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DE: Newton's Law of Cooling

  1. Oct 4, 2008 #1
    So I have a slight problem since I'm getting a negative for time.

    A murder victim is discovered at midnight and the temperature of the body is recorded at 31C. One hour later, the temperature of the body is 29C. Assume that the surrounding air temperature remains constant at 21C.

    a) Find k

    [tex]T=A+(T_0-A)e^{-kt}[/tex]

    [tex]T(1)=21+(31-21)e^{-k}=29[/tex]

    [tex]k=\ln{\frac 5 4}[/tex]

    b) Solve for t

    [tex]37=21+(31-21)e^{-t\ln{\frac 5 4}}[/tex]

    [tex]16=10e^{\ln{\frac 4 5}^t}[/tex]

    [tex]t=\frac{\ln{\frac 8 5}}{\ln{\frac 4 5}}[/tex]

    But this gives me a negative t. Would that be okay since I'm trying to use this t value to compute an earlier time?
     
  2. jcsd
  3. Oct 4, 2008 #2

    Redbelly98

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    Yes.

    t=0 when the body is discovered and is at 31 C. So it would have been 37 C at an earlier (negative) time.
     
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