Newton's Law of Cooling

Question

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?

Answer 1

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.

Answer 2

What is the answer if the coroner arrives at the scene at 1 a.m. and immediately takes the temperature of the body and is 94.6F. She then takes it 1 hour later and it is 93.1F. The outside temperature is 60F