According to Newton’s law of cooling, the rate at which a cup
of coffee cools is proportional to the difference between its temperature and that of
the room it is in. A certain cup of coffee cools from $164^{\circ}$ to $140^{\circ}$ (all temperatures
Fahrenheit) in five minutes, and then from $140^{\circ}$ to $122^{\circ}$ in the next five minutes. What
is the temperature of the room?
Let $C$ be the room temperature, $f(t)$ be the temperature of the coffer at the time of $t$. Hence, we have
$$f'(t) = k(f(t)-C)$$
where $k$ is a constant. Let $g(t)=f(t)-C$. Then $g'(t)=f'(t)$ and
$$g'(t)=kg(t)$$
This equation can be solved by
$$g(t)=\lambda e^{kt}$$
where $\lambda$ is a constant. This means that
$$f(t) = \lambda e^{kt} +C$$
Let $t$ measured in minutes and $t=0$ indicates the moment when the observation starts. Therefore, we have
$$\left\{\begin{array}{rll} f(0)&=\lambda + C &= 164\\ f(5)&= \lambda e^{5k} +C &= 140 \\ f(10)&= \lambda e^{10k} +C &= 122 \end{array}\right.\implies \left(e^{5k}, \lambda, C\right)=\left(\frac{3}{4}, 96, \boxed{68}\right)$$