17 In an industrial process, a container initially contains 1000 litres of liquid. Liquid is drawn from the bottom of the container at a rate of 5 litres per minute. At the same time, salt is added to the top of the container at a constant rate of 10 grams per minute. After \(t\) minutes the mass of salt in the container is \(x\) grams, and you are given that \(x = 0\) when \(t = 0\).
In modelling the situation, it is assumed that the salt dissolves instantly and uniformly in the liquid, and that adding the salt does not change the volume of the liquid.
- Show that the concentration of salt in the liquid after \(t\) minutes is \(\frac { \mathrm { X } } { 1000 - 5 \mathrm { t } }\) grams per litre.
- Hence show that the mass of salt in the container is given by the differential equation
$$\frac { d x } { d t } + \frac { x } { 200 - t } = 10$$
- Show by integration that \(\mathrm { x } = 10 ( 200 - \mathrm { t } ) \ln \left( \frac { 200 } { 200 - \mathrm { t } } \right)\).
- Hence determine the mass of salt in the container when half the liquid is drawn off.
- Determine also the time at which the mass of salt in the container is greatest.
- When the process is run, it is found that the concentration of salt over time is higher than predicted by the model.
Suggest a reason for this.