# how many hours elapsed when the body was found...

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• February 21st 2011, 10:13 AM
slapmaxwell1
how many hours elapsed when the body was found...
a dead body was found within a closed room of a house where the temperature was a constant 70 F. at the time of discovery the core temperature of the body was determined to be 85 F. one hour later a second measurement showed that the core temperature of the body was 80 F. assume that the time of death corresponds to t=0 and that the core temp at that time was 98.6 F, how many hours elapsed before the body was found?

ok so my equation that i want to use will be T = Tm + Ce^(kt)
so can i say that t(0)= 98.6 F. if that is true then i can find my C value and have the actual problem set up. but im not sure where to go after this. I know i need to find k and then eventually t...which my book says is 1.6 hours

well any help would be appreciated..thanks in advance
• February 21st 2011, 10:31 AM
running-gag
Hi
The temperature of the body is $T(t) = T_m + Ce^{kt}$, k being negative
We know that :
- at infinite time T(t) is the temperature of the room, therefore Tm=70
- T(0)=98.6=Tm+C therefore we know the value of C

The temperature of the body at the time it was found is $T_1 = T_m + Ce^{kt} = 85$
One hour later $T_2 = T_m + Ce^{k(t+1)} = 80$

Now you just have to solve
• February 21st 2011, 10:32 AM
Ackbeet
Don't confuse capital T with lowercase t in Newton's Law of Cooling problems! Highly dangerous. Might get a negative citation from the Citation Writing Subgroup of the Committee for the Prevention of Notational Abuse.

I would agree that $T(0) = 98.6$.

The target variable is the time at which the temperature is 85 degrees, call it $t_{d},$ for the time of discovery. You know that

$T(t_{d})=85,$ and you also know that

$T(t_{d}+1)=80.$

$T=70+Ce^{kt},$

you should be able to find what you're looking for. Make sense?
• February 21st 2011, 01:47 PM
slapmaxwell1
yeah that does make sense, that was the trick that i missed! that was the missing link the Td and then Td+1....

thanks!
• February 21st 2011, 05:39 PM
Ackbeet
You're welcome!
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