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Math Help - Law of cooling problem

  1. #1
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    Law of cooling problem

    I am going to use Uo for u nought

    A cup of coffee has a temperature of Uo when freshly poured. Then it has cooled to 60 C after 1 min and to 40 C after 2 min. Also the room temperature is 20 C. The temperature u(t) of the coffee at time t changes at a rate proportional to the difference of its temperature and the room temperature: du/dt = -k (u-20) for a constant k>0

    Treating k and Uo as fixed parameters, calculate the constant k

    I have done similar problems but the initial temperature was given so then I could use seperable equations. Not sure how to go about this I keep running into two variables without a system to solve.

    Any help would be greatly appreciated.
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  2. #2
    Super Member Deadstar's Avatar
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    Quote Originally Posted by billbarber View Post
    I am going to use Uo for u nought

    A cup of coffee has a temperature of Uo when freshly poured. Then it has cooled to 60 C after 1 min and to 40 C after 2 min. Also the room temperature is 20 C. The temperature u(t) of the coffee at time t changes at a rate proportional to the difference of its temperature and the room temperature: du/dt = -k (u-20) for a constant k>0

    Treating k and Uo as fixed parameters, calculate the constant k

    I have done similar problems but the initial temperature was given so then I could use seperable equations. Not sure how to go about this I keep running into two variables without a system to solve.

    Any help would be greatly appreciated.
    You don't need initial conditions. You have the conditions...

    u(1) = 60 and
    u(2) = 40.

    Can you post up what you've done.

    From a quick look I got u(t) = Ce^{-kt} + 20.
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  3. #3
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    Yeah. I got that, but then using initial conditions I would have:

    60 = ce^-k + 20 and 40 = ce^-2k + 20

    I am still left with c and a k in both of them. If I was given initial temp I would lose the e^k and just be left with the c. That's where my confusion is.
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  4. #4
    Super Member Deadstar's Avatar
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    Quote Originally Posted by billbarber View Post
    Yeah. I got that, but then using initial conditions I would have:

    60 = ce^-k + 20 and 40 = ce^-2k + 20

    I am still left with c and a k in both of them. If I was given initial temp I would lose the e^k and just be left with the c. That's where my confusion is.
    It's just two equations two unknowns.

    Ce^{-k} = 40
    Ce^{-2k} = 20

    Ce^{-k} = 40
    2Ce^{-2k} = 40

    e^{-k} = 2e^{-2k}

    => \frac{e^{-2k}}{e^{-k}} = \frac{1}{2}

    => e^{-k} = \frac{1}{2}

    => -k = \ln(2^{-1})

    => -k = -\ln(2)

    => k = \ln(2)
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  5. #5
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    Can't believe I didn't see that one......

    Thanks.
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  6. #6
    Super Member Deadstar's Avatar
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    Quote Originally Posted by billbarber View Post
    Can't believe I didn't see that one......

    Thanks.
    Lol no worries. Earlier I forgot how to differentiate x^n. Too much maths around exam time...
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