Question

A body was found at 10 a.m. outdoors on a day when the temperature was 40oF. The medical examiner found the temperature of the body to be 80oF. What was the approximate time of death? Use Newton's law of cooling, with k = 0.1947.

Answer 1

can anyone help me out on this??

Answer 2

We have:
\[
T=S+(T_0-S)e^{-kt}
\]Right?
Let's solve for \(t\):
\[
\frac{T-S}{T_0-S}=e^{-kt}
\]So we have:
\[
\ln\left(\frac{T-S}{T_0-S}\right)=-kt
\]And, finally:
\[
-\frac{1}{k}\ln\left(\frac{T-S}{T_0-S}\right)=t
\]A body's normal temperature is: \(T_0\approx98.6\). Try using that.

Answer 3

right i know the formula

Answer 4

just having trouble with times

Answer 5

I have 5am, 8am, 9:30, and 9:45 am

Answer 6

How many hours and minutes have you calculated have passed?

Answer 7

I'm thinking 8am

Answer 8

What did you get for \(t\)?

Answer 9

around 2

Answer 10

hours

Answer 11

I would agree. So, then 10-2 is 8AM.

Answer 12

you agree with? t=2hours?

Answer 13

Yep.

Answer 14

cool

Answer 15

thanks!!

Answer 16

Sure thing.