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Unread 06-29-2003, 11:37 PM   #43
Alchemy
Cooling Savant
 
Join Date: Oct 2002
Location: Boston
Posts: 238
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Quote:
Originally posted by Since87
I believe that what Dave was trying to convey, was that given two rods of equal length and mass, one made of copper and one of aluminum, the aluminum rod would conduct more heat for a given temperature differential between the ends of the rods, than the copper rod. Is this not correct?
That's what I got, too. It's not implicitly obvious, but it can be checked pretty easily.

The density of Al is 2700 kg/m^3 and the density of Cu is 8920 kg/m^3.

The thermal conductivity of Al is 2.37 W/cm-C and Cu is 4.01 W/cm-C

Writing out Fourier's law for two cylindrical bars of equal length, equal mass, and equal heat load, I get:

delta-T2 / delta-T1 = k1/k2 * rho2/rho1

So if "2" is aluminum and "1" is copper, you get:

delta-T (Al) / delta-T (Cu) = 4.01/2.37 * 2700/8920 = 0.512

So, for cylindrical rods of the same mass, you're going to get around twice the transfer of heat (or half the temperature differential) using aluminum.

Wow, this really takes a while without my texts. Maybe I should keep this computer in my office . . .

Alchemy

Edit - adjusted thermal conductivities. Thanks, flyingass!

Last edited by Alchemy; 06-30-2003 at 01:37 AM.
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