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Newtons Cooling problem/ possible error in book.

  1. Jun 6, 2013 #1
    I have a book called REA 's problem solvers: Differential Equations Year 2004 and the problem 12-14 on page 244 might have problems with it and I want to ask to see if anyone else thinks so.

    The problem is basically Newton's cooling where they ask to find

    T - T_o = (T_i - T_o) e^(-kt) (1)

    where T_i = T when t = 0.

    Given dT/dt = -k(T - T_o) (2)

    where T_o is the temperature of the surrounding medium.

    The part that seems incorrect to me is where they integrate after rearranging (2) to get

    LN(T - T_o) = -kt + LN(C)

    1. Shouldn't the left side be -LN(T_o - T) after integration?
    2. How do they get LN(C) instead of just C and then solving for t = 0, C = -LN(T_o - T)?
     
  2. jcsd
  3. Jun 6, 2013 #2

    dextercioby

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    To put C or ln C is the same, the constant is arbitrary. Ln C is more convenient. As for the first question, you're basically computing

    [tex] \int \frac{dT}{T-T_0} [/tex]
     
  4. Jun 6, 2013 #3
    Would C = T - T_o instead? That would make more sense.
    And then LN C (or LN(T - T_o) ) at t = 0 would be the integrating factor?

    Right, I know I'm computing [tex] \int \frac{dT}{T-T_0} [/tex] but is it correct in the book or am I right. I used Derive 6 and I got a different answer after integration.

    EDIT: Nevermind on the integration, it is correct. I made a mistake in Derive 6. But I'm still not sure about the constant C.
     
    Last edited: Jun 6, 2013
  5. Jun 7, 2013 #4

    Integral

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    C is a constant, so is lnC, or C2 or any other operation done to C. The constant is arbitrary, so the exact form of it is not important. Though the form can make succeeding steps more transparent.
     
  6. Jun 7, 2013 #5
    https://www.youtube.com/watch?v=xq-JiYMQydc
     
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