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Math Help - [Calculus] Newton's Law of Cooling

  1. #1
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    [Calculus] Newton's Law of Cooling

    In my assignment, my Calculus teacher was murdered.

    The three suspects are:
    An English Teacher (no alibi from 9:15-10:00AM)
    The Principal (no alibi from 10:05-10:50AM)
    A History Teacher (no alibi from 10:55-11:40AM)

    The temperature of the body was 87 degrees at 11:55AM, and 80 degrees at 12:45PM. The room was 75 degrees.

    My job is to identify the murderer and support my answer with a written paragraph and some math.

    I know I need to use (dy/dk)=K(y-Ts), e^(kt+c)=y-Ts, and y=Ce^(kt)+Ts
    Assuming that Y = the temperature of an object, t = time, Ts = Room Temperature, and C = y - Ts. I have an example in my notes, but I am having trouble following the different equations and still get lost at the various transitions between e and ln.

    Thank you in advance.
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  2. #2
    Member arpitagarwal82's Avatar
    Joined
    Feb 2009
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    Just at death the body temperature was 98 degrees Fahrenheit (approx Normal body temperature)

    It was 87 degree at 11:55
    and 75 degree at 12:45
    Consider y=Ce^(kt)+Ts


    Ts = 75 degree

    let t_1 be time of death. temp 98
    t_2 be 11:55 temp 87
    t_3 be 12:45 temp 80

    So 98 = Ce^(kt_1) + 75
    87 = Ce^(kt_2) +75
    80 = Ce^(kt_3) + 75

    i.e.

    23 = Ce^(kt_1 ) ----------eq1
    12 = Ce^(kt_2)  ----------eq2
    5 = Ce^(kt_3) ----------eq3

    Now do eq3 -eq2 and eq2 -eq1 and use t_3 -t_2 = 50 minutes

    So a bit calculation will give you
    t_1 = t_2 - 50 (\ln 23/11)(\ln 12/5) ( t_2 is in minutes here)

    Find t_1 and you can take the guess of murderer on basis of t_1
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