A tank contains 120 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 3 L/min; the well-mixed solution is pumped out at the same rate. Find the number A (t) of grams of salt in the tank at time t.

A (t) = 120 - 100 e^ (-t/40)

Step-by-step explanation:

Since The Fluid is been pumped both in and out, there will always be a 120 Liters of fluid in a tank

The amount of salt at time t in tank is = A (t) (in grams)

the amount of salt in a tank at time t is given as = A (t) grams

Level of concentrations of salt is = A/120 g/L

The brine quantity that is pumped in the tank = 3L (of salt concentration = 1 g/L)

The salt pumped in: 3L x 1g/L = 3g rams

The level of Amount of brine that is pumped out at time t: 3L (of salt concentration = A/120 g/L)

The quantity of salt pumped out at time t: 3L x A/120 g/L = A/40 g

Therefore, dA/dt = 3 - A/40

40 dA/dt = 120 - A

40 / (120 - A) dA = dt

-40 ln|A-120| = t + C₀

ln|A-120| = - t/40 + C₁, where C₁ = - C₀/40

A - 120 = Ce^ (-t/40) ... where C = e^C₁

A = 120 + C e^ (-t/40)

Recall that, the initial brine contains 20 grams of salt

so,

A (0) = 20

120 + C e^0 = 20

C = - 100 or 100

Then

A (t) = 120 - 100 e^ (-t/40)