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Thermodynamics: Enthralpy of body

  1. Feb 28, 2010 #1
    1. I meant entropy of body. According to Debye's law, the molar heat capacity at constant volume of a diamond varies with the temperature as follows:
    cv = 3R(4(pi)4/5)(T/Θ)3

    What is the entropy change in units of R of a diamond of 1.2g mass when it is heated at constant volume from 10K to 350K? The molar mass of diamond is 12g and Θ = 2230K

    2. ΔS = (integral)dQ/T
    dQ = cvdT (heat capacity at constant volume)
    so: ΔS = (integral from T1 to T2)(cvdT/T)

    3. So before doing the integral I replaced cv with the Debye's law. I then moved all the constants outside the integral so I had:
    ΔS = 3R(4(pi)4/5)(integral from T1 to T2)(T23dT/T)
    Then I integrated and replaced T1 with 10K and T2 with 350K. I also multiplied by .012kg since that's the molar mass.
    So I end up getting .00362R for my answer.

    I'm just really new at all this (enthralpy), so if someone can let me know if I'm doing this correctly or way off that would be great. I feel pretty good about it though.
    Last edited: Mar 1, 2010
  2. jcsd
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