Question

A tank contains 100 L 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 in the tank after 6 minutes?

Answer 1

Let $y(t)$ be the amount of salt (in kg) after $t$ minutes. Then $y(0) = 0$. The amount of liquid in the tank is 100 L at all times, so the concentration at time $t$ (in minutes) is $y(t) / 1000\ \mathrm{kg/L)$ and

$\dfrac{dy}{dt} = \left( 0.1\dfrac{\mathrm{kg}}{\mathrm{L}}\right) \left( 10 \dfrac{\text{L} }{\text{ min}}\right)- \left[ \dfrac{y(t)}{100} \dfrac{\mathrm{kg}}{\mathrm{L}}\right] \left( 10 \dfrac{\text{L} }{\text{ min}}\right) \\ \\ \dfrac{dy}{dt} = 1 \dfrac{\mathrm{kg}}{\mathrm{min}}-\dfrac{y(t)}{10}\dfrac{\mathrm{kg}}{\mathrm{min}} \\ \\ \dfrac{dy}{dt} = 1 - \dfrac{y}{10} = \dfrac{10 - y}{10} \\ \displaystyle\int \dfrac{dy}{10 - y} = \int \frac{1}{10} dt \\ -\ln|10-y| = \frac{t}{10} + C$

and $y(0) = 0$ ⇒ $-\ln|10-0| = \frac{0}{10} + C\ \Rightarrow\ C = -\ln 10$

$-\ln|10-y| = \frac{t}{10} - \ln 10 \\ \Rightarrow \ln|10-y| = \ln 10 - \frac{t}{10} \\ \Rightarrow |10 - y| = e^{\ln 10 - t/10} \\ \Rightarrow |10 - y| = 10e^{- t/10} \\ \Rightarrow 10 - y = \pm 10e^{- t/10} \\ \Rightarrow y = 10 \pm 10e^{- t/10}$

Choose (-) because that satisfies $y(0) = 0$

$y(t) = 10 - 10e^{- t/10} \\ y(6) = 10 - 10e^{-6/10} \approx 4.512\ \mathrm{kg}$

4.512 kg