Find Solution (1 point) A tank contains loo kg of salt and 2000 L of water. Pure

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(1 point) A tank contains loo kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the rate 3 L/min. (a) What is the amount of salt in the tank initially? amount = (kg) (b) Find the amount of salt in the tank after 1 hours. amount= (kg) (c) Find the concentration of salt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.) concentration = (kg/L)

Explanation / Answer

(a) Initially Amount of salt in the tank = 100 kg

(b) Let y(t) be the amount of salt in the tank. Then we have y(0) = 100
We have to first get the function y(t)
We use the fact that dy/dt = Rate at which salt enters the tank - Rate at which salt leaves the tank
Since pure water enters the tank, Rate at which salt enters the tank = 0
Volume of water in the tank at time t = 2000 + (6t - 3t) = 2000 + 3t
Rate at which salt leaves the tank = [y/(2000 + 3t) kg/L] * 3 L/min = 3y/(2000 + 3t) kg/min
=> dy/dt = 0 - [3y/(2000 + 3t)] = -3y/(2000 + 3t)
dy/y = -3dt/(2000 + 3t)
dy/y = -3dt/(2000 + 3t)
ln y = -ln(2000 + 3t) + C
Using the fact that y(0) = 100, we get
ln 100 = -ln 2000 + C
C = ln 2000 + ln 100 = ln 200000
=> ln y = -ln(2000 + 3t) + ln 200000 = ln[200000/(2000 + 3t)]
=> y(t) = 200000/(2000 + 3t)
1 hour = 60 min
y(60) = 200000/(2000 + 3 * 60) = 91.74
Amount of salt in the tank after 1 hour = 91.74 kg.

(c) y(t) = 200000/(2000 + 3t)

Plugging t -> Infinity we get y -> 0

=> concentration of salt in the solution as time approaches infinity becomes equal to 0 kg/L

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