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Homework Help: Thermal conductivity question

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data
    A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other. The rod consists of a 1.00m section of copper (with one end in the boiling water) joined end-to-end to a length L_2 of steel (with one end in the ice water). Both sections of the rod have cross-sectional areas of 4.00 cm^2. The temperature of the copper-steel junction is 65.0 C after a steady state has been reached. How much heat per second, H, flows from the boiling water to the ice-water mixture?
    k_copper = 385 W/(m*K)
    k_steel = 50.2 W/(m*K)

    2. Relevant equations
    H = kA(T_H - T_C)/L

    3. The attempt at a solution
    I'm positive I know how to do this question. The thing is the site keeps rejecting my answer. I first find the heat of the copper end, because all values are known and I get a value for H, which comes to be 5.39W. Because it is at steady state, we know this heat is also equal to the heat of the steel side. if we plug in 5.39W into the equation again, we can solve for L_2 of the steel end, which I have, resulting in a length of 0.242152133m. Now we have can calculate the total heat flow of the system, which is H = H_copper + H_steel. I have done so and resulted in a total heat of 10.8W, which is supposedly wrong. Am I missing anything here? It says the cross sectional area of both rods, A, is 4.00cm^2, which I converted to 0.0004m^2. I can't think of what I'm doing wrong!!!

    Last edited: Oct 7, 2007
  2. jcsd
  3. Oct 8, 2007 #2
    anybody please?
  4. Sep 4, 2008 #3
    Hi Anthony,
    you have already solved this problem by calculating the length of the steel bar. You used the total energy of 5.39watts to ascertain the length of steel as apx 242mm. The assumption is that there is no loss of energy across the two materials therefore the answer is 5.30watts!
    Hope this helps
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