Question

The police arrive at the crime scene at 9:00 AM. They immediately measure that the body temperature of the deceased is 83 F. After they finish gathering evidence, a process which took exactly one hour, they measure the body again and the temperature is 81,F. The room temperature is 68 F. When was the murder committed? (Assume his body temperature normally was 98, as opposed to 98.6.)

Answer 1

Let T(t) = temperature of the body t hours after the temperature was first recorded

By Newton's Law of cooling,

T(t) = Ts + (T0 - Ts)e-kt

where Ts is the temperature of the surroundings, T0 is the temperature the first time that it is recorded, and k is a constant to be determined.

So, we have T(t) = 68 + (83 - 68)e-kt = 68 + 15e-kt

Next, we need to find k:

We are given that the temperature after 1 hour from 9 AM was 81 F. So, we get:

T(1) = 68 + 15e-1k = 81

=> 15e-1k = 13

=> e-1k = 0.8666

=> -1k = ln(0.8666)

=> k = 0.143

So, T(t) = 68 + 15e-0.143t

We are trying to find the time of death. In other words, since it is assumed that the body temperature was normal at the time of death, we need to find t so that T(t) = 98 F.

68 + 15e-0.143t = 98

=> 15e-0.143t = 30 =>  e-0.143t = 2 => -0.143t = ln(2) => -0.143t = 0.693 => t = -4.8461 hours

t = -4.8461 hours , Here the negative sign indicates that the time of death was 4.8461 hours prior to the time that the temperature of the body was first taken.

Thus, the time of death was 4.8461 hours before 9 AM.

4.8461 hours = 4.8461(60) minutes = 291 minutes (approx.)

Time of death = 9 AM - 291 minutes = 4:09 AM