74. According to Newton's law, the cooling rate of a body in air is proportional to the difference between the body's temperature $T$ and the air temperature $T_{0}$. Determine the law of temperature change of the body over time, if the experiment is conducted at $T_{0}=20^{\circ} \mathrm{C}$, and the body cooled from $100^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 20 minutes. (see problem 24).
Solution. The cooling rate of a body (the rate of change of its temperature) is equal to the derivative $\frac{d T}{d t}$, and thus Newton's law can be expressed by the equation

$$
\frac{d T}{d t}=k(T-20)
$$

where $k-$ is the proportionality coefficient.

Solving the obtained differential equation with separable variables, we have

$$
\frac{d T}{T-20}=k d t ; \int \frac{d T}{T-20}=k \int d t ; \ln (T-20)=k t+\ln C
$$

from which

$$
T=20+C e^{k t}
$$

To determine $C$ and $k$, we use the conditions of the problem: $T=100^{\circ}$ at $t=0 ; T=60^{\circ}$ at $t=20$.

Substituting these values into equation (4), we get

$$
\left\{\begin{aligned}
100 & =20+C \\
60 & =20+C e^{20 k}
\end{aligned}\right.
$$

From this, $C=80, e^{20 k}=\frac{1}{2}, e^{k}=\left(\frac{1}{2}\right)^{1 / 20}$. Thus,

$$
T=20+80\left(\frac{1}{2}\right)^{1 / 20}
$$

This is the law of temperature change $T$ of the body over time $t$ under the given conditions.