Posted by Chris on Thursday, May 5, 2011 at 4:02pm.
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?

College Algebra Application  MathMate, Friday, May 6, 2011 at 12:32pm
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.
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