A tank contains 100L of pure water. Brine that contains... DE problem.?

Question

A tank contains 100L of pure water. Brine that contains .1 kg of salt per liter enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains form the tank at the same rate. How much salt is in the tank after 6 minutes?

I think you set it up as follows: dy/dt = (rate in) - (rate out). dy/dt = (.1kg/L x 10L/min) - (10L/min x y(t)/100L).

can u show how you get that?

Answer 1

Your follow-up comment is correct. The differential equation is

dy/dt = 1 - y/10

and since at time t=0 there is no salt in the tank, we have

y(0) = 0

You should get y(t) = 10(1 - e^(-t/10))

Note that as t -> infinity, y -> 10. That is, after a very long time nearly all of the water in the tank is replaced by brine.