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Thread: Application of first order differential equation-tomorrow test

  1. #1
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    Application of first order differential equation-tomorrow test

    I have a problem of this type of question.
    An object thrown into a large body of water cools at a rate proportional to the difference between its temperature and the water temperature. Suppose know that the the water is at a temperature of 27 degrees Celsius. After 4 minutes the object's temperature is 67 degrees, and after 9 minutes the object's temperature is 47 degrees Celsius. What was the temperature of the object when it was thrown into the water?


    Differential equation is dQ/dt = -k(Q-Qs)


    Q is the temperature of an object
    Qs is the water temperature.

    Not like other question, i can find K (constant) but in this question K ( constant ) i even cant find it out.
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  2. #2
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    Quote Originally Posted by sanikui View Post
    I have a problem of this type of question.
    An object thrown into a large body of water cools at a rate proportional to the difference between its temperature and the water temperature. Suppose know that the the water is at a temperature of 27 degrees Celsius. After 4 minutes the object's temperature is 67 degrees, and after 9 minutes the object's temperature is 47 degrees Celsius. What was the temperature of the object when it was thrown into the water?


    Differential equation is dQ/dt = -k(Q-Qs)


    Q is the temperature of an object
    Qs is the water temperature.

    Not like other question, i can find K (constant) but in this question K ( constant ) i even cant find it out.
    $\displaystyle \frac{dQ}{dt} = k(Q - 27)$

    $\displaystyle \frac{dQ}{Q-27} = k \, dt$

    $\displaystyle \ln(Q-27) = kt + C$

    $\displaystyle
    Q = 27 + Ae^{kt}
    $

    at $\displaystyle t = 4$, $\displaystyle Q = 67$

    $\displaystyle 67 = 27 + Ae^{4k}$

    $\displaystyle
    40 = Ae^{4k}
    $

    at $\displaystyle t = 9$, $\displaystyle Q = 47$

    $\displaystyle
    47 = 27 + Ae^{9k}
    $

    $\displaystyle
    20 = Ae^{9k}
    $

    $\displaystyle \frac{40}{20} = \frac{Ae^{4k}}{Ae^{9k}}$

    $\displaystyle
    2 = e^{-5k}
    $

    solve for k , then solve for A and finish.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    haha, too fast for me. I just finished the problem. The work above is correct...or at least the same as what I got.
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  4. #4
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    Quote Originally Posted by skeeter View Post
    $\displaystyle \frac{dQ}{dt} = k(Q - 27)$

    $\displaystyle \frac{dQ}{Q-27} = k \, dt$

    $\displaystyle \ln(Q-27) = kt + C$

    $\displaystyle
    Q = 27 + Ae^{kt}
    $

    at $\displaystyle t = 4$, $\displaystyle Q = 67$

    $\displaystyle 67 = 27 + Ae^{4k}$

    $\displaystyle
    40 = Ae^{4k}
    $

    at $\displaystyle t = 9$, $\displaystyle Q = 47$

    $\displaystyle
    47 = 27 + Ae^{9k}
    $

    $\displaystyle
    20 = Ae^{9k}
    $

    $\displaystyle \frac{40}{20} = \frac{Ae^{4k}}{Ae^{9k}}$

    $\displaystyle
    2 = e^{-5k}
    $

    solve for k , then solve for A and finish.


    Okay, i get it...hope tomorrow i can do this type of question.
    Thanks everyone. V^^
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