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Newton's Law of Cooling on cup of coffee

  1. Mar 18, 2005 #1
    A cup of coffee at 174 degrees is poured into a mug and left in a room at 76 degrees. After 5 minutes, the coffee is 134 degrees. Assume that the differential equation describing Newton's Law of Cooling is (in this case) dT/dt = k(T-76)

    here's what i done:

    [tex]y(0) = 98e^{kt}[/tex]

    y(5) = 134-76 = 58, so...

    98e^{5k} = 58
    k = -0.104904
    [tex]y(15) = 98*e^{15*-0.104904} + 76[/tex]
    and got 159.8448 which is wrong. anyone know where i went wrong?
     
  2. jcsd
  3. Mar 18, 2005 #2

    Integral

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    Go back to the general solution of the DE, you have made some mistakes in how you have handled the constants.
     
  4. Mar 18, 2005 #3

    xanthym

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    All your work is basically correct (although the presentation could improve a bit). You've made a careless mistake computing the final equation (in RED above). That equation is correct and evaluates to:
    {Temperature at t=(15 min)} = (96.316 deg)


    ~~
     
    Last edited: Mar 18, 2005
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