192. A metal ball, the temperature of which at the beginning of the experiment was $12^{\circ} \mathrm{C}$, is cooled by a stream of water having a temperature of $0^{\circ}$. After 8 minutes, the ball cooled down to $9^{\circ}$. Assuming the cooling rate is proportional to the difference between the temperature of the body and the temperature of the cooling medium, determine how long it took for the ball to cool down to $7^{\circ}$.
Solution. $1^{\circ}$. Let $T$ denote the temperature of the sphere, and $t-$ the time elapsed since the start of the experiment. Then the cooling rate of the sphere is the derivative $\frac{d T}{d t}\left(T^{\prime}\right)$. According to the condition, $T^{\prime}=k(T-0)=k T$, where $k$ is the proportionality coefficient.

$2^{0}$. This is a separable differential equation. Separating the variables, we have $\frac{d T}{d t}=k T, \frac{d T}{T}=k d t$, from which $\ln T=k t+\ln C$, or $\ln T=\ln e^{k t}+\ln C$. Finally, after exponentiating, we get

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

$3^{0}$. To find the constants $C$ and $k$, we use the data from the problem: at $t=0$ the temperature of the sphere $T=12^{\circ}$, and at $t=8$ it is $T=9^{\circ}$. Substituting the values of $t$ and $T$ into equation (1), we get $12=C e^{0}$ and $9=C e^{8 k}$. From the first equation, we have $C=12$, and from the second, $e^{8 k}=0.75$. To find $e^{k}$, we take the eighth root of both sides of the equation:

$$
e^{k}=\sqrt[8]{0.75}=0.75^{1 / 8}
$$

Raising both sides of this equation to the power of $t$, we have

$$
e^{k t}=0.75^{t / 8}
$$

Substituting the found values of $C$ and $e^{k t}$ into equation (1), we get

$$
T=12 \cdot(0.75)^{1 / 8}
$$

To answer the question of the problem, we take the logarithm base 10 of equation (2):

$$
\lg T=\lg 12+\frac{t}{8} \lg 0.75
$$

and set $T=7$ in the obtained equation:

$$
\lg 7=\lg 12+\frac{t}{8} \lg 0.75
$$

From this, we finally find

$$
t=\frac{8(\lg 7-\lg 12)}{\lg 0.75}=\frac{8(0.8451-1.0792)}{\overline{1}, 8751}=\frac{-8 \cdot 0.2341}{-0.1249}=\frac{1.8728}{0.1249} \approx 15 \text{ min }
$$