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Homework Help: Heat flow problem- what did I do wrong?

  1. Apr 27, 2008 #1
    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 View Figure . The rod consists of 1.00-m section of copper (one end in steam) joined end-to-end to a length L2 of steel (one end in ice). Both sections of the rod have cross-section areas of 4.00 cm^2. The temperature of the copper-steel junction is 65.0 deg C after a steady state has been set up.

    View Figure at http://session.masteringphysics.com/problemAsset/1042082/4/YF-17-70.jpg

    1. How much heat per second flows from the steam bath to the ice-water mixture?

    2. What is the length L2 of the steel section?

    Now, here is my solution:


    ∇q = 0
    Because there is only heat flux along the rod this simplifies to:
    dq/dx = 0 => q = constant
    along the rod.

    Heat flux is defined as
    q = k·dT/dx
    k is thermal conductivity

    For constant q and k you find:
    q = - k·ΔT/Δx

    So the heat flow through the rod is
    Q = A·q = -A·k·ΔT/Δx

    The heat flow is constant , That means heat flow through copper section as through steel section as through the whole rod.

    consider copper section:
    k = 400 W/Km for copper

    Q = -A·k·ΔT/Δx
    = -4.00×10-4m² · 400W/Km · (65°C - 100°C) / 1m
    = 5.6W

    I put in 5.6, and it told me it was the wrong answer.

    What did i do wrong?

  2. jcsd
  3. Apr 27, 2008 #2


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    Steel and copper have different thermal conductivities, so the temperature differentials will be different. The boiling water at 1 atm is at 100°C, which is sat temp at 1 atm, and the ice is at 0°C.

    The interface is at 65°C, and the heat flux on both sides is equal, but the dT/dx is not.

    Find expressions for the heat flux in the copper and steel at the interface and set them equal.
  4. Apr 27, 2008 #3
    How would you do that? I am not good with heat flux.

    Please help me get started.
    Last edited: Apr 27, 2008
  5. Apr 27, 2008 #4
    How would I set up the expression?
  6. Apr 27, 2008 #5


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    Q = A·q = -A·k·ΔT/Δx is correct.

    Heat flux = q = -k ΔT/Δx. One knows the ΔT's and the length of copper.

    So q (copper) = q (steel), and the k's are different.

    The 5.6 seems correct, but for 1 sec the heat (energy) should be J. W is J/s, which is power.

    Does the answer input require units?
    Last edited: Apr 27, 2008
  7. Apr 27, 2008 #6
    It requires W. I put in 5.6 and got "try again."

    I don't know what to do. Shouldn't I add both q's?
  8. Apr 27, 2008 #7
    Please! help me like a comdemned animal.
  9. Apr 28, 2008 #8


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    In order to have steady-state, what goes in one end must go out the other, i.e. the rate of heat transfer is constant along the length of the bars. The formula one used seems correct.
  10. May 1, 2008 #9
    Thermal Conductivity

    Looks like you used a Thermal Conductivity value of 400 which is close to the value that wikipedia has. However, my book (and yours if same book p 664) uses the value of 385.0 for the thermal conductivity of copper.

    ...so you did it correctly. However, the answer using a thermal conductivity of 385 would be 5.39W.
  11. May 5, 2008 #10
    Length of L_2

    How do you actually find the length?

    Thanks in advance
  12. May 5, 2008 #11


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    One would use the heat flux equation and solve for [itex]\Delta x[/itex].
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