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Homework Help: Specific Heat/Energy problem

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    The specific heat of a substance varies with temperature according to:
    c = 0.20 + 0.14T + 0.023T[tex]^{2}[/tex],​
    with T in deg-C and c in cal/gK. Find the energy required to raise the temperature of 2.0g of this substance from 5.0C to 15C.

    2. Relevant equations
    K= C + 273.15

    3. The attempt at a solution
    m= 2.0g

    Q=m[tex]\int[/tex](0.2 + 1.4T + .023T[tex]^{2}[/tex])dt

    Convert C to Kelvin:
    T1: K = 5.0 + 273.15 = 278.15K
    T1: K = 15 + 273.15 = 288.15K

    Solve for heat:
    Q = [2.0g(0.2 + 1.4(288.15) + 0.23(288.15[tex]^{2}[/tex])] -
    [2.0g(0.2 + 1.4(278.15) + 0.23(278.15[tex]^{2}[/tex])]


    I don't think I did the math correctly can anyone please check it for me and tell me what I did wrong? Thank you in advanced.

  2. jcsd
  3. Mar 1, 2009 #2
    Looking at this it doesn't look like you did the integral for the heat capacity? You should have 0.2T+1.4T^2/2 etc.
  4. Mar 1, 2009 #3
    Why divide by 2? And I thought I did take the integral? I couldn't figure out how to add the 15 and 5 onto the integral sign, but I thought I did solve it in that manner? Maybe not? Now I've confused myself lol.
  5. Mar 1, 2009 #4
    This is the value of the heat capacity?
    This is your formula for Q

    Here you have just subbed the value of c into [tex]Q=mc\delta T[/tex] I see no evidence that c has been integrated?
  6. Mar 1, 2009 #5
    You put the limits on like this (just click on the equation to see the source),


    and you divide by two because,

    [tex]\int x dx=\frac{x^{2}}{2}[/tex]
  7. Mar 1, 2009 #6
    Oh ok, I see thanks alot. Didn't notice I skipped that part. Thanks again.
  8. Mar 1, 2009 #7
    Your welcome. Does the answer come out ok? If you need anymore help just ask.
  9. Mar 1, 2009 #8
    I did it again and I got Q = 9894.36. I don't think that's right. This question is so simple I don't know why I'm not getting it.
  10. Mar 1, 2009 #9
    Don't worry, it's only easy once you know it. Check your calculation again and post it here before you enter it, that way I can check it and hopefully you won't lose another attempt. I seem to be getting quite a big number.

    Edit: Why don't you think it's right? Also I got confused with someone else, thought you were using a mastering physics package (me tired, sorry).
  11. Mar 1, 2009 #10
    It's probably not clicking in my head since I'm studying for organic chemistry II and doing this at the same time, but this is what I did:

    c = [0.20 + 1.4(288.15) + 0.023(288.152))] - [0.20 + 0.14(278.15) + 0.023(278.152)]
    c = 494.78


    Q = 9894.36
  12. Mar 1, 2009 #11
    Or is Q negative, where T1 = 278.15 and T2 = 288.15?
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