- #1
xortan
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Homework Statement
Hello everyone I am studying for my finals and this question has me stumped.
The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?
We know that Q/t = 10 cal/s
Homework Equations
P = Q/t = kAT/L
k = thermal conductivity constant
A= surface area of object
T= change in temperature
L = length that heat through
k for steel is 0.111 cal/gC
The Attempt at a Solution
I tried saying that 10 cal/s = k*2A*T/2L and since the area of the cylinder is 2*pi*(d/2)*L, I subbed that into the 2A and canceled off some stuff then tried pulling out the coefficients and diving that by 10 but I'm not to sure if this is the right direction