# Need some help with a hard question

• Jan 22nd 2007, 05:18 PM
Skor
Need some help with a hard question
Ok the question is

Coroners estimate the time of death from body temperature using the simple rule that a body cools about 1C in the first hour after death and about 1/2C for each addictional hour.

Assuming an air temperature of 20C and a living body temperature of 37C, the temperature, T(t), in degrees Celcius is given by T(t) = 20 + 17e^-kt, where t=0 is the instant that death occurred..

a) For what value of k will the body cool by 1C in the first hour?
b) Using the value of k found in part a) after how many hours will the temperature of the body be decreaing at a rate of 1/2 C per hour?
c) Using the value of k found in part a) Show that 24h after death, the coroner's rule gives approx the same temp as the forumula.

I know I need an equation to do this but Im not sure what it is, I understood my other log study questions but this one stumps me. Any help would be appreciated greatly.
• Jan 23rd 2007, 06:05 AM
CaptainBlack
Quote:

Originally Posted by Skor
Ok the question is

Coroners estimate the time of death from body temperature using the simple rule that a body cools about 1C in the first hour after death and about 1/2C for each addictional hour.

Assuming an air temperature of 20C and a living body temperature of 37C, the temperature, T(t), in degrees Celcius is given by T(t) = 20 + 17e^-kt, where t=0 is the instant that death occurred..

a) For what value of k will the body cool by 1C in the first hour?
b) Using the value of k found in part a) after how many hours will the temperature of the body be decreaing at a rate of 1/2 C per hour?
c) Using the value of k found in part a) Show that 24h after death, the coroner's rule gives approx the same temp as the forumula.

I know I need an equation to do this but Im not sure what it is, I understood my other log study questions but this one stumps me. Any help would be appreciated greatly.

a) You have:

$T(t) = 20 + 17e^{-kt}$

Now if we measure time in hours we have if $1$ is the time to cool by 1 degree C

$
36=20+17 e^{-k}
$

or:

$
e^{-k}=16/17
$

or:

$
k=-\ln(16/17)\approx 0.060625
$

b) The rate of fall of temprature is:

$\frac{dT}{dt}=-k \times 17 e^{-kt}$

If the temprature is falling at a rate of 1/2 a degree per hour we have:

$-k \times 17 e^{-kt}=-1/2$

so:

$e^{-kt}=0.5/(k \times 17)$

or:

$-kt=\ln(0.5/(k \times 17))$

so:

$t=-(1/k)\ln(0.5/(k \times 17))\approx 11.28 \mbox{ hours}$

RonL