# thermal physics

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hey i just wanted to check if i was going the right way about doing this question i am not sure whether, after what i have already done here, i should then use the radiation equation (the rate at which any object radiates heat)

anyway, here is the problem and my answer

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this is my solution

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my quesion, is whether i should use this equation (radiation one):

but i can't see how i would exactly...hmmm , anyways yep thats all guys

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my quesion' date=' is whether i should use this equation (radiation one):

but i can't see how i would exactly...hmmm , anyways yep thats all guys [/quote']

Yes, you should. One hint is that they gave you the emissivity. But you have to use the correct form of the Stefan-Boltzmann equation, which I pint out in the human cooling thread - the temperature of the reservoir matters.

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I think, to be more accurate, you have to equate heat loss thru conduction and radiation to calculate the outside glass temperature.

That means, k(T1-T2)/l = sigma*emissivity*area*power((T2- T3),4)

you know T1, T3 and all other variable. So calculate T2. Then use q/t= K(T1-T2)/l to calculate heat loss.

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I think' date=' to be more accurate, you have to equate heat loss thru conduction and radiation to calculate the outside glass temperature.

That means, k(T1-T2)/l = sigma*emissivity*area*power((T2- T3),4)

you know T1, T3 and all other variable. So calculate T2. Then use q/t= K(T1-T2)/l to calculate heat loss.[/quote']

i think the problem with this is that you end up with a 4th degree polynomial, and so yeah makes things a little tricky....

or is it meant to be that tricky?

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the temperature of the reservoir matters.

reservoir?

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hmmm i can't quite see how to do this using both formulas...

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or do you simply add the two formulas together? (by that i mean add the energy lost through conducion + energy lost through radiation)...but then i suppose...aghhh head imploding!!!

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Although I for some reason cannot see the images, judging from the exchanges in the posts, I would say yes you need to account for both conduction and radiation.

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ok here is my solution now, i really hope its right , well here goes....how is it swansont? (and others )

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reservoir?

The surroundings.

You may notice in some areas that, on a cloudless night you get frost, even when the temperature does not drop to 0 C. That's because the open sky at night is at ~3K, and radiation is a much more important heat-loss mechanism. An object that doesn't "see" the sky (e.g. under a carport or tree) doesn't get frost on it, since it's radiating to a much warmer reservoir, >273K.

Your calculation looks OK to me, but I haven't actually checked the math.

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oh ok well i am not worried about the math being right (thats my fault if its wrong), but the method is right, the thinking is right, in that you add the two different methods of heat loss together....?

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well i am pretty sure this method is correct, as it gives a feesible answer

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thanks for you help everyone

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No, sarah, that's not correct. The correct answer will be what you get from following some nerd's suggestion (post #5). Unforunately, this does give you a 4th order polynomial. If you're clever, you can use a Taylor expansion and throw away high order terms and make life easier for yourself.

Conduction through the pane and radiation off its surface are processes in series not in parallel. You can not add them up.

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yeah thats what i thought, but that would be rather complex , hmm ... oh well i thought the problem was supposed to be simple guess not, huh?

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yeah thats what i thought, but that would be rather complex , hmm ... oh well i thought the problem was supposed to be simple guess not, huh?

I think it was supposed to be simple, and for the material in question, adding the heat loss mechanisms is a reasonable approximation.

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I think it was supposed to be simple, and for the material in question, adding the heat loss mechanisms is a reasonable approximation.
I can't say I agree with this - it is completely unphysical. If the outside of the wall is assumed to be at T(out) - this is the assumption you make in justifying the conduction equation as written by Sarah - there will be NO radiation. And if the outside wall is assumed to be at T(in) - this is the assumption made in writing the radiation equation in that form - there will be no conduction.

As, I said before, conduction and radiation are not processes that happen in parallel here. "Adding them up" makes no sense to me (and this answer will be an overestimate; albeit by a fraction comparable to P(rad)/P(cond) as calculated by Sarah).

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I can't say I agree with this - it is completely unphysical. If the outside of the wall is assumed to be at T(out) - this is the assumption you make in justifying the conduction equation as written by Sarah - there will be NO radiation. And if the outside wall is assumed to be at T(in) - this is the assumption made in writing the radiation equation in that form - there will be no conduction.

As' date=' I said before, conduction and radiation are not processes that happen in parallel here. "Adding them up" makes no sense to me (and this answer will be an overestimate; albeit by a fraction comparable to P(rad)/P(cond) as calculated by Sarah).[/quote']

I don't buy it. How do you explain all the IR cameras that show radiation through windows with thermal imaging? Radiation is obviously present.

The question isn't about heat loss from a window, it's about heat loss through a window. You make the radiation go away by making it opaque so the inside can't "see" the cold surroundings - it sees the warm window instead. If it's transparent, the temperature of the window doesn't matter as far as radiation is concerned.

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Okay, I get what you're saying now. But the transmission of radiation by a transparent medium is not given by the Stefan-Boltzmann Law. That only talks about radiation from a heated surface. The energy transmitted through a transparent medium is a function of the transmission coefficient, not the emissivity.

Oh, wait, you're treating the inside surface of the glass window as the radiator ? I'm not sure I see how exactly that works, but I haven't really given it much thought.

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I was assuming the emissivity referred to whatever was inside the room.

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