a) Show that by Newton’s Law of Cooling, the time required to cool an object from temperature A to temperature B is
$$t = \frac{1}{k} \ln\frac{B-T}{A-T}$$
where $T$s is the temperature of the surroundings and $k$ is the cooling constant ($k < 0$).
(b) During the early morning following Halloween you just happened to come across a fully functioning locked freezer in an alley. Curious about its contents you broke open the door and found a dead body inside. When you found the body the temperature was $30^\circ$C. Assuming a human corpse obeys Newton’s Law of Cooling with cooling constant $-0.012 \text{ hour}^{-1},$ and the temperature in the freezer is $-15^\circ$C, estimate how long ago the person died? (Assume the temperature of a living body is $98.6^\circ\text{F} = 37^\circ\text{C}.$) (Hint: Use the result in part (a).)
Suppose $S$ is the temperature of the object at time $t.$ The Newton's law of cooling says $$ S-T = (S_0 - T) e^{-kt} $$ where $S_0$ is the temperature when $t=0.$ Let $t_A$ and $t_B$ be the times at which the temperature is $A$ and $B$ respectively. Then you have \begin{align} & A-T = (S_0-T)e^{-kt_A} \\[5pt] & B-T = (S_0-T)e^{-kt_B} \\[8pt] \text{Therfore } & \frac{B-T}{A-T} = e^{-k(t_B-t_A)} \\[8pt] \text{So } & \ln \frac{B-T}{A-T} = -k(t_B-t_A) = k(t_A-t_B) \end{align} etc.