A tank contains 200 gal of salt water with 150 lbs of dissolved salt. Salt water with 5 lbs of salt per gal. flows into the tank at the rate of 3 gal/min. and the mixture kept uniform by stirring flows out at the same rate. How long will it be before there are 150 lbs of salt in the tank.
Question
A tank contains 200 gal of salt water with 150 lbs of dissolved salt. Salt water with 5 lbs of salt per gal. flows into the tank at the rate of 3 gal/min. and the mixture kept uniform by stirring flows out at the same rate. How long will it be before there are 150 lbs of salt in the tank.
Solution
This problem can be solved using the concept of differential equations.
Step 1: Understand the problem The tank initially contains 200 gallons of salt water with 150 lbs of salt. Salt water with 5 lbs of salt per gallon is flowing into the tank at a rate of 3 gallons per minute. The mixture is kept uniform by stirring and flows out at the same rate. We need to find out how long it will take for there to be 150 lbs of salt in the tank.
Step 2: Set up the differential equation Let's denote the amount of salt in the tank at any time t by S(t). The rate of change of S(t) is given by the rate at which salt is flowing in minus the rate at which salt is flowing out.
The rate at which salt is flowing in is 5 lbs/gal * 3 gal/min = 15 lbs/min.
The rate at which salt is flowing out is (S(t)/200 lbs/gal) * 3 gal/min = 3S(t)/200 lbs/min.
So, the differential equation is dS/dt = 15 - 3S/200.
Step 3: Solve the differential equation This is a first-order linear differential equation. Its general solution is S(t) = Ce^(-3t/200) + 100, where C is a constant.
Step 4: Find the initial condition At t = 0, S(0) = 150. So, 150 = Ce^(0) + 100, which gives C = 50.
Step 5: Find when S(t) = 150 We need to solve the equation 150 = 50e^(-3t/200) + 100. This gives e^(-3t/200) = 1, so -3t/200 = 0, which gives t = 0.
So, the amount of salt in the tank remains at 150 lbs and does not change over time.
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