CSI: Time of Death

(Due Friday)

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.

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 T0 = 98.6 to T. From the first temperature reading, the coroner could obtain kt:

0

lnT S

ktT S

85.7 70ln

98.6 70kt

ln 0.44755kt

0.5997kt

Solution:

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

1 82.8 70ln

2 98.6 70k t

1ln 0.44755

2k t

0.5 0.8040k t

Solution:Part 3: Using the two systems of equations solve for k and t:

0.5997

0.5 0.8040

kt

k t

0.5 0.8040kt k

0.5 0.2043k 0.4086k

Distribute the k

Substitute kt and solve for k

Now substitute k into the first equation

0.4086 0.5997t

1.4677t

0.5997 0.5997 0.5997 0.5 0.8040k

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.

1.4677t , 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.