Crazedwraith wrote:
Thanks for the detailed response, Feil. That's great. Sort of proves how very rusty my physics is, since I used Stefan-Boltzmann stuff in relation to stars in my degree (only just scraped it though)
So I take those terms are only equal when there is no temperature change.
The terms are equal all the time - it's what changes force what other changes that tell us interesting things about what's going on. If the temperature drops, and area is held constant, then we know the power being radiated will drop, too. If we increase the power that our object has to dissipate, then either the temperature has to increase or the surface area has to increase.
The way we could handle that without a ship that changes size or cooks its passengers is, first, by pumping heat to the outside while keeping the inside cool by means of an air conditioning system. This makes our temperature at the surface (which is the only one we care about) higher, while keeping the inside at the same temperature. We can do the same thing on the surface itself by pumping the heat to a radiator with a very high emissivity and large area, away from the surface of our living component, which has a low emissivity and a small area. If we do that, then our equation changes slightly to account for the fact that we no longer have an even distribution: e becomes the function e(r(s,t)) and T becomes the function T(r(s,t)), so our equation changes to
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Ss*e(r(s,t))T(r(s,t))^4 dS = P
for the purposes of our equation. (The average over the surface given by the function r(s,t) of eT^4, times the area, times the stephan-boltzmann constant, equals the power radiated.) Solve piecewise if the shape of the tin can can't be described with a single-valued function, or if piecewise solution makes the integral easy.
Note that P is the net power that the object has to emit to avoid changing temperature. That is to say, it is the sum of all power produced by and absorbed by the object. This is why it is easy to keep warm when it is warm out, but hard when it is cold out: You are receiving more power to your skin on a warm day than on a cold day.
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And its basically an surface area vs amount of energy being radiated? A ship with a lot of higher volume compared to its surface area is going to cool slower than something with a higher surface area/volume ratio?
Correct. Heat capacity C is directly proportional to volume, and if we take the time derivative of the definition of heat capacity, CdT/dt = dQ/dt = P. Power is directly proportional to area, as given by the Stephan-Boltzmann eq, so shapes with higher volume to surface area ratios lose temperature more slowly in the same environment. (This is why frigid climates lack insects and most small animals, for instance: they have small volumes for their surface area, so it is comparatively costly for them to maintain a temperature different from their environment.)