Algebra 2 problem

Question

The question:

 

At 1:00 am, a homicide detective found the reading of a corpse's body temperature to be 88 degrees F, One hour later the body temperature is 83 degrees F. If the body has been in a 74 F room since its death, what is the time of death assuming the body was 98.6 F at the time of death?

 

I am having trouble figuring out the steps to solve this and equations like it.

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 = 74 F

First, we need to find k.

Let 1:00 AM be the t=0 time.  Then T0 = 88 F, A = 74 F

Then at 2:00 AM, t = 1 hour, T = 83 F, A = 74 F

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

 

83 = 74 + (88 - 74)*e^[-k(1)]

 

9 = 14*e^(-k)

 

k = -ln(9/14) = 0.44183 hours^(-1)

 

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

 

T(t) = 74 + (T0 - 74)*e^(-0.44183*t), with t in hours and T in F

 

Now let the time of death be the t=0 time.  Then T0 = 98.6 F.  T at 1:00 AM is 88 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 when the 88 F temperature was read.

 

88 = 74 + (98.6 - 74)*e^(-0.44183*t)

 

14 = 24.6*e^(-0.44183*t)

 

t = ln(14/24.6) / (-0.44183) hours = 1.27581 hours = 1 hour + 16.5486 minutes

 

So time of death was about 1 hour and 17 minutes before 1:00 AM.

 

We could have also used the data from 2:00 AM:

 

The time of death is still the t=0 time. Then T0 = 98.6 F. T at 2:00 AM is 83 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 2:00 when the 83 F temperature was read.

83 = 74 + (98.6 - 74)*e^(-0.44183*t)

9 = 24.6*e^(-0.44183*t)

t = ln(9/24.6) / (-0.44183) hours = 2.27581 hours = 2 hours + 16.5486 minutes

 

So time of death was about 2 hours and 17 minutes before 2:00 AM, which is the same as 1 hour and 17 minutes before 1:00 AM.