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
and


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

and
`y(0) = 0` ⇒
`-\ln|10-0| = (0)/(10) + C\ \Rightarrow\ C = -\ln 10`


`-\ln|10-y| = (t)/(10) - \ln 10 \\ \Rightarrow \ln|10-y| = \ln 10 - (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