Question

a body was found at 10 a.m. outdoors on a day when the temperature was 40°f. the medical examiner found the temperature of the body to be 80°f.
what was the approximate time of death? use newtons law of cooling, with ( k = 0.1947 ).
( t(t)=t_{a}+(t_{0}-t_{a})e^{-kt} )
9:45 a.m.
5 a.m.
9:30 a.m.
8 a.m.

Explanation:

Step1: Substitute the known values into the formula

We know that \(T(t) = 80^{\circ}F\), \(T_{A}=40^{\circ}F\), \(T_{0} = 98.6^{\circ}F\) (normal body - temperature) and \(k = 0.1947\).
The formula \(T(t)=T_{A}+(T_{0}-T_{A})e^{-kt}\) becomes \(80 = 40+(98.6 - 40)e^{-0.1947t}\).
First, simplify the equation:
\(80-40=(98.6 - 40)e^{-0.1947t}\)
\(40 = 58.6e^{-0.1947t}\)
Then, \(\frac{40}{58.6}=e^{-0.1947t}\)
\(0.6826 = e^{-0.1947t}\)

Step2: Take the natural - logarithm of both sides

Since \(y = e^{x}\) implies \(x=\ln(y)\), for \(0.6826 = e^{-0.1947t}\), we have \(\ln(0.6826)=- 0.1947t\)
We know that \(\ln(0.6826)\approx - 0.382\)
So, \(-0.382=-0.1947t\)

Step3: Solve for \(t\)

\(t=\frac{-0.382}{-0.1947}\approx2\) hours

Answer:

The body was found at 10 a.m. and \(t\approx2\) hours. So the time of death was approximately 8 a.m.