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(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