<p>A tank has pure water flowing into it at 10 litres/min. The contents of the tank are kept thoroughly mixed, and the contents flow out at 10 litres/min.</p>
<p>Initially, the tank contains 10 kg of salt in 100 litres of water. How much salt will there be in the tank after 30 minutes?</p>

A tank has pure water flowing into it at 10 litres/min. The contents of the tank are kept thoroughly mixed, and the contents flow out at 10 litres/min.

Initially, the tank contains 10 kg of salt in 100 litres of water. How much salt will there be in the tank after 30 minutes?

Mixing Tank Problem

Differential Equations

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

Worked Example 2

Let x kg be the amount of salt at time t.

**Given:**
• Volume = 100 L
• \(\frac{\mathrm{d}V}{\mathrm{d}t}_{\text{in}} = 10\) L/min
• \(\frac{\mathrm{d}V}{\mathrm{d}t}_{\text{out}} = 10\) L/min

**Setting up the DE:**

\(\frac{\mathrm{d}x}{\mathrm{d}t} = \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)_{\text{in}} - \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)_{\text{out}}\)

\(= 0 - \left(\frac{\mathrm{d}x}{\mathrm{d}V}\right)_{\text{out}} \times \frac{\mathrm{d}V}{\mathrm{d}t}_{\text{out}}\)

Change in mass per unit volume: \(\frac{\mathrm{d}x}{\mathrm{d}V}_{\text{out}} = \frac{x}{100}\)

\(\therefore \frac{\mathrm{d}x}{\mathrm{d}t} = -\frac{x}{100} \times 10\)

\(\frac{\mathrm{d}x}{\mathrm{d}t} = -\frac{x}{10}\)

**Solving the DE:**

\(\displaystyle\displaystyle\int \frac{1}{x} \, \mathrm{d}x = -\frac{1}{10} \displaystyle\displaystyle\int 1 \, dt\)

\(\ln|x| = -\frac{t}{10} + C\)

\(x = \pm e^{-\frac{t}{10} + C} = \pm e^C \cdot e^{-\frac{t}{10}}\)

\(x = Ae^{-\frac{t}{10}}\); \(A = \pm e^C\)

**Applying initial conditions:**

\(t = 0\), \(x = 10\): \(A = 10\)

\(x = 10e^{-\frac{t}{10}}\)

\(t = 30\):
\(x = 10e^{-\frac{30}{10}} = 10e^{-3} = \frac{10}{e^3}\) kg

\(\approx 0.498\) kg (3 s.f.)

Worked Example 2

Worked Example 2

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