Question

In a murder investigation, the temperature of a corpse was 32.5 degrees C at 1:30 pm and 30.3 degrees C an hour later. Normal body temperature is 37.0 degrees C and the temperature of the surroundings was 20.0 degrees C. When did the murder take place?

Answer 1

Let $T(t)$ be the temperature of the body $t$ hours after 1:30 PM. Then $T(0) = 32.5$ and $T(1) = 30.3$.

Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$, we have $\dfrac{dT}{dt} = k(T-20)$. Now let $y = T - 20$, so
$y(0) = T(0) - 20 = 32.5 - 20 = 12.5$, so $y$ is a solution to the initial value problem $dy/dt = ky$ with $y(0) = 12.5$.
By separating and integrating, we have $y(t) = y(0)e^{kt} = 12.5e^{kt}$.

$y(1) = 30.3 - 20\ \Rightarrow\ 12.5e^{k(1)} = e^{k} = \frac{10.3}{12.5}\ \Rightarrow \\ k = \ln \frac{10.3}{12.5}$

$y(t) = 37 - 20\ \Rightarrow\ 12.5e^{kt} = 17\ \Rightarrow\ e^{kt} = \frac{10.3}{12.5} \ \Rightarrow \\ \\ kt = \ln \frac{10.3}{12.5}\ \Rightarrow\ t = \left( \ln \frac{17}{12.5} \right)/ \frac{10.3}{12.5} \approx -1.588 \mathrm{\ h}$

≈ 95 minutes. Thus the murder took place about 95 minutes before 1:30 PM, or 11:55 AM.