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

Answer:

a) $y(t)=0.65\frac{Kg}{min}(tmin)$

b) $y(40)=26Kg$

Step-by-step explanation:

Data

Brine a (Ba)

$V_{Ba}=5\frac{Lt}{min}\\ Concentration(Bca)=0.05\frac{Kg}{Lt}$

Brine b (Bb)

$V_{Bb}=10\frac{Lt}{min}\\ Concentration(Bcb)=0.04\frac{Kg}{Lt}$

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: $y=V_{Ba}*B_{ca}+V_{Bb}*B_{cb}=5\frac{Lt}{min}*0.05\frac{Kg}{Lt}+10\frac{Lt}{min}*0.04\frac{Kg}{Lt}=\\0.25\frac{Kg}{min}+0.4\frac{Kg}{min}=0.65\frac{Kg}{min}$

Finally:

a) $y(t)=0.65\frac{Kg}{min}(tmin)$

b) $y(40)=0.65\frac{Kg}{min}(40min)=26Kg$

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