1. A large tank contains 8000 liters of water in which 410 grams of salt has bee
ID: 2943390 • Letter: 1
Question
1. A large tank contains 8000 liters of water in which 410 grams of salt has been dissolved. At time t = 0 , water containing 4.6 grams of salt per liter begins flowing into the tank at a rate of 49 liters per hour. The well-mixed solution flows out of the tank at the same rate (49 liters per hour).Let represent the amount of salt in the tank hours after this process begins. Write an IVP that models this situation and solve it:
A. y(t) =?
B. How long will it take until the tank contains exactly 307.5 liters of salt? (Round your answer to the nearest minute.)
C. At time t = 98 another worker flips the switch back to its original position, stopping the flow of pure water and once again starting the flow of salt water into the tank. How much salt will be in the tank at time t= 99 ? (Round your answer to the nearest gram.)
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Explanation / Answer
V(0)=8000 L S(0)=410 g dV/dt = dV(in)/dt-dV(out)/dt = 0 dS/dt = dS(in)/dt - dS(out)/dt dS/dt= (4.6g/L)*(49L/h) - C(t)*dV(out)/dt, with C(t) defined as the concentration, or S(t)/V(t)...check the units for ds/dt (g/h) = g/L*(L/h) - (g/L)* (L/h)...CHECKS ALRIGHT. I write dV(out)/dt = w from now on since it is merely a constant value of liters. I write 4.6g/L*49L/h = A, merely a constant. dS/dt = A - w*S(t)/V(t) ... V(t) is constant because dV/dt = 0 for all time. dS/dt = A - (w/V)*S. This is a separable ODE, solve for S(t) [y(t) in your answer space] Solve for t, for S(t)=307.5g...I assume the question was a mistake to ask for liters. I can't help on (c) for it seems to contradict that a salt-mixture was flowing in for the entire time...at no point in the question was freshwater flowing in.
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