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Differential Eqns: Newton's Law of Cooling

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    A hot solid body is immersed into a large body of water
    which is a 4 degrees C. After one hour the body's temperature is measured
    at 40 degrees C, after 2 hours at 15 degrees. Assuming that Newton's law of
    cooling applies and that the water body is mixed constantly and its temperature stays at 4 degrees, what was the temperature of the body initially?


    2. Relevant equations

    Newton's Law of Cooling
    dT/dt=κ(T-Tm)

    3. The attempt at a solution
    Tm=4°C
    T(1h)=40°C
    T(2h)=15°C
    T(0h)=?

    dT/dt=k(T-Tm)
    dT/dt=k(T-4)
    separating variables and integrating yields
    ln|T-4|= kt+c
    T=ce^(kt)+4 .............eq1

    T(1)=40 → 40=4+ce^k .....eq 2
    T(2)=15 → 15=4+ce^2k ......eq3

    So I solved for k using eq2 & eq3
    36e^(2k)=11e^k
    11/36=e^k
    k=ln(11/36)

    plugging k back into eq2
    c=1296/11

    eq 3 becomes :
    T=(1296/11)e^(kt)+4

    finding the initial temperature
    T(0)=121.8°C


    This answer just doesn't seem logical to me, a lot higher than I expected. I can't seem to figure out where I went wrong or if I did? Thank you! :)
     
  2. jcsd
  3. Oct 2, 2013 #2
    I'm not sure on the math, but your answer seems logical... If the water started at 4 degrees C and after an hour got to 40, the starting temperature had to be really high (since the temperature of the solid body is decreasing too).

    But I'm not sure.
     
  4. Oct 2, 2013 #3

    Office_Shredder

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    Science Advisor
    Gold Member

    I double checked for you and the final function that you got definitely matches the two data points they gave you, I think this is all correct.

    If anything that number sounds way lower than I would have guessed actually, so that goes to show you how good intuition is at this kind of thing.
     
  5. Oct 2, 2013 #4

    ehild

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    Homework Helper
    Gold Member

    Your result is correct.

    Newton's Law of Cooling can be written in the form

    T(t)-Tm=(T(0)-Tm)e-kt.

    T(t) is the temperature of the body at time t, T(0) is the initial temperature, Tm is the temperature of the surroundings. You are given T(1) and T(2) and need to find T(0).


    T(1)-Tm=(T(0)-Tm)e-k
    T(2)-Tm=(T(0)-Tm)e-2k

    square the first equation and divide by the second one.

    ehild
     
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