Question

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L/min.

(a) How much salt is in the tank after t minutes? y = kg
(b) How much salt is in the tank after 40 minutes? (Round the answer to one decimal place.) y(40) = kg

Answer 1

a)
`y(t)=0.65(Kg)/(min)(tmin)`

b)
`y(40)=26Kg`

Explanation:

Data

Brine a (Ba)

`V_(Ba)=5(Lt)/(min)\\  Concentration(Bca)=0.05(Kg)/(Lt)`

Brine b (Bb)

`V_(Bb)=10(Lt)/(min)\\  Concentration(Bcb)=0.04(Kg)/(Lt)`

we have that per every minute the amount of solution that enters the tank is the same as the one that leaves the tank (15 Lt / min)

, then the amount of salt (y) left in the tank after (t) minutes:
`y=V_(Ba)*B_(ca)+V_(Bb)*B_(cb)=5(Lt)/(min)*0.05(Kg)/(Lt)+10(Lt)/(min)*0.04(Kg)/(Lt)=\\0.25(Kg)/(min)+0.4(Kg)/(min)=0.65(Kg)/(min)`

Finally:

a)
`y(t)=0.65(Kg)/(min)(tmin)`

b)
`y(40)=0.65(Kg)/(min)(40min)=26Kg`

being y(t) the amount of salt (y) per unit of time (t)