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Equilibrium Temperature.

  1. Jun 15, 2008 #1
    I am not a novice at thermodynamics but still the following question made me think over my status more than twice.

    I have to find the equilibrium temperature when a 50 Kg block of copper at 80C is placed in a container which as 120L of water at 25C.

    I made simple equations of heat lost = heat gained. Used Specific Heat for water = 4.18 and for copper 0.285 (and 0.391) and used mcdelT. But the answer I get is below 25C. Where as it should be between 25 and 80.
     
  2. jcsd
  3. Jun 15, 2008 #2

    tiny-tim

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    Hi Altairs! :smile:

    You know you should show us your working … otherwise how can we see where you went wrong? :confused:
     
  4. Jun 16, 2008 #3
    Oh I forgot the PF SOP :rolleyes:

    Heat Lost (by copper) = Heat Gained (by water)

    [tex]50 * 0.385 (T_{e} - 80) = 0.12 * 1000 *4.18 (T_{e} - 25)[/tex]
     
  5. Jun 16, 2008 #4

    tiny-tim

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    hmm … that's why we ask to see your working!

    no wonder they both got colder! :rolleyes:

    It would be less confusing if you said "total heat change = 0, therefore:"

    [tex]50 * 0.385 (T_{e} - 80)\,+\,0.12 * 1000 *4.18 (T_{e} - 25)\,=\,0[/tex] :smile:
     
  6. Jun 18, 2008 #5
    Ooh ! Thanks :blushing:
     
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