Question

A tank contains 1000 L of brine (that is, salt water) with 15 kg of dissolved salt.
Pure water enters the top of the tank at a constant rate of 10 L / min. The solution
is thoroughly mixed and drains from the bottom of the tank at the same rate so
that the volume of liquid in the tank is constant. Find a differential equation expressing the rate at which salt leaves the tank.

Answer 1

The answer is\[\frac{ds}{dt}=-\frac{s}{100}\]

Answer 2

I don't get why the rate of change depends on the amount of salt in the tank.

Answer 3

Let "s" kg of salt be in the tank at time "t".
At delta_t minutes later, delta_s kg of salt would have left the tank and so the amount of salt in the tank will be less by delta_s.
The concentration of salt in the tank at time t is: s / 1000 kg/liter
In delta_t minutes, 10 * delta_t liters of liquid would have left the tank and the amount of salt that leaves with it = 10 * delta_t * s / 1000 kg. = delta_t * s / 100
delta_s = - delta_t * s / 100
ds/dt = -s / 100

Answer 4

Thanks, for some reason I didn't realize that the salt would spread into the pure water too which means that there would be less and less salt leaving over time.