Suppose a brine containing 0.3 kilogram (kg) of salt per litter (L) runs into a tank initially filled with 400 L of water containing 2 kg of salt. If the brine enters at 10 L/min. the mixture is kept uniform by stirring, and the mixture flows out at the same rate. Find the mass of salt in the tank after 10 min.

Use the fact,

rate of increase in A = rate of input - rate of exit.

Let x(t) is amount of salt at time in kg.

\(\begin{array}{c}{\bf{x'}}\left( {\bf{t}} \right){\bf{ = 3 - }}\left( {\frac{{{\bf{x}}\left( {\bf{t}} \right)}}{{{\bf{400}}}}} \right)\left( {{\bf{10}}} \right)\\{\bf{x'}}\left( {\bf{t}} \right){\bf{ = 3 - }}\frac{{{\bf{x}}\left( {\bf{t}} \right)}}{{{\bf{40}}}}\\{\bf{x'}}\left( {\bf{t}} \right){\bf{ = }}\frac{{{\bf{ - 1}}}}{{{\bf{40}}}}\left( {{\bf{x}}\left( {\bf{t}} \right){\bf{ - 120}}} \right)\, \cdot \cdot \cdot \cdot \cdot \cdot \left( {\bf{1}} \right)\end{array}\)

Cross multiplication on both sides in the equation \(\left( 1 \right)\)

\(\frac{{{\bf{x'}}\left( {\bf{t}} \right)}}{{\left( {{\bf{x}}\left( {\bf{t}} \right){\bf{ - 120}}} \right)}}{\bf{ = }}\frac{{{\bf{ - 1}}}}{{{\bf{40}}}}\,\,\)

Taking integration both sides in the above equation

\(\int {\frac{{{\bf{x'}}\left( {\bf{t}} \right)}}{{\left( {{\bf{x}}\left( {\bf{t}} \right){\bf{ - 120}}} \right)}}} {\bf{dt = }}\int {\frac{{{\bf{ - 1}}}}{{{\bf{40}}}}{\bf{dt}}\,} \)

Solve the above equation,

\(\begin{array}{l}{\bf{ln}}\left( {{\bf{x}}\left( {\bf{t}} \right){\bf{ - 120}}} \right){\bf{ = }}\frac{{{\bf{ - t}}}}{{{\bf{40}}}}{\bf{ + c}}\\{\bf{x}}\left( {\bf{t}} \right){\bf{ - 120 = }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{\frac{{{\bf{ - t}}}}{{{\bf{40}}}}}}\\{\bf{x}}\left( {\bf{t}} \right){\bf{ = }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{\frac{{{\bf{ - t}}}}{{{\bf{40}}}}}}{\bf{ + 120}}\, \cdot \cdot \cdot \cdot \cdot \cdot \left( {\bf{2}} \right)\end{array}\)

Here c is the integration constant.

Substituting t = 0 in the equation\(\left( 2 \right)\)

\({\bf{x}}\left( {\bf{0}} \right){\bf{ = }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{\frac{{{\bf{ - 0}}}}{{{\bf{40}}}}}}{\bf{ + 120}}\,\,\)

Here, given\(x\left( 0 \right) = 2kg\)

\(\begin{array}{c}{\bf{2 = }}{{\bf{c}}_{\bf{1}}}{\bf{ + 120}}\,\\{{\bf{c}}_{\bf{1}}}{\bf{ = - 118}}\end{array}\)

Substituting\({{\bf{c}}_{\bf{1}}}{\bf{ = - 118}}\)in the equation\(\left( 2 \right)\)

\(\begin{array}{l}{\bf{x}}\left( {\bf{t}} \right){\bf{ = - 118}}{{\bf{e}}^{\frac{{{\bf{ - t}}}}{{{\bf{40}}}}}}{\bf{ + 120}}\,\\{\bf{x}}\left( {\bf{t}} \right){\bf{ = 120 - 118}}{{\bf{e}}^{\frac{{{\bf{ - t}}}}{{{\bf{40}}}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( {\bf{3}} \right)\end{array}\)

Substituting t = 10 in the equation\(\left( 3 \right)\)

\({\bf{x}}\left( {{\bf{10}}} \right){\bf{ = 120 - 118}}{{\bf{e}}^{\frac{{{\bf{ - 10}}}}{{{\bf{40}}}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\)

Simplify the above equation,

\(\begin{array}{c}{\bf{x}}\left( {{\bf{10}}} \right){\bf{ = 120 - 118}}\left( {{\bf{0}}{\bf{.7788}}} \right)\\{\bf{ = 28}}{\bf{.1016}}\;{\bf{kg}}\end{array}\)

\({\bf{x}}\left( {{\bf{10}}} \right){\bf{ = 28}}{\bf{.1}}\;{\bf{kg}}\)

Hence,the mass of salt in the tank after 10 min is 28.1 kg.