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Math Help - newtons law of cooling

  1. #1
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    newtons law of cooling

    Question given: newtons law of cooling is given by
    (Θ=theta)

    Θ=Θoe^-kt

    where the excess of tempreature at 0 deg = Θo deg and the time T is Θ deg.
    given that Θo=15 deg and K=-0.02 find the time when the rate of change of tempture is 1 deg


    my work

    1=15e^-0.02XT
    i assume u must use log to rearence the forumla in oder to get a value then differentiate it to find the rate of change of T.
    is what iv done so far correct?
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  2. #2
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    Quote Originally Posted by batman121 View Post
    Question given: newtons law of cooling is given by
    (Θ=theta)

    Θ=Θoe^-kt

    where the excess of tempreature at 0 deg = Θo deg and the time T is Θ deg.
    given that Θo=15 deg and K=-0.02 find the time when the rate of change of tempture is 1 deg


    my work

    1=15e^-0.02XT
    i assume u must use log to rearence the forumla in oder to get a value then differentiate it to find the rate of change of T.
    is what iv done so far correct?
    What's that X in there? (And please start using parenthesis.)

    1 = 15e^{-0.02t}

    Yes, take the ln of both sides:
    0 = ln(15) - 0.02t
    and solve for t.

    -Dan
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    -0.02t= ln(15)
    t= ln(15)/-0.02

    t=58.804sec

    I dont think thats right, im not sure hoe to rearange when ln is in the equation.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by batman121 View Post
    -0.02t= ln(15)
    t= ln(15)/-0.02

    t=58.804sec

    I dont think thats right, im not sure hoe to rearange when ln is in the equation.
    Quote Originally Posted by topsquark View Post
    What's that X in there? (And please start using parenthesis.)

    1 = 15e^{-0.02t}

    Yes, take the ln of both sides:
    0 = ln(15) - 0.02t
    and solve for t.

    -Dan
    A minor sign mistake here
    0.02t = ln(15)

    t = \frac{ln(15)}{0.02} \approx 135.403

    I'm not sure why you made the numerical mistake.

    ln( ) is like any other function: treat ln(15) as a constant and you'll do fine with it.

    -Dan
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    Thanks for the help greatly appreciated
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