(1 point) A tank contains 1000L of pure water. Brine that contains 0.02kg of sal
ID: 2856394 • Letter: #
Question
(1 point) A tank contains 1000L of pure water. Brine that contains 0.02kg of salt per liter enters the tank at a rate of 5L/min. Also, brine that contains 0.08kg of salt per liter enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15L/min. Answer the following questions.
1. How much salt is in the tank after t minutes?
Answer (in kilograms): S(t)=
2. How much salt is in the tank after 2 hours?
Answer (in kilograms)..................
Note: Please show clearly the final answer
Explanation / Answer
Let S kg be Quantity of salt in 1000Litre tank at any time t
Volume of the tank is 1000L of Pure water
==> Initially salt concentration in the tank is 0 ==> S(0) = 0
0.02 kg / litre enters at 5 litres / min and 0.08 kg / litre enters at 10 litres / min
==> Rate of salt entering tank = 0.02 (kg / litre) 5 (litres / min) + 0.08 (kg / litre) 10 (litres / min)
==> Rate of salt entering tank = 0.1 kg/min + 0.8 kg/min = 0.9 kg/min
Salt water from tank is drained at 15 litres/ min
==> Rate of salt leaving the tank = S (kg / (1000) litres) * 15 (litres/ min) = 3S/200 kg/ min
Volume of the tank remains same since (5 + 10) litres are added and 15 litres are drained
==> Rate of change of salt quantity in tank = (Rate of salt entering tank) - (Rate of salt leaving tank)
==> dS/dt = 0.9 - 3S/200 = (-3/200)(S - 60)
==> dS/(S - 60) = (-3/200) dt
Integrating on both sides
==> dS/(S - 60) = (-3/200) dt
==> ln(S - 60) = -3t/200 + c
==> (S - 60) = e-3t/200 + c
==> S = 60 + Ce-3t/200
salt concentration when t = 0 is 0 kg/litre
==> 0 = 60 + Ce-3(0)/200
==> C = -60
==> S = 60 - 60e-3t/200
==> S(t) = 60 [e-3t/200 - 1]
at 2 hours
==> t = 120 minutes
==> S(120) = 60 [e-3(120)/200 - 1]
==> S(120) = 60 [0.1653 - 1]
==> S(120) = -50.08
Hence Quantity of salt after 2 hours = -50.08 kg / litre
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