You are a medical forensic expert and have been called to the scene of a homicia
ID: 3341294 • Letter: Y
Question
You are a medical forensic expert and have been called to the scene of a homiciade. The following information was given to you by police. The murdervictim was found at 4:30am and the tempture of the body was 30 degrees C. The room in which the murder victim lay was a constant tempature of 22 degrees C. An hour later you determine the temperature of the body is 28 degrees C. We know the normal tempature of a human body when alive is 37 degrees C. Use this information to determine the approximate time that the murder occured.
Hint: use newtons law of cooling
dT/dt= -k(T-Ta)
T(t)= Ta +(To-Ta)e^-kt
T(t)= temp of object at time t
Ta=ambient temp
To = initial temp
Explanation / Answer
We know that T(t) = Ta + (To-Ta)e^-kt. Additionally, we know that the ambient temperature is the room temperature, or 22 degrees Celsius, and that the initial temperature of the body (right after it was killed) is 37 degrees Celsius. Therefore, we have the formula T(t) = 22 + 15e^-kt. Then if t corresponds to 4:30 AM, we know that T(t) = 30 = 22 + 15e^-kt, so e^-kt = 8/15. Moreover, an hour later, the temperature is 28 degrees, so we know that T(t+1) = 28 = 22 + 15e^-k(t+1) so e^-k(t+1) = 6/15. Then by taking the ratio, we get that e^-k(t+1)/e^-kt = e^-k = (6/15)/(8/15) = 3/4. Then since e^-kt = 8/15, this is the same as (3/4)^t = 8/15, so t = log(8/15)/log(3/4) = 2.185 hours. Therefore, the murder occurred around 2.185 hours before 4:30 AM, which corresponds to around 2 hours and 11 minutes before, or 2:19 AM.
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