A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L/min.
(a) How much salt is in the tank after tminutes?(b) How much salt is in the tank after 60 minutes? (Round theanswer to one decimal place.)

(a) After time (t), the amount of salt in the tank is s(t) = 130 − 130e ∧−3t/200 / 3

(b) After 60 minutes, the salt in the tank is s(60) ≈ 25.7153

Step-by-step explanation:

To start with,

Let s(t) = amount of salt in kg of salt at time t.

Then we have:

dt/ds =  (rate of salt into tank) − (rate of salt out of tank)

= (0.05 kg/L · 5 L/min) + (0.04 kg/L · 10 L/min) − ( s kg/ 1000 L x 15 L/min)

= 0.25 kg/min + 0.4 kg/min − 15s / 1000 kg/min

So we get the differential equation

dt/ds = 0.65 − 15s / 1000

= 65 / 100 − 15s / 1000

dt/ds = 130 - 3s / 200

We separate s and t to get

1 / 130 - 3s ds = 1 / 200 dt

Then we Integrate,

Thus we have

∫1 / 130 - 3s ds = ∫1 / 200 dt

- 1/3  · ln |130 − 3s| =1 / 200 t + C1

ln |130 − 3s| = - 3 /200 t + C2

|130 − 3s| = e− ³⁺²⁰⁰ ∧ t+C2

|130 − 3s| = C3e − ∧3t/200

130 − 3s = C4e  ∧−3t/200

−3s = −130 + C4e  ∧−3t/200

s = 130 - C4e ∧−3t/200 / 3

Since we begin with pure water, we have s(0) = 0. Substituting,

0 = 130 − C4e  ∧−3·0/200 / 3

0 = 130 − C4

C4 = 130

So our function is

s(t) = 130 − 130e ∧−3t/200 / 3

After one hour (60 min), we have

s(60) = 130 − 130e ∧−3·60/200 / 3

s(60) ≈ 25.7153

Thus, after one hour there is about 25.72 kg of salt in the tank.

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