Question

a coroner arrives at a murder scene at 11 p.m. she finds the temperature of the body to be 80.4°f. she waits one hour, takes the temperature again, and finds it to be 75.3°f. she notes that the room temperature is 77°f. assuming that the temperature of the body was 98.6°f when the murder occurred, when was the murder committed?
the murder was committed at approximately approximately
(do not round until the final answer. then round to the nearest hour)

Explanation:

Step1: Recall Newton's Law of Cooling

The formula for Newton's Law of Cooling is $T(t)=T_s+(T_0 - T_s)e^{-kt}$, where $T(t)$ is the temperature of the object at time $t$, $T_s$ is the surrounding - temperature, $T_0$ is the initial temperature of the object, and $k$ is a constant. Here, $T_s = 77^{\circ}F$, $T_0=98.6^{\circ}F$.

When the coroner arrives at $t = 0$ (in hours), $T(0)=80.4^{\circ}F$. So, $80.4 = 77+(98.6 - 77)e^{-k\times0}$, which simplifies to $80.4=77 + 21.6$. This is wrong. Let's start over. The general formula is $T(t)=T_s+(T_1 - T_s)e^{-kt}$, where $T_1$ is the temperature at time $t = 0$.

When the coroner arrives at $t = 0$, $T(0)=80.4^{\circ}F$, so $T(0)=77+(80.4 - 77)e^{-k\times0}=80.4$. After 1 hour ($t = 1$), $T(1)=85.3^{\circ}F$. Substitute into the formula $T(t)=77+(80.4 - 77)e^{-kt}$, we get $85.3=77+(80.4 - 77)e^{-k\times1}$.

Step2: Solve for $k$

First, simplify the equation $85.3=77 + 3.4e^{-k}$.
Subtract 77 from both sides: $85.3−77=3.4e^{-k}$, so $8.3 = 3.4e^{-k}$.
Then, $e^{-k}=\frac{8.3}{3.4}\approx2.4412$. Taking the natural - logarithm of both sides, $-k=\ln(2.4412)$, so $k=-\ln(2.4412)\approx - 0.893$.

Step3: Find the time when the body was at $98.6^{\circ}F$

We want to find $t$ when $T(t)=98.6^{\circ}F$. Using the formula $T(t)=77+(80.4 - 77)e^{-kt}$, substitute $T(t)=98.6$, $T_1 = 80.4$, $T_s = 77$, and $k\approx0.893$.
$98.6=77+(80.4 - 77)e^{-0.893t}$.
First, simplify: $98.6−77 = 3.4e^{-0.893t}$, so $21.6=3.4e^{-0.893t}$.
Then, $e^{-0.893t}=\frac{21.6}{3.4}\approx6.3529$.
Take the natural - logarithm of both sides: $-0.893t=\ln(6.3529)$.
$t=\frac{\ln(6.3529)}{- 0.893}\approx\frac{1.846}{-0.893}\approx - 2.07$.

Answer:

The murder was committed approximately 2 hours before the coroner arrived, so at 9 p.m.