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Homework Help: Final temperature of diamond using Cv

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data

    At low temperatures, Cv of solids is given by:

    Cv = D([itex]\frac{T}{\Theta}[/itex])3


    D = 1.94*10^3
    [itex]\Theta[/itex] = Debye Temperature, for diamond 1860K

    If 100J of heat is transferred to 1mole of diamond, initially at 10K. Calculate the final temperature.

    2. Relevant equations

    Cv = D([itex]\frac{T}{\Theta}[/itex])3
    Q = mCv(Tf-Ti)

    3. The attempt at a solution

    This is a straight forward question as the equation becomes:

    Q = mCv(Tf-Ti)

    Tf = [itex]\frac{Q}{mC}[/itex]+Ti

    So the real problem I'm having is the fact that Cv is a function of T and T is changing as heat is being put in.
    So how do I go about this question?

    Thanks in advance
  2. jcsd
  3. Jun 2, 2012 #2
    Oh and I'm guessing mass is just 1 molar mass of carbon
  4. Jun 2, 2012 #3
    dQ = CvdT

    Plug in your expression for Cv and integrate from 10 to T:

    Q = 1000J = ∫CvdT' from [10, T]

    then solve for T.
  5. Jun 2, 2012 #4
    Oh... that would do it... *Face Palm*

    Can't BELIEVE I didn't see that.:uhh:

  6. Jun 2, 2012 #5
    Actually how do you propose to do that?

    Cv is the only variable changing with Temperature so:

    Cv = D[itex]\int[/itex][itex]\frac{T^3}{θ^3}[/itex]

    Cv = [itex]\frac{D}{θ^3}[/itex][itex]\int[/itex]T3 (from 10 to Tf)

    Cv = [itex]\frac{D}{4θ^3}[/itex] (T4f-104)

    Q =mCv (Tf-10)

    Q =[itex]\frac{mD}{4θ^3}[/itex] (T4f-104)

    Let A = [itex]\frac{mD}{4θ^3}[/itex]

    Q =A (Tf-10) (T4f-104)

    At this point it becomes pretty much impractical to solve. Am I doing something wrong?
  7. Jun 2, 2012 #6
    Q is the heat you added to the system, 1000J. It's just simple algebra from there.
  8. Jun 3, 2012 #7
    If we expand it (Q = 100J btw)

    [itex]\frac{T^5}{10}[/itex] -T4-103T-104=[itex]\frac{10}{A}[/itex]

    I suppose you're going to tell me I have to do long division? No easier way?
  9. Jun 3, 2012 #8
    No, your integration is wrong here.

    Q = ∫CvdT = A(T4 -104)

    Q/A + 104 = T4

    T = (Q/A + 104)1/4
  10. Jun 3, 2012 #9
    Wow... I should go back to school. Thanks a lot.
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