College Algebra, Chapter 5, 5.5, Section 5.5, Problem 24

To determine the time of death in homicide investigations, Newton's Law of Cooling is used. The normal body temperature is $98.6^{\circ} F$.
Immediately following death, the body begins to cool. It has been determined experimentally that the constant in Newton's Law of
Cooling is approximately $k = 0.1497$, assuming that the time is measured in hours. Suppose that the temperature of the surroundings is $60^{\circ} F$.

(a) Determine a function $T(t)$ that models the temperature $t$ hours after death.

(b) How long ago was the time of death, if the temperature of the body is now $72^{\circ} F$?



Recall the formula for Newton's Law of Cooling is

$T(t) = Ts + D_0 e^{-kt}$

where

$T(t)$ = temperature of the object at time $t$

$T(s)$ = surrounding temperature

$D_0$ = initial temperature difference between the object and its surroundings

$k$ = positive constant, depending on the object

$t$ = time

a.) If the temperature of the object and the surrounding is $98.6^{\circ} F$ and $60^{\circ} F$ respectively, then

$D_0 = 98.6 - 60 = 38.6^{\circ} F$

Thus, the model is

$T(t) = 60 + 38.6 e^{-0.1947 t}$

b.) If $T(t) = 72$, then


$
\begin{equation}
\begin{aligned}

72 =& 60 + 38.6 e^{-0.1947 t}
&& \text{Subtract each side by } 60
\\
\\
12 =& 38.6 e^{-0.1947 t}
&& \text{Divide each side by } 38.6
\\
\\
\frac{12}{38.6} =& e^{-0.1947 t}
&& \text{Take $\ln$ of each side}
\\
\\
\ln \left( \frac{12}{38.6} \right) =& -0.1947 t
&& \text{Divide each side by } -0.1947
\\
\\
t =& - \frac{\displaystyle \ln \left( \frac{12}{38.6} \right)}{0.1947}
&& \text{Solve for } t
\\
\\
t =& 6 \text{ hours}

\end{aligned}
\end{equation}
$



It shows that the body has been dead after 6 hours.