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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

Respuesta :

prozehus

Answer:

a) [tex]y(t)=0.65\frac{Kg}{min}(tmin)[/tex]

b) [tex]y(40)=26Kg[/tex]

Step-by-step explanation:

Data

Brine a (Ba)

[tex]V_{Ba}=5\frac{Lt}{min}\\  Concentration(Bca)=0.05\frac{Kg}{Lt}[/tex]

Brine b (Bb)

[tex]V_{Bb}=10\frac{Lt}{min}\\  Concentration(Bcb)=0.04\frac{Kg}{Lt}[/tex]

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: [tex]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}[/tex]

Finally:

a) [tex]y(t)=0.65\frac{Kg}{min}(tmin)[/tex]

b) [tex]y(40)=0.65\frac{Kg}{min}(40min)=26Kg[/tex]

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

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