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Newton's Law of Cooling of pizza

  1. Mar 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Bob loves vegetarian pizza. How long will it take Bob's pizza to cool to 110 degrees if he bakes it at 450 degrees for 20 minutes? The temperature in his house is a balmy 70 degrees.

    2. Relevant equations
    T(t) = Ts + Do*e^(-k*t) where...

    Do = initial temperature difference (Initial Temperature - Ts)
    Ts = surrounding temperature
    t = time
    k = constant (cooling rate)

    3. The attempt at a solution

    We know the following...

    > Ts := 70;

    The initial temperature difference (Do) is the pizza temperature (when it came out of the oven) minus the surrounding temperature...

    > Ti := 450;

    > Do := Ti-Ts; ** 450 - 70 = 380 degrees **

    We also know the temperature at time "t"... 110 degrees.

    > T := 110;

    Therefore our formula is now:

    110 = 70 + 380 exp(-k*t)

    ** The problem is that I need to know what the cooling constant "k" is, in order to solve for the real question, "t". **

    The 20 minutes given in the problem must be useful, but I'm not sure how. Can you please help?
  2. jcsd
  3. Mar 4, 2007 #2
    I guess we could object to the question since we don't know the temp of the pizza before it enters the oven. Do you think that matters here?
  4. Mar 4, 2007 #3
    I honestly don't think this was meant to be a "trick" question... but you're right. We don't know the temperature of the pizza *before* we put it in the oven. Would this matter? I assumed (maybe wrongly) that after 20 minutes in the oven @ 450 degrees, that it would come out at that temperature. That's why I used 450 as my initial temperature (after heating in the oven).
    Any other ideas?
  5. Mar 4, 2007 #4
    If the question does not state the initial temp before the oven, I would imagine you that you are to assume it comes out at 450 degrees.
  6. Mar 4, 2007 #5
    Still don't know what the heck "k" is... I have one equation and two unknowns... The formula simplifies to:

    0.105 = e^(-k*t)

    and I need to solve for t...
  7. Mar 4, 2007 #6
    I agree, had we one other point we could solve for both k and t obviously.
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