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Newton's law of cooling and thermometer

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi, I need some help in solving Newton's law of cooling problem.

    Use Newton’s law of cooling to determine the
    reading on a thermometer 5 min after it is taken from an oven
    at 72 ◦ C to the outdoors where the temperature is 20 ◦ C, if
    the reading dropped to 48 ◦ C after one min.


    2. Relevant equations

    [tex]\frac{dT}{dt}=k(T-R)[/tex]

    3. The attempt at a solution
     
    Last edited: Nov 24, 2009
  2. jcsd
  3. Nov 24, 2009 #2

    Mark44

    Staff: Mentor

    What have you tried to do? You have to make an attempt before anyone will give you any help - them's the rules.
     
  4. Nov 24, 2009 #3
    OK, we let T be the temperature of the object. Then, by Newton's Law of Cooling, [tex]\frac{du}{dt}=k(u-20)[/tex], for some constant k. Let [tex]y=T-20[/tex]. Then [tex]\frac{dy}{dt}=\frac{du}{dt}=ky[/tex]. Thus we are now using the formula [tex]y=y_{0}e^{kt}[/tex]. Since T is initially 72 ◦ C , [tex]y_{0}=72-20=52[/tex]. So, [tex]y=52e^{kt}[/tex].
    This is what I can do, so I'm asking you for some further help.
     
    Last edited: Nov 24, 2009
  5. Nov 24, 2009 #4

    Mark44

    Staff: Mentor

    You have thrown in an extra variable that you don't need.

    Then, by Newton's Law of Cooling, [tex]\frac{dT}{dt}=k(T-20)[/tex], for some constant k. Let [tex]u=T-20[/tex]. Then [tex]\frac{du}{dt}=ku[/tex], and the solution to this differential equation is [tex]u=u_{0}e^{kt}[/tex]. Since T is initially 72 ◦ C , [tex]u_{0}=72-20=52[/tex]. So, [tex]u=52e^{kt}[/tex].

    Reverting to T (we really want to know the temperature), we have T - 20 = 52ekt, or T(t) = 20 + 52ekt.

    You're given that T(1) = 48, so use this fact to solve for k. After you have k, evaluate T(5) and you're done.
     
  6. Nov 24, 2009 #5
    this helped me a lot, thank you! :)
     
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