Thermal conductivity twice that of brass

Dx

Hiya!

The thermal conductivity al Al is twice that of brass. Two rods (1 Al and the other brass) are joined together end to end in excellent thermal contact. The rods are of equal lengths and radii. The free end of the brass rod is maintained at 0 degrees C and the Al free end is heated to 200 degree C. If no heat escapes from the sides of the rods, what is the temperatur at the interface between the two metals.

This sounds somewhat like a difference equation problem and i think i must find the thermal conductivity of Al and brass which i think one is .11 and the other .84 or something, ill have to recheck it. My question is what formula do i use to solve for this?

Thanks!
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Tom Mattson

Staff Emeritus
Gold Member
Originally posted by Dx
The thermal conductivity al Al is twice that of brass. Two rods (1 Al and the other brass) are joined together end to end in excellent thermal contact. The rods are of equal lengths and radii. The free end of the brass rod is maintained at 0 degrees C and the Al free end is heated to 200 degree C. If no heat escapes from the sides of the rods, what is the temperatur at the interface between the two metals.
At what time? That bit of information is crucial!

This sounds somewhat like a difference equation problem and i think i must find the thermal conductivity of Al and brass which i think one is .11 and the other .84 or something, ill have to recheck it. My question is what formula do i use to solve for this?
A difference equation? No, you are going to run into a partial differential equation, namely the heat equation.

Dx

Re: Re: Thermal conductivity

Originally posted by Tom
At what time? That bit of information is crucial!
You know it doesnt say, interesting? It must be a trick question then.

A difference equation? No, you are going to run into a partial differential equation[/i], namely the heat equation.
Your right, a partial.

Thanks!
Dx Tom Mattson

Staff Emeritus
Gold Member
Yes, the heat equation is (for constant conductivity):

c&rho;&part;T/&part;t=&kappa;[nab]2T

Since there is a time dependence, we need to know something about that.

Integral

Staff Emeritus
Gold Member
Go for the time independent solution. If both ends of the system are held at a constant temp a non time dependent solution will develop.

dQ/dt = - kAdT/dx
A = cross sectional area
k = thermal conductivity

This holds in each material. Further, since no heat is lost from the system, we have dQ1/dt = dQ2/dt, also given in the problem A1 = A2

let dT/dx = (T2 - T1)/L in each material

You need to draw a picture of the sytem, define T1, T2 & T3 for your system, recall that the boundry will be a common point. Assemble the pieces of the puzzel I have presented and do some algebra.

gnome

i think i must find the thermal conductivity of Al and brass which i think one is .11 and the other .84 or something
Don't look up the thermal conductivities. In this problem you are given that "the thermal conductivity of Al is twice that of brass" (which is approximately correct) so that is the relationship you should use to solve it (i.e. let k=thermal conductivity of brass and 2k=thermal conductivity of Al).

I don't disagree with Integral, but you may find this easier (it is based on exactly the same approach as Integral's answer):
Assuming that the temperature gradient is uniform,
P (the rate of energy transfer) is proportional to the cross-sectional area A and the temperature difference, and inversely proportional to the length.
k, the thermal conductivity, is the proportionality constant, so (for EACH rod):
P = kA(T2 - T1)/L

Once a steady state is reached, the rate of heat transfer through both rods will be equal.

Using this, you should be able to calculate the steady-state temperature at the interface.

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