherent rants and talk about washing people's balls, you could've written down the equation and values describing the enclosed source temperature. In fact, you could've done that many times over. Instead, you cuss in ALL CAPS. Are you really going to give a copy to your grandchildren?
If you can't write down that equation because you're not sure which one is correct, just let me know and I'll try to help. If you're not sure which values to plug in, I'll try to show you where we've already agreed on those values. If you're still not sure these values are correct, just let me know and I'll try to help you understand. Again, don't worry about all that confusing algebra and arithmetic. I'll help you take care of all that. But you need to take a baby step on your own, by showing that you're capable of writing down an equation that can be used to solve for the enclosed source temperature.
Once again, we'll obviously have to agree to disagree about the net heat transfer between two gray surfaces.
Again, you seem to be asserting that Jane's equation should be used instead of Wikipedia's equation. Is that the case? If so, all you need to do to catch up is to list the values you'll plug into that equation, like I did. This would only take a few minutes. If you're confused and need help, just ask.
... I will judge your method based on its physics, when I see your answer.
... [Jane Q. Public, 2014-09-09]
No, that's my entire point. I already described my physics. If you need to see my final numerical answer before you can judge my method, then you're not actually judging my method based on its physics.
Ironically, you actually are judging my method based on its physics, which is actually a step forward:
As before, that net radiative power is described by Wikipedia’s equation which accounts for areas and view factors.
I just want to make it very clear why I object to the way you ask for agreement, all the while throwing in ambiguities. You say you're using the equation for radiative power, when you're linking to the equation for heat transfer. We already know what the equation for radiative power is: (epsilon)(sigma)T^4. [Jane Q. Public, 2014-09-09]
Once again, your equation is only for net radiative power (or net heat transfer) to a 0K blackbody. But I've obviously failed to explain net radiative power (or net heat transfer) between two gray surfaces, so we'll have to agree to disagree.
But this is good. You're actually judging my method based on its physics! I'm proud of you, Jane!
You seem to be asserting that Jane's equation should be used instead of Wikipedia's equation. Is that the case? If so, all you need to do to catch up is to list the values you'll plug into that equation, like I did. This would only take a few minutes. If you're confused and need help, just ask.
... Where is your method in action? Is there an answer in there somewhere? You told me you were going to calculate the temperature of the heat source at steady-state. This is utter nonsense. I simply asked you for an explanation of how you calculated the figure you stated (long) before, after we agreed on the nature of the problem and the initial conditions.
... [Jane Q. Public, 2014-09-08]
Once again, you've seen my method in action from start to finish. I've repeatedly asked if we can agree on that method before posting my final numerical answer. That's because I think you deserve a chance to show that you're capable of judging my method based on its physics, as opposed to reflexively objecting if my numerical answer contradicts the PSI Sky Dragon Slayers.
I haven't even tried to calculate an answer yet. I won't know if I agree with your method until I see it. In action, that is. [Jane Q. Public, 2014-09-08]
You've seen my method from start to finish. If you're capable of judging my method based on its physics, why won't you know if you agree with my method until you see my final numerical answer? For instance, suppose I told you that my final numerical answer agrees with the PSI Sky Dragon Slayers. Would that make you agree with my method's physics? In that case, would you really be agreeing with my method, or agreeing with the answer you want to hear?
If you won't know if you agree with my method until you see my final numerical answer, you're depriving yourself of this chance to demonstrate your intellectual integrity.
Alternatively, you could finally explain your own method of solving for the enclosed source temperature.
... I haven't calculated a solution yet. And THAT is largely due to what I clearly stated before: I have been busy, and don't have a lot of time to devote to this right now. I've been trying to squeeze in what I could, around work and other obligations.
... [Jane Q. Public, 2014-09-08]
Since it's important to agree on the equations before plugging in values, all you have to do to describe your method is to state the equation you're using, and state the values you'll plug in. This would only take about five minutes. I know because that's what I did below.
... I have explained several times now that these Sage equations are not exactly straightforward and easy to read. I have been doing my own calculations in a clear and straightforward manner, making them as easy to read as possible. You really expect me to read this stuff?
... [Jane Q. Public, 2014-09-08]
Once again, I'm sorry. I take full responsibility. I've changed the formatting so that each value being plugged in is on its own line. Does that make it more readable? I've also added some comments to the code which might help you understand it:
Now that we've agreed on the inner shell temperature of ~149.9F, let's take the last step. Calculate the enclosed source temperature.
Draw a boundary just inside the inner surface of the enclosing shell. Because nothing in the boundary is changing with time, power in = power out. The same constant electrical power flows in as before the shell was added. Net radiative power flows out from the source to the enclosing shell's inner surface.
As before, that net radiative power is described by Wikipedia’s equation which accounts for areas and view factors.
#Completely surrounded by shell with finite conductivity.
var('sigma T_c T_h A_c A_h F_hc power epsilon_h epsilon_c')
eq1 = power == sigma*(T_h^4 - T_c^4)/((1-epsilon_h)/(epsilon_h*A_h) + 1/(A_h*F_hc) + (1-epsilon_c)/(epsilon_c*A_c))
soln4 = solve(eq1.subs(
T_c=338.629929346551,
power=15028.4258648090,
sigma=5.670373e-8,
epsilon_h=0.11,
epsilon_c=0.11,
F_hc=1,
A_h=510.064471909788,
A_c=511.185932522526) #End of constant definitions.
soln4[0].rhs().n()
The first line "var('sigma..." declares my variables.
The line "eq1 = power == sigma..." is my "power in = power out" equation using Wikipedia's equation for net radiative power.
The next line "soln4 = solve..." plugs in all the relevant variables and solves it for the enclosed source temperature T_h.
The last line displays the answer.
So I've described my method for calculating the enclosed source temperature from start to finish. Before I post that final answer, can we agree with my method? If not, could you please describe your method?
Again, all you have to do to describe your method is to state the equation you're using, and state the values you'll plug in. That's what I did above, and it only takes about five minutes. If you don't know how to derive the equation, just let me know and I'll try to help. If you don't know which values to plug in, let me know which ones you're confused about. Since you're busy, don't worry about solving the equation. I'll solve it for you.
I haven't even tried to calculate an answer yet. I won't know if I agree with your method until I see it. In action, that is. [Jane Q. Public, 2014-09-08]
Once again, you've already seen my method. I just described my entire method start to finish once again because that's what you demanded:
... Create a realistic scenario, draw yourself a diagram, and run some actual numbers on them rather than just tossing equations around without seeing how they fit together in the real world.
... [Jane Q. Public, 2014-08-29]
See? Same shit different day. You won't sit down and do the calculations start-to-finish, instead you do one small part, then start indulging in your hallmark game of out-of-context he-said, she-said, toss in a straw-man, then claim it's all proved.
... It's simply another illustration of the depths of hand-waving you will go to, rather than actually doing all the calculations on the actual experiment from start to finish. All you're doing is tossing in more straw-men and irrelevancies. You won't do the actual experiment. The only reasonable conclusion to be drawn here is that you won't do it because you know you're wrong. [Jane Q. Public, 2014-08-30]
I was worried that Jane was just trolling, and had no intention of ever acknowledging my method even if I described them from start-to-finish. Now that I've described my method from start-to-finish and Jane is pretending that he hasn't seen my method "in action" it seems like my worries came true.
Jane, if you won't do a single, solitary calculation of your own, could you at least please stop pretending that you haven't seen my method from start to finish? Here's my last step again:
Now that we've agreed on the inner shell temperature of ~149.9F, let's take the last step. Calculate the enclosed source temperature.
Draw a boundary just inside the inner surface of the enclosing shell. Because nothing in the boundary is changing with time, power in = power out. The same constant electrical power flows in as before the shell was added. Net radiative power flows out from the source to the enclosing shell's inner surface.
As before, that net radiative power is described by Wikipedia’s equation which accounts for areas and view factors.
#Completely surrounded by shell with finite conductivity.
var('sigma T_c T_h A_c A_h F_hc power epsilon_h epsilon_c')
eq1 = power == sigma*(T_h^4 - T_c^4)/((1-epsilon_h)/(epsilon_h*A_h) + 1/(A_h*F_hc) + (1-epsilon_c)/(epsilon_c*A_c))
soln4 = solve(eq1.subs(T_c=338.629929346551,power=15028.4258648090,sigma=5.670373e-8,epsilon_h=0.11,epsilon_c=0.11, F_hc=1, A_h=510.064471909788, A_c=511.185932522526),T_h)
soln4[0].rhs().n()
... Please explain what calculations you are using where, because I find it hard to tell the Sage-formatted calculations apart. [Jane Q. Public, 2014-09-08]
The first line "var('sigma..." declares my variables.
The line "eq1 = power == sigma..." is my "power in = power out" equation using Wikipedia's equation for net radiative power.
The next line plugs in all the relevant variables and solves it for the enclosed source temperature T_h.
The last line displays the answer.
So I've described my method for calculating the enclosed source temperature from start to finish. Before I post that final answer, can we agree with my method? If not, could you please describe your method?
Now that we've agreed on the inner shell temperature of ~149.9F, let's take the last step. Calculate the enclosed source temperature.
Draw a boundary just inside the inner surface of the enclosing shell. Because nothing in the boundary is changing with time, power in = power out. The same constant electrical power flows in as before the shell was added. Net radiative power flows out from the source to the enclosing shell's inner surface.
As before, that net radiative power is described by Wikipedia’s equation which accounts for areas and view factors.
#Completely surrounded by shell with finite conductivity.
var('sigma T_c T_h A_c A_h F_hc power epsilon_h epsilon_c')
eq1 = power == sigma*(T_h^4 - T_c^4)/((1-epsilon_h)/(epsilon_h*A_h) + 1/(A_h*F_hc) + (1-epsilon_c)/(epsilon_c*A_c))
soln4 = solve(eq1.subs(T_c=338.629929346551,power=15028.4258648090,sigma=5.670373e-8,epsilon_h=0.11,epsilon_c=0.11, F_hc=1, A_h=510.064471909788, A_c=511.185932522526),T_h)
soln4[0].rhs().n()
... Please explain what calculations you are using where, because I find it hard to tell the Sage-formatted calculations apart. [Jane Q. Public, 2014-09-08]
The first line "var('sigma..." declares my variables.
The line "eq1 = power == sigma..." is my "power in = power out" equation using Wikipedia's equation for net radiative power.
The next line plugs in all the relevant variables and solves it for the enclosed source temperature T_h.
The last line displays the answer.
... I don't see why you keep asking if I agree with your methods.
... [Jane Q. Public, 2014-09-08]
Because you've seemed to disagree with my method. That's why I've described my method for calculating the enclosed source temperature from start to finish. Before I post that final answer, can we agree with my method? If not, could you please describe your method?
Now that we've agreed on the inner shell temperature of ~149.9F, let's take the last step. Calculate the enclosed source temperature.
Draw a boundary just inside the inner surface of the enclosing shell. Because nothing in the boundary is changing with time, power in = power out. The same constant electrical power flows in as before the shell was added. Net radiative power flows out from the source to the enclosing shell's inner surface.
As before, that net radiative power is described by Wikipedia’s equation which accounts for areas and view factors.
#Completely surrounded by shell with finite conductivity.
var('sigma T_c T_h A_c A_h F_hc power epsilon_h epsilon_c')
eq1 = power == sigma*(T_h^4 - T_c^4)/((1-epsilon_h)/(epsilon_h*A_h) + 1/(A_h*F_hc) + (1-epsilon_c)/(epsilon_c*A_c))
soln4 = solve(eq1.subs(T_c=338.629929346551,power=15028.4258648090,sigma=5.670373e-8,epsilon_h=0.11,epsilon_c=0.11, F_hc=1, A_h=510.064471909788, A_c=511.185932522526),T_h)
soln4[0].rhs().n()
... Please explain what calculations you are using where, because I find it hard to tell the Sage-formatted calculations apart. [Jane Q. Public, 2014-09-08]
The first line "var('sigma..." declares my variables.
The line "eq1 = power == sigma..." is my "power in = power out" equation using Wikipedia's equation for net radiative power.
The next line plugs in all the relevant variables and solves it for the enclosed source temperature T_h.
The last line displays the answer.
So I've described my method for calculating the enclosed source temperature from start to finish. Before I post that final answer, can we agree with my method? If not, could you please describe your method?
On general principle, yes. When all factors are considered, this is true. I haven't disagreed with this general principle, and at this point I'm only really interested in seeing the rest of your calculations. Please explain what calculations you are using where, because I find it hard to tell the Sage-formatted calculations apart. [Jane Q. Public, 2014-09-08]
In order to explain what calculations I'm using, we have to first agree on the fundamental principle all my calculations are based on.
I'm glad we agree that power going in minus power going out through some boundary equals the rate at which energy inside that boundary changes.
Notice that this general principle applies to all systems, even if they're at different temperatures or out of (thermal/radiative) equilibrium.
Now suppose that nothing inside that boundary is changing with time. Since this includes the energy inside that boundary, the rate at which energy inside the boundary changes is zero. This means power in = power out through any boundary where nothing inside that boundary is changing with time.
If we can agree so far, just say "yes" and ignore the rest of this comment. Then we can move on to the final step, which is calculating the enclosed source temperature.
If we can't agree, here's why we first need to agree that power in = power out through any boundary where nothing inside that boundary is changing with time.
... a simple power-in = power-out view is not always the right answer.
... it shows how power-in = power-out calculations can easily mislead. [Jane Q. Public, 2014-09-07]
... your "energy conservation just inside the surface" won't work.
... [Jane Q. Public, 2014-09-07]
How could it mislead? Why won't it work? As long as nothing inside the boundary is changing, a simple power in = power out view is always the right answer.
... The problem is that an analysis of this kind, based on the assumption that power-in = power-out, is doomed to fail except in coincidental cases. Even conservation of energy can give very misleading results.
... power-in = power-out is not necessarily true, and in fact that is probably a very rare exception. Therefore, you aren't going to prove anything with this approach. I wanted to stop you before you wasted more of your time. [Jane Q. Public, 2014-09-07]
How is it doomed to fail? How could it give very misleading results? As long as nothing inside the boundary is changing, power in = power out is necessarily true.
... it does not translate directly into power in = power out at a boundary just inside the cavity surface. It most certainly does not if the bodies are not in thermal equilibrium, which again I must point out this system is not in.
... [Jane Q. Public, 2014-09-07]
No, energy is conserved even when the bodies aren't in thermal equilibrium. As long as nothing inside the boundary is changing, power in = power out.
... energy does not have to be conserved between two bodies at different temperatures. That was what Incorpora was saying in his book.
... [Jane Q. Public, 2014-09-07]
No, energy is conserved even between two bodies at different temperatures. As long as nothing inside the boundary is changing, power in = power out.
Can we agree that power in = power out through any boundary where nothing inside that boundary is changing with time? If so, then let's move on to the final step. Calculate the enclosed source temperature.
"Who alone has reason to *lie himself out* of actuality? He who *suffers* from it." -- Friedrich Nietzsche
