College Algebra Application

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Newton's Law of Cooling
T(t)=T0+(T1-T0)e^-kt

The police discover the body of a murder victim. Critical to solving the crime is determining when the murder was committed. The coroner arrives at the murder scene at 12:00pm. She immediately takes the temperature of the body and finds it to be 94.6 degrees F. She then takes the temperature 1 hour later and finds it to be 93.4 degrees F. The temperature of the room is 70 degrees F. When was the murder committed?

  • College Algebra Application -

    Let t=0 at time of death.

    We have two temperatures, 94.6 at t=t1, and 93.4 at t=t1+1.
    Also, normal body temperature is 98.6, so T1=98.6.
    T0=ambient temperature = 70.
    Substitute in given formula:
    T(t)=T0+(T1-T0)e^-kt

    94.6=70+(98.6-70)e^(k*t1)...(1)
    93.4=70+(98.6-70)e^(k*(t1+1))...(2)

    We try to solve for t1 and k.
    Subtract (2) from (1):
    Rearrange (1) and (2)
    e^(k*t1)=0.86014 ...(1a)
    e^(k*(t1+1))=0.81818 ...(2a)

    Divide (2a) by (1a)
    e^(k*(t1+1)-k*t1)=0.81818/0.86014
    e^k=0.95122
    take natural log on both sides,
    k=-0.0500

    Substitute k in (1a):
    e^(-0.05t1)=0.86014
    take natural log on both sides
    -0.05t1 = -0.15066
    t1=-0.15066/(-0.05)
    =3.01

    So the victim was dead 3 hours before the coroner arrived, or at 9 a.m.

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