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Entropy integration problem

  1. Dec 8, 2004 #1
    Find a) the energy absorbed as heat and b) the change in entropy of a 2 kg block of copper whose temperature is increased reversibly from 25 degrees C to 100 degrees C. The specefic heat of copper is 386 J/kg * K.

    a) Q = 386 J/kg *K * (373 K - 298 K) * 2 kg = 57900 J

    b) the change in entropy = integral(dQ / T). I'm unsure of how to integrate to be able to have a usable formula???
     
  2. jcsd
  3. Dec 9, 2004 #2

    Andrew Mason

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    Since [itex]\Delta Q = cm\Delta T[/itex]:

    [tex] \Delta S = \int_{Ti}^{Tf }\frac{\delta Q}{T} = \int_{Ti}^{Tf } \frac{cm\delta T}{T} [/tex]

    Note:
    [tex]\int_{Ti}^{Tf }\frac{\delta T}{T} = ln(\frac{T_f}{T_i})[/tex]

    AM
     
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