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Math Help - Newton's Cooling

  1. #1
    Member jacs's Avatar
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    Newton's Cooling

    Hi, I have tried this many times and just can't seem to figure it out.

    A body, initially at room temperature 20C, is heated so that its temperature would rise by 5C/min if no cooling took place. Cooling does occur in accordance with Newton's Law of Cooling and the maximum temperature the body could attain is 120C. How long would it take to reach a temperature of 100C?

    using dt/dt = -k(T - P) where T is the current temp and P is the temp of the surrounding medium.

    any help wouuld be appreciated, thanks

    the answer provided is 32.2 mins
    Last edited by jacs; January 15th 2006 at 07:36 AM.
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  2. #2
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    The general form for Newton's Law of Cooling is...

    T=(T_{0}-T_{s})e^{-kt}+T_{s}

    if my memory serves my correctly.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by jacs
    Hi, I have tried this many times and just can't seem to figure it out.

    A body, initially at room temperature 20C, is heated so that its temperature would rise by 5C/min if no cooling took place. Cooling does occur in accordance with Newton's Law of Cooling and the maximum temperature the body could attain is 120C. How long would it take to reach a temperature of 100C?

    using dt/dt = -k(T - P) where T is the current temp and P is the temp of the surrounding medium.

    any help wouuld be appreciated, thanks

    the answer provided is 32.2 mins
    The rate of change of temperature is the sum of the heating component and
    the cooling component.

    So in this case:

    \frac{dT}{dt}=5-k(T-20)

    The 5 on the RHS is the contribution to the rate of change of
    temp from the heating, the rest is the contribution to the rate of change of
    temp due to the cooling (the 20 is the temp of the surrounding
    medium).

    At a temp of 120 the rate of change of temp is zero, so:

    0=5-k(120-20)\Rightarrow k=0.05.

    Hence:

    \frac{dT}{dt}=6-0.05T,

    which is a DE of variables separable type, so:

    \int_{20}^{T(t)} \frac{1}{6-0.05T}dT=\int_0^t dt.

    Hence:

    \left[ \frac{-\ln (6-0.05T)}{0.05} \right]_{20}^{T(t)}=t ,

    so:

    \ln (6-0.05T)-\ln (5)=-0.05t

    again:

    \ln \left(\frac{6-0.05T}{5} \right) = -0.05t

    Putting T=100 into this will allow you to find the corresponding
    t.

    RonL
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  4. #4
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    Quote Originally Posted by Jameson
    The general form for Newton's Law of Cooling is...

    T=(T_{0}-T_{s})e^{-kt}+T_{s}

    if my memory serves my correctly.
    Yes it is correct. Let me come to that conclusion, this is similar to what CaptainBlack did.

    Law of cooling:
    \frac{dT}{dt}=k(T-H) H is air-temperature.
    Diffrencial equation is seperable,
    \frac{dT}{T-H}=kdt
    Integrating,
    \ln(T-H)=kt+C_1
    Under base e,
    T-H=e^{kt+C_1}=e^{kt}e^{C_1}=Ce^{kt}
    Thus,
    T=H+Ce^{kt}
    when t=0 then T=T_o
    thus,
    T=H+(T_0-H)e^{kt}
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  5. #5
    Member jacs's Avatar
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    Thanks!!!!

    Captain Black, you are a legend, it was the 5 that was really throwing me and just didn't know where to put it in, and that works out just nicely.

    my gratitude


    jacs
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