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Heat flow through metal bars

  1. Feb 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A steel bar 10cm long is welded end-to-end to a copper bar 20cm long. Both bars are insulated perfectly along their sides. Each bar has a radius of 2.0cm. The free end of the steel bar is maintained at 100 degrees C and the free end of the copper bar is maintained at 0 degrees C. Find the temperature at the junction between the two bars and the total rate of flow of heat.

    k(steel) = 50.2 Wm[tex]^{-1}[/tex]K[tex]^{-1}[/tex]
    k(copper) = 385.0 Wm[tex]^{-1}[/tex]K[tex]^{-1}[/tex]

    2. Relevant equations

    H = kA(T1-T2)/L

    3. The attempt at a solution

    I substituted the given values into the above equation for copper and again for steel and set them equal to each other (with the temperature in Kelvin):

    kA(T1-T2)/L (steel) = kA(T2-T1)/L (copper)

    and I ended up with an answer of 284.5K

    My tutor, however, ended up with an answer of 293.7K
    I used the different surface areas of the steel and copper parts of the tube in each side of the equation whereas my tutor used the area of the full length of the bar (i.e. an area of 2pi*rL with L = 30cm.
    Besides that our workings are the same.

    I don't understand why the the total area of the bar should be used instead of the area of each respective material.
    I hope that makes sense...

    Please help!
     
    Last edited: Feb 2, 2008
  2. jcsd
  3. Feb 2, 2008 #2

    Hootenanny

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    That isn't the correct area. Surely [itex]A = \pi r^2 = 0.0004\pi[/itex].
     
    Last edited: Feb 2, 2008
  4. Feb 2, 2008 #3
    Are you sure it isn't an area?
    2pi*rL in units is cm*cm = cm^2.

    As the bars are cylindrical wouldn't your way give the volume when you introduce the length of the bar?

    Aren't I looking for surface area as opposed to cross-sectional area?
     
    Last edited: Feb 2, 2008
  5. Feb 2, 2008 #4

    Hootenanny

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    Why are you looking for the surface area when the sides are insulating? Surely the heat conduction would occur through the cross-sectional area?
     
  6. Feb 2, 2008 #5
    Ooh hang on, it is cross-sectional area isn't it!
    So that would be why my tutor's method cancels out the areas =p

    Sorry about that!
     
  7. Feb 2, 2008 #6
    Yeah sorry, normally I'm ok at these. My exam's coming up and I think it's pre-exam nerves =/

    Sorry again!
     
  8. Feb 2, 2008 #7

    Hootenanny

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    Hey don't worry about it, you spotted the mistake so you know what your talking about. Listen if your revising, you don't need to be stressing, it those that don't revise that should be worried. Good luck in your exam :biggrin:
     
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