1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    where:

    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:

    Thanks!
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Final temperature of diamond using Cv
  1. Final temperature (Replies: 0)

  2. Final temperature (Replies: 1)

  3. Final temperature? (Replies: 3)

Loading...