Newton's Law

Printable View

June 19th 2006, 07:17 PM
distance

Newton's Law

i have a problem, which confused me a lot.

Suppose a murder has been committed and the corpse cools according to the Newton's law of heating(cooling). At the time of the muder, let's assume that the core temperture of the victim (who had a touch of the flu) was 99F. We will aso asume that the ambient temperature has been constant 60F since the henious act. ReSEARCH has produced table of core temperatures. and tables availble to forenic investigations show that a body of this victim's weight would cool to 90F after three hours. If the body is discovered at 10:00PM and its core temperature is 78F, at what time, did the murder occur ?

need help...as soon as possible. Final coming up...hic....! thanks a lot
June 20th 2006, 05:11 AM
CaptainBlack

Quote:

Originally Posted by distance

i have a problem, which confused me a lot.

Suppose a murder has been committed and the corpse cools according to the Newton's law of heating(cooling). At the time of the muder, let's assume that the core temperture of the victim (who had a touch of the flu) was 99F. We will aso asume that the ambient temperature has been constant 60F since the henious act. ReSEARCH has produced table of core temperatures. and tables availble to forenic investigations show that a body of this victim's weight would cool to 90F after three hours. If the body is discovered at 10:00PM and its core temperature is 78F, at what time, did the murder occur ?

need help...as soon as possible. Final coming up...hic....! thanks a lot

Newton's law of cooling is:

$ \frac{dT}{dt}=-K(T-T_s) $ ,

where $T(t)$ is the temprature at time $t$ , $T_s$ is the tempreture of the
surroundings, and $K$ is a constant.

Now this has solution (see here , for derivation):

$ T(t)=T_s+(T_0-T_s)e^{-Kt} $

where $T_0$ is the temprature at $t=0$ .

Now we are told that after 3 hours the temprature should have dropped from 99F to 90F when $T_s$ is 60F, so:

$ T(3)=90=60+(99-60)e^{-3K} $

so:

$ e^{-3K}=30/39=10/13 $

hence:

$ -3K=\ln (10/13) $

$ K \approx 0.08746 $

Now the body was found at 10:00 pm at a temprature of 78F, so
if it had been cooling for $t$ hrs then:

$ 78=60+(99-60)e^{-0.08746 t} $ ,

solving for $t$ gives:

$t \approx 8.84$ hrs.

which is 8hrs 50m and 24s, so the murder occured at 01:09:36pm
(check the sums there could always be an error).

RonL
June 20th 2006, 09:06 AM
distance

thank you

Last night, i found a problem similar to that, but different methods to solve. However, i check my answer is same with your answer, i got 8.84 hours, then i don't why falling asleep made me did not read the question carefully, this is my fault anyway. After i found 8.84 hours then i convert to minutes. The question, when did the murder occur ? and my answer is for how long does the murder occur. That did not make any sense, why i didn't i think about that. Really stupid me...! anyway, thanks for answering me. If anyone need something, just post on the web. If i think i can solve it, i will do it.