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Math Help - newton's law of cooling

  1. #1
    Newbie
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    dubai
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    Post newton's law of cooling

    hey everyone.
    i hve a doubt in dis question: A body at temperature T higher than its surroundings cools down according to the formula T=(T-T(with base 0))e(raise to the power of -kt), (newton's law of cooling) where T(with base 0) is the temperature of the surroundings, t is the time in hours, and k is a constant. It is found that the temperature of the body falls from 80degrees celsius to 60 degrees celsius in 2 hours. If T(with base 0)=20 degress celsius, how long does it take for the temperature of the body to fall to 30 degrees celsius?....well is the answer coming out to be 8 hrs 50 mins...i just wana confirm...
    thankx for seeing dis qs....hope 2 c a reply soon...
    tc pple
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  2. #2
    MHF Contributor
    Joined
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    France
    Posts
    1,458
    Hi

    Newton's law is
    T = T_0 + (T_i - T_0) e^{-kt}
    where T_0 is the temperature of the surroundings and T_i the initial temperature of the body

    Let T_1 be the temperature at t_1
    and T_2 the temperature at t_2

    Then
    T_1 - T_0 = (T_i - T_0) e^{-kt_1}
    T_2 - T_0 = (T_i - T_0) e^{-kt_2}

    Dividing the 2 expressions

    k = \frac{1}{t_2-t_1}\,ln \frac{T_1 - T_0}{T_2 - T_0}
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