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Specific Heat

  1. Sep 25, 2008 #1
    A 215-g sample of a substance is heated to 330°C and then plunged into a 105-g aluminum calorimeter cup containing 165 g of water and a 17-g glass thermometer at 12.5°C. The final temperature is 35.0°C. What is the specific heat of the substance? (Assume no water boils away.)
    Please, can anybody tell me if I'm doing this right? I got to the certain moment and then I got stock. Heelp! c(glass)=0.84, c(aluminum)=0.900, c(water)=4.2 J/g*C

    Tinitial (aluminum)=Tinitial (water)=Tinitial (glass thermometr)=12.5 C?
    And Tfinal (al)=Tf(water)=Tf(glass)=35C?
    Q(substance)=Q(aluminum)+Q(water)+Q(glass)
    Q=mc*(Tf-Ti)
    But what is the Tinitial for the substance? We know only 330C and this is final?
    215*c*(330-Ti)=105*0.900*(35-12.5)+165*4.2*(35-12.5)*17*0.84*(35-12.5)
     
  2. jcsd
  3. Sep 26, 2008 #2

    hage567

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    The 330 degrees is the initial temperature of the substance. When it is placed in the calorimeter its temperature will come to 35 degrees since the system is in equilibrium.
     
  4. Sep 26, 2008 #3
    So, everything else is right? Then change of temperature is equal to 330-35? Or 35-330? And then I just calculate c from the equation?
    thanks for help!!
     
  5. Sep 27, 2008 #4

    hage567

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    If you define the change in temperature as ∆T = Tf - Ti , which do you think it is?
    I think your terms are OK, but just remember that the heat lost by the substance equals the heat gained by the water+aluminum+glass.
     
  6. Sep 27, 2008 #5
    Thank you! That's why it's going to be minus! I just got it!
     
  7. Sep 27, 2008 #6

    hage567

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    Well it shouldn't be minus, specific heat capacity can't be negative! What I meant was that
    (heat lost by the substance) = - (heat gained by water+aluminum+glass).
    This is the proper way to write your equation (conservation of energy applies). So if you put the minus in your answer will come out positive, as it should.

    You're welcome, by the way. :smile:
     
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