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Salt tank mixing problem

Short answer

Let x(t) be grams of salt. The rate equation dx/dt = 750 - x/200 with x(0)=0 solves to x=150000(1-e^(-t/200)). Setting x=5000 gives t = 200 ln(30/29) ≈ 6.78 minutes.

A tank contains 1000 liters of fresh water. Salt water which contains 150 gms of salt per liter runs into it at the rate of 5 liter per minute and well-stirred mixture runs out of it at the same rate. When will the tank contain 5000 gms of salt?

Let x(t) be the amount of salt (in gm) in the tank after t minutes.

Rate equation

The volume stays at 1000 litres because inflow = outflow = 5 L/min.

  • Salt in: 150 gm/L × 5 L/min = 750 gm/min.
  • Salt out: concentration x/1000 gm/L × 5 L/min = x/200 gm/min.

dx/dt = 750 − x/200

Solve

dx/dt + x/200 = 750, IF = e^(t/200)

x e^(t/200) = 750 · 200 e^(t/200) + C = 150000 e^(t/200) + C

x = 150000 + C e^(−t/200)

At t = 0, x = 0 (fresh water): C = −150000.

x(t) = 150000(1 − e^(−t/200)) gm

Time to reach 5000 gm

5000 = 150000(1 − e^(−t/200))

e^(−t/200) = 1 − 1/30 = 29/30

t = 200 ln(30/29) = 200 × 0.033902

t ≈ 6.78 minutes (about 6 min 47 s).

(The salt content tends to 150000 gm = 1000 L × 150 gm/L, the concentration of the incoming brine.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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