# Math Help - Newton's law of cooling

1. ## Newton's law of cooling

I was looking over a past exam paper today and came across this longer question. I'm unsure of which values to use, and where to go once I find these values.

A murder victim has been shot through the head with a revolver. A suspect is found standing over the victim at 10:15PM holding the revolver. At 10:30PM, the coroner records a body temperature of 26°C and it was also recorded that the room temperature was 21°C. An hour later a second reading of the body temperature yielded 24°C. The suspect was seen in public from 6PM to 9PM. Given that the normal body temperature of a living person is 37°C, use Newton’s law of cooling to determine whether or not the suspect has an alibi.

Thanks

2. With Newton's Law of Cooling, you can either remember the formula or remember the idea behind it and derive it. It is kind of a pain to derive under pressure if you haven't done it a few times, so maybe you should just memorize it when you have to use it.

The end result is: $T(t) = T_a+(T_0-T_a)e^{-kt}$

You are given T_0 and T_a, the starting temp for the body and the ambient, or outside temperature. The only thing left to find is k, the decay rate. Since you are given information about the temperature after a certain time, t, you can plug everything in except for the missing k. Solve for k.

3. Would t be in minutes or hours? It doesn't say in the question which one to use.

4. Originally Posted by Danog
Would t be in minutes or hours? It doesn't say in the question which one to use.
Either one will work as long as you are consistent. I would use hours.

5. Wait... Would you start taking t values from 10:15? That wouldn't give the suspect an alibi. I'm confused.

6. Originally Posted by Danog
Wait... Would you start taking t values from 10:15? That wouldn't give the suspect an alibi. I'm confused.
No. The first temperature reading for the body was at 10:30. Then the next one was an hour later. What is not making sense? I want to help but need you to be more explicit.