Question

A tank contains 1000 L of brine (salt water) with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 L/min. The solution is kept throughly mixed and drains from the tank at the same rate. How much salt is in the tank after t minutes

Answer 1

`y=15e^{e^{- (t)/(100)}}kg`

Explanation:

Let y(t) represent the amount of salt in the tank after t minutes. Therefore:

`(dy)/(dt)=rate\ in-rate\ out`

Pure water is entering at a rate of 10 L/min, therefore rate in = 0

The tank contains 1000 L of brine (salt water) with 15 kg of dissolved salt, and a mixed solution leaving at 10 L/min hence;

`rate\ out = (y(t)\ kg)/(1000\ L)*10\ L/min= (y(t)\ kg)/(100\ min)\\\\Therefore:\\\\(dy)/(dt)=0- (y(t))/(100) \\\\(dy)/(dt)=- (y(t))/(100)\\\\(dy)/(y)=- (1)/(100)dt\\\\\int\limits{(dy)/(y)} =\int\limits- (1)/(100)dt\\\\lny=- (t)/(100)+C\\\\Taking\ exponetial\ of\ both\ sides:\\\\y=e^{- (t)/(100)+C}\\\\y=e^C.e^{- (t)/(100)}\\\\y=Ae^{- (t)/(100)}(A=e^C)\\\\at\ y=15kg, t=0\\\\15=Ae^{e^{- (0)/(100)}}\\\\A=15\\\\`

`y=15e^{e^{- (t)/(100)}}kg`