1. CSI: Time of Death (Due Friday)

2. Example Problem: A coroner was called to the home of a person who had died during the night. In order to estimate the time of death the coroner took the person’s body temperature twice. At 9:00a.m. the temperature was 85.7°F and at 9:30a.m. the temperature was 82.8°F. The room temperature stayed constant at 70°F. Find the approximate time of death assuming the body temperature was normal at 98.6°F at the time of death.

3. Solution: Part 1: To estimate the time of death, the coroner used Newton’s law of cooling, where k is a constant, S is the temperature of the surrounding air, and t is the time it takes for the body temperature to cool from T 0 = 98.6 to T. From the first temperature reading, the coroner could obtain kt:

4. Solution: Part 2: From the second temperature reading, the coroner could obtain k(t + time since 1 st temperature reading):

5. Solution: Part 3: Using the two systems of equations solve for k and t: Distribute the k Substitute kt and solve for k Now substitute k into the first equation

6. Finding the time of death: Since t represents time in hours, use the whole number for hours and find the number of minutes that passed by multiply the decimal amount by 60., which is one hour and some minutes. Convert the decimal amount to find the number of minutes. 0.4677(60) = 28.062 This gives us one hour and 28 minutes that had passed since the death occurred. The first temperature was taken at 9:00am, so counting backwards one hour and 28 minutes the time of death was approximately 7:32am.