193. A vessel with a capacity of 100 liters is filled with brine containing 10 kg of dissolved salt. In 1 minute, 3 liters of water flow into it, and the same amount of the mixture is pumped into another vessel of the same capacity, initially filled with water, from which the excess liquid is drained. At what moment in time will the amount of salt in both vessels be the same?
Let at time $t$ (min) the first vessel contains $x$ (kg) of salt, and let in the subsequent small time interval $d t$ the amount of salt in this vessel decreases by $d x$.

During the time $d t$, $3 d t$ (l) of brine will flow out of the vessel. The concentration of the brine (the amount of salt in one liter of solution) at time $t$ will be $\frac{x}{100}(kg / l)$. If we assume that the concentration of the brine remains constant during the small time interval $d t$, then the amount of salt will decrease by $-d x=\frac{x}{100} \cdot 3 d t$ (since $d x<0$). Separating variables in this equation and integrating, we get

$$
\frac{d x}{x}=-\frac{3}{100} d t ; \ln x=-0.03 t+C_{1}
$$

From the initial condition $x=10$ when $t=0$, we determine the value of the constant $C_{1}=\ln 10$.

Therefore, the dependence of the amount of salt $x$ in the first vessel on time $t$ is expressed by the equation

$$
\ln x=-0.03 t+\ln 10 \text { or } x=10 e^{-0.03 t}
$$

Now let's find the dependence of the amount of salt $y$ on time for the second vessel. In the second vessel, the concentration of the brine at time $t$ is $\frac{y}{100}(kg / l)$. During the time $d t$, $3 d t(l)$ of brine will flow into it, containing $\frac{3 x}{100} d t$ (kg) of salt, and $3 d t$ (l) of brine will flow out, containing $\frac{3 y}{100} d t$ (kg) of salt, i.e., during the time $d t$ the amount of salt in the second reservoir will change by the amount

$$
d y=0.03 x d t-0.03 y d t \text { or } d y=0.03(x-y) d t
$$

Replacing $x$ in this equation with its expression from formula (3), we get a first-order linear equation

$$
d y=0.03\left(10 e^{-0.03 t}-y\right) d t ; y^{\prime}+0.03 y=0.3 e^{-0.03 t}
$$

the general integral of which has the form

$$
y=e^{-0.03 t}\left(C_{2}+0.3 t\right)
$$

The value of the constant $C_{2}$ is determined from the initial conditions; since $y=0$ when $t=0$, then $C_{2}=0$. Therefore, the dependence of the amount of salt $y$ in the second vessel on time $t$ is expressed by the equation

$$
y=0.3 t e^{-0.03 t}
$$

The sought moment of time, at which the amount of salt in both vessels will become the same, we find by setting $x=y$:

$$
10 e^{-0.03 t}=0.3 t e^{-0.03 t} ; 10=0.3 t ; t=33 \frac{1}{3} \text{ min}
$$

At this moment, there will be approximately $10 / e \approx 3.68$ kg of salt in each vessel.