Word problems involving tempature.

Question

A longer problem that I'm having trouble figuring out.

 

Police arrive at a murder scene at 1: 00 AM and immediately record the body's temperature which was 90
degrees F At 3:00 AM.Later, they again took the temperature of the body which had dropped to 84 degrees F. The temperature of the crime scene has remained at a constant 80 degrees F.
Determine when the person was murdered.? (Assume that the temperature of the body at the time of death was 98.6
degrees? F.)

Answer 1

This problem is an application of Newton's law of cooling:

T(t) = A + (T0 - A)*e^(-kt)

T0 = temperature of body at time t = 0

t = time elapsed from moment when T = T0

k = constant with units of time^(-1)

A = constant environment temperature = 80 F

First, we need to find k.

Let 1:00 AM be the t=0 time. Then T0 = 90 F, A = 80 F
Then at 3:00 AM, t = 2 hours, T = 84 F, A = 80 F

Plug this into the T(t) equation and solve for k:

84 = 80 + (90 - 80)*e^[-k(2)]

4 = 10*e^(-2k)

k = -ln(0.4) / 2 = 0.458145 hours^(-1)

So for this problem, the equation for Newton's law of cooling is

T(t) = 80 + (T0 - 80)*e^(-0.458145*t), with t in hours and T in F

Now reset the clock and let the time of death be a new t=0 time. Then T0 = 98.6 F. T at 1:00 AM is 90 F. Plug these numbers into T(t) and solve for t. The value of t will be the number of hours elapsed from time of death until 1:00 AM when the 90 F temperature was read.

90 = 80 + (98.6 - 80)*e^(-0.458145*t)

10 = 18.6*e^(-0.458145*t)

t = ln(10/18.6) / (-0.458145) hours = 1.3545 hours ≅ 1 hour + 21 minutes

 

So time of death was about 1 hour and 21 minutes before 1:00 AM, which puts it at about 11:39 pm.