A 500-gallon tank originally contains 350 gallons of pure water. A 2-pound per gallon brine solution is pumped into the tank at 10 gallons per minute and the well-stirred mixture is pumped out at one-half that rate. How much salt is in the tank at th emoment of overflow?
Answer: 510 lb
dA/dt = [(rate in)(Concentration in)] - [(rate out)(concentration out)]
The Attempt at a Solution
Here is what I did.
dA/dt = (2lb/gal * 10 gal/min) - [A/(350+5t) * 5]
dA/dt = 20 - A/(70+t)
dA/dt + A/(70+t) = 20
Initial condition: A (0) = 0
eintgral[1/(70+t)] = 70 + t
A(t) = [1/(70+t)] * [10* integral(t+70)dt + c
A(t) = 5(t+70) + c/(t+70)
A(0) = 0;
c = -24,500
Thus; A(t) = 5(t+70) - 24500/(t+70)
A (10)/500 = 400 - 24500/80
= 93.75 lb/gal
This answer is incorrect, as the answer should be 510 lb.