A tank contains 500 liters of water with 50 kg of salt dissolved. Water with 2.5 kg/L salt enters at 5 L/min, and well-mixed solution leaves at the same rate. Find the salt amount S(t).

Let $S(t)$ be the amount of salt (in kg) in the tank at time $t$ (in minutes).
The volume of the solution in the tank is constant, $V = 500$ L, because the inflow rate equals the outflow rate.
Initial amount of salt: $S(0) = 50$ kg.

Incoming concentration = 2.5 kg/L
Incoming flow rate = 5 L/min
Rate in = $(2.5 \text{ kg/L}) \times (5 \text{ L/min}) = 12.5 \text{ kg/min}$

Concentration of salt in the tank at time $t$ = $\frac{S(t)}{V} = \frac{S(t)}{500}$ kg/L
Outgoing flow rate = 5 L/min
Rate out = $\left(\frac{S(t)}{500} \text{ kg/L}\right) \times (5 \text{ L/min}) = \frac{5S(t)}{500} = \frac{S(t)}{100} \text{ kg/min}$

The rate of change of salt in the tank is $\frac{dS}{dt}