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If I have a line of copper wire (lets say 1 meter long, 1mm thick) and one end is a flattened disk of copper about the size of a quarter, and I apply a lot of heat to it (I'm talking 800 Celsius) will the entire line be heated to the same degree? I mean what temperature will the unheated end be after, say, a minute? Can it too reach 800 degrees over time? how much time? And lets assume it's in a vacuum so no air removes any of the heat the copper holds.

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I have TIG wielded with #12AWG copper wire. One end is liquid, several inches away I hold it in my gloved hand. The glove is for UV protection. I have done the same with aluminum. –  Optionparty Dec 31 '13 at 2:45

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Heat transport here obeys the heat diffusion equation $\partial_t T = \chi \partial^2_{xx} T$ where T(x,t) is the temperature and $\chi$ is the heat diffusivity. For copper $\chi \sim 10^{-4}$ m$^2$/s, according to Neglecting heat losses (thermal radiation in vacuum) we can estimate that in time $\tau \sim L^2/\chi$ where $L\sim$1 m is the wire length, the temperature will be pretty well equilibrated along the wire. Putting the numbers in, $\tau \sim 1/10^{-4} = 10^{4}$ s $ \sim $ 1 hour. An exact solution (in the form of a series) can be easily obtained if radiation is neglected. In particular it will show that the temperature at the unheated end will reach 800 C asymptotically in time (but will get close to it in $\tau \sim L^2/\chi$). With radiation, the problem becomes nonlinear so a numerical solution will be needed. But I would expect for this relatively low temperature and assuming a realistic diameter wire the effect of radiation losses will be pretty small, probably it will produce a correction below 1% .

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The surface area may be small in absolute terms, but the volume is smaller still. The temperature of a thin wire should be affected more by radiation than that of a thick wire. –  Douglas Zare Feb 17 at 10:09
You are certainly right, the relative role of surface losses (radiation and others) should be larger for a smaller size body; correcting the misleading statement in the answer. –  Maxim Umansky Feb 17 at 18:58

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