College Algebra Application
posted by Chris on .
Newton's Law of Cooling
T(t)=T0+(T1T0)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?

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+(T1T0)e^kt
94.6=70+(98.670)e^(k*t1)...(1)
93.4=70+(98.670)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.