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Homework Help: Newton's Law of Cooling (diff eq. - seperation of variables)

  1. Apr 6, 2010 #1
    Newton's Law of Cooling (diff eq. -- seperation of variables)

    1. The problem statement, all variables and given/known data

    Fresh coffee sitting in a room cooling...you know the routine.
    Anyhow T(0) = 90degreesCelcius.
    Room temp=25degrees Celcius

    find k.
    Then he asks us to use Euler's method to estimate coffee temp after five mins. (using step size h=1).

    2. Relevant equations

    T(t) satisfies the equation: dT/dt=k(T-Troom).
    we know that at T=65 degrees (for coffee) dT/dt (or the rate of cooling, as the problem states)= 1 degree per minute

    3. The attempt at a solution

    Well thought hey look here the ole plug and chug MAY work, lets see...:
    1=k(65-25) brings k=.025. but then I thought, hey for the crap to cool k needs to be negative (this is the only way the limit as t tends to infinity for e^(kt) to tend to 0). So this can't be the appropriate way of going about this. So, without thinking (again) I integrated to see if that would bring about new light. I got

    T=65e^(kt)+25.

    No real help even if the integration is right because I'm missing a value for t to help me solve for k. So I'm a little stuck here.

    I'm confused even more in that the blank where the answer to the second part goes, the Euler's method answer, has a blank that reads T(10)=____________. But I know (thought) the question asked for T(5). Am I missing somthing here?? Thank you, Ian.
     
  2. jcsd
  3. Apr 6, 2010 #2

    LCKurtz

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    Re: Newton's Law of Cooling (diff eq. -- seperation of variables)

    Since the temperature is decreasing 1 degree per minute, that means T' = -1, not +1.
     
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