Question

A tank contains 100 liters of pure water. Brine that contains 0.1 kg of salt per liter enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is the tank after 6 minutes?

Answer 1

Answer on Question #83852 – Math – Differential Equations

Question

A tank contains 100 liters of pure water. Brine that contains $0.1\,\mathrm{kg}$ of salt per liter enters the tank at a rate of $10\,\mathrm{L/min}$. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is the tank after 6 minutes?

Solution

Let $x(t)$ be the mass of salt, in kilograms, that is in the tank at time $t$. Since the tank is initially filled with fresh water we know that $x(0) = 0$.

An expression for $x'(t)$ will be given as the rate salt enters the tank minus the rate salt leaves the tank (in kilograms per minute).

$$\begin{array}{l} x'(t) = \frac{dx}{dt} = (\text{rate in}) - (\text{rate out}) = \\ = \left(\frac{0.1\,\mathrm{kg}}{1\,\mathrm{L}}\right) \left(10\,\frac{\mathrm{L}}{\mathrm{min}}\right) - \left(\frac{x(t)\,\mathrm{kg}}{100\,\mathrm{L}}\right) \left(10\,\frac{\mathrm{L}}{\mathrm{min}}\right) \\ \frac{dx}{dt} = \frac{10 - x}{10} \\ \frac{dx}{10 - x} = \frac{1}{10} \, dt \\ \int \frac{dx}{10 - x} = \frac{1}{10} \int dt \\ - \ln |10 - x| = \frac{1}{10} t + C \\ x(0) = 0: C = -\ln 10 \\ \ln |10 - x| = \ln 10 - \frac{1}{10} t \\ |10 - x| = 10 e^{-t/10} \\ \text{Since } x < 10 \\ 10 - x = 10 e^{-t/10} \\ x = 10 - 10 e^{-t/10} = 10 \left(1 - e^{-t/10}\right) \\ \text{Then} \\ x(6) = 10 \left(1 - e^{-6/10}\right) = 10 \left(1 - e^{-0.6}\right) \, (\mathrm{kg}) \\ 10 \left(1 - e^{-0.6}\right) \, \mathrm{kg} \approx 4.512\,\mathrm{kg} \\ \text{Answer: } 10 \left(1 - e^{-0.6}\right) \, \mathrm{kg} \approx 4.512\,\mathrm{kg} \end{array}$$

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