A tank orignally contains 100gal of fresh water.Then water containing 1/2 lb of

Question

A tank orignally contains 100gal of fresh water.Then water containing 1/2 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate.After 10min the process is stopped, and fresh water is poured into the tank at a rate of 2gal/min,with the mixture again leaving at the same rate.Find the amount of salt in the tank at the end of an additional 10min.

Explanation / Answer

(1) First 10 minutes: Q(t) = the amount of salt in the tank at time t dQ / dt = ratein - rateout ratein = [ 0.5 lb/gal ] [ 2 gal/min ] = 1 lb/min rateout = [ (Q lb) / (100 gal) ] [ 2 gal/min] = (Q / 50) lb/min dQ / dt = 1 - Q / 50 = 50 /50 - Q / 50 = (50 - Q) /50 => dQ / (50 - Q) = dt / 50 => Integral [ dQ / (50 - Q) ] = Integral [ dt/ 50 ] => -ln(50 - Q) = t / 50 + C1 => ln(50 - Q) = -t / 50 + C => 50 - Q = e-t / 50 + C = eCe-t /50 = ke-t / 50=> Q = 50 - ke-t / 50 Q(0) = 0 => 50 - k = 0 => k = 50 => Q = 50 - 50e-t / 50 Q(10) = 50 - 50e-10 / 50 = 9.06346235 lb (2) Next 10 minutes: ratein = 0 lb/min rateout = (Q / 50) lb/min dQ/dt = -Q / 50 => dQ / -Q = dt / 50 => Integral [ dQ / -Q ] = Integral [ dt / 50 ] => -ln(Q) = t / 50 + C1 => ln(Q) = -t / 50 + C => Q = e-t / 50 + C = eCe-t / 50 = ke-t /50 Q(0) = 9.06346235 => k = 9.06346235 => Q = 9.06346235e-t / 50 Q(10) = 9.06346235e-10 / 50 = 7.42053536 g ˜ 7.421 lb

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