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Newton's Law of Cooling of porridge

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data
    A smaller bowl of porridge served at 200 degres F cools to 160 degres in 1 min. What tempature (too cold) will this porridge be when the bowl of exercise 27 has reached 120 degres F (just right)?


    2. Relevant equations
    y(prime)(t)=k((y(t)-T sub a )
    y(t)=Ae^(kt)+T sub a


    3. The attempt at a solution
    I really don't get this.
    Thanks for your help, and if you can explain that would be great.
     
  2. jcsd
  3. Oct 17, 2007 #2
    Well, this is just a simple differential equation:

    [tex] y' = \frac{dT}{dt}=k (T-T_a)[/tex]
    [tex]\frac{dT}{(T-T_a)}=k dt[/tex]
    [tex]\int{\frac{dy}{(T-T_a)}},\int{k dt}[/tex]
    [tex]ln(T-T_a) = k t + C[/tex]
    [tex]T=Ae^{kt}+T_a[/tex]
    So that's how the two equations are related.
    Now:
    [tex]T(0)=Ae^{k 0}+T_a=T_0[/tex] Where [tex]T_0[/tex] is the temperature the soup starts at (initial T) and [tex]T_a[/tex] is the temp of the air around the soup(ambient T). Use 200 degrees as the starting temp.
    [tex]T(t)=(T_0-T_a)e^{kt}+T_a[/tex]
    So now you need the constant k. Plug in 1 min for t, 160 for T(t) and then the values for the initial T and the ambient T and solve for k. After all that you can plug the time from execercise 27 and find the temp of the soup.
     
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