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Math Help - newtons cooling law. I need help with this problem. Thank you.

  1. #1
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    newtons cooling law. I need help with this problem. Thank you.

    A thermometer reading 70F is placed in an oven preheated to
    a constant temperature. Through a glass window in the oven
    door, an observer records that the thermometer read 110F after
    .5 minute and 145F after 1 minute. How hot is the oven?

    using the following:
    dT/dt = k(T- Tm)
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    Re: newtons cooling law. I need help with this problem. Thank you.

    Hey ALAIN971.

    This is a separable DE which means you move the T's to one side and the t's (and the differentials) to the other. Can you separate these variables and their differentials and integrate both sides?
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    Re: newtons cooling law. I need help with this problem. Thank you.

    I have done this up to this step.

    T(t)=Tm + Ce^(kt)

    i get 2 equations with 2 variables but i am stuck.

    145= Tm + Ce^(kt)

    110= Tm + Ce^(0.5kt)

    C and k are constants.
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    Re: newtons cooling law. I need help with this problem. Thank you.

    sorry i meant :

    I have done this up to this step.

    T(t)=Tm + Ce^(kt)

    i get 2 equations with 2 variables but i am stuck.

    145= Tm + Ce^(k)

    110= Tm + Ce^(0.5k)

    C and k are constants.
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  5. #5
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    Re: newtons cooling law. I need help with this problem. Thank you.

    What about the information at t = 0? Can you use this information to get another piece of information?
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    Re: newtons cooling law. I need help with this problem. Thank you.

    at t=o temperature is 70
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    Re: newtons cooling law. I need help with this problem. Thank you.

    So now you have three equations for Tm, C and k.

    70 = Tm + C.
    145= Tm + Ce^(k)
    110= Tm + Ce^(0.5k)

    C = 70 - Tm
    k = ln(145 - Tm) - ln(C)
    k = 2*ln(110 - Tm) - 2*ln(C)

    ln(145 - Tm) = 2*ln(110 - Tm) - ln(C)

    145 - Tm = 70 + C
    110 - Tm = 40 + C which means

    ln(70 + C) = 2*ln(40 + C) - ln(C) which means

    ln([70 + C]C) = ln([40 + C]^2) taking exponentials we get:

    C[70+C] = [40+C]^2. and how you have a quadratic equation for C.
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    Re: newtons cooling law. I need help with this problem. Thank you.

    ok how do i proceed then.
    i found C=-160
    pit it back in 145-Tm=70+C
    and i found Tm=235 Farenheights

    the answer should be 390
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  9. #9
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    Re: newtons cooling law. I need help with this problem. Thank you.

    I made a mistake, 145 - Tm = 75 + C.

    You should be checking what I do not just following blindly. I am only trying to supplement the thinking process: not do the whole thing.
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  10. #10
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    Re: newtons cooling law. I need help with this problem. Thank you.

    thank you for your help. i returned back to school this fall and calculus is far away for me i have not done any for years.
    i dont know if i did it correctly but i get 230 degrees now.
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