Question

a coroner arrives at a murder scene at 9 p.m. she finds the temperature of the body to be 88.4°f. she waits one hour, takes the temperature again, and finds it to be 83.7°f. she notes that the room temperature is 69°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.

We know that $T_s = 69^{\circ}F$, $T_0=98.6^{\circ}F$. At $t = 0$ (when the coroner arrives at 9 p.m.), $T(0)=88.4^{\circ}F$, so $88.4 = 69+(98.6 - 69)e^{-k\times0}$, which is just a starting - point check. After $t = 1$ hour (10 p.m.), $T(1)=83.7^{\circ}F$.

Substitute into the formula: $T(1)=69+(98.6 - 69)e^{-k\times1}$.
So, $83.7=69+(98.6 - 69)e^{-k}$.

Step2: Solve for $k$

First, simplify the equation:
$83.7-69=(98.6 - 69)e^{-k}$
$14.7 = 29.6e^{-k}$
$e^{-k}=\frac{14.7}{29.6}\approx0.4966$
Take the natural logarithm of both sides: $-k=\ln(0.4966)$
$k=-\ln(0.4966)\approx0.70$.

Step3: Find the time $t$ when the body started cooling

We want to find $t$ when $T(t)=88.4^{\circ}F$. Using the formula $T(t)=69+(98.6 - 69)e^{-kt}$ and substituting $T(t)=88.4$, $T_s = 69$, $T_0 = 98.6$, and $k\approx0.70$:
$88.4=69+(98.6 - 69)e^{-0.7t}$
$88.4 - 69=29.6e^{-0.7t}$
$19.4 = 29.6e^{-0.7t}$
$e^{-0.7t}=\frac{19.4}{29.6}\approx0.6554$
Take the natural logarithm of both sides: $-0.7t=\ln(0.6554)$
$t=-\frac{\ln(0.6554)}{0.7}\approx0.6$ hours.

Since 0.6 hours is approximately $0.6\times60 = 36$ minutes, the murder was committed about 36 minutes before 9 p.m., which is approximately 8:24 p.m. Rounding to the nearest hour, the murder was committed at approximately 8 p.m.

Answer:

8 p.m.