Question

A tank initially holds 10 lit. of freshwater. At \[t = 0\], a brine solution containing \[\dfrac{1}{2}kg\] of salt per lit. is poured into tank at a rate 1lit/min while the well-stirred mixture leaves the tank at the same rate. Find the concentration of salt in a tank at a particular time.
A. \[ - 5{e^{ - t}} + 5\]
B. \[5{e^t} + 5\]
C. \[5{e^{ - t}} - 5\]
D. \[ - 5e - 5\]

Answer 1

Hint: We use the concept of mass balance of salt for time dt. That is \[\dfrac{{dC}}{{dt}} = Q - c\], where \[Q\] amount of salt is contained in solution. \[c\] is the concentration of the salt that we need to find. \[\dfrac{{dC}}{{dt}}\] is the rate of change of concentration with time \[t\]. Using the integration method we can find the concentration of the salt at a particular time.

Complete step by step answer:
We have, \[\dfrac{{dC}}{{dt}} = Q - c\]
Where \[Q = \dfrac{1}{2}kg\], then above becomes:
\[ \Rightarrow \dfrac{{dC}}{{dt}} = \dfrac{1}{2} - c\]
This can be rewritten as
\[ \Rightarrow \dfrac{{dC}}{{dt}} = 0.5 - c\]
We have a differential equation, it can be solved by a variable separable method.
Now separating the variables we have,
\[ \Rightarrow \dfrac{{dC}}{{0.5 - c}} = dt\]
Integrating on both sides, we have
\[ \Rightarrow - \int {\dfrac{{dC}}{{0.5 - c}} = - \int {dt} } \]
(Multiply by negative sign on both sides)
We know that \[\int {\dfrac{1}{x}} = \ln x + k\], and integration is the reverse processes of differentiation,
Then,
\[ \Rightarrow \ln (0.5 - c) = - t + k{\text{ - - - - - - (1)}}\]
Where, \[k\] is the integration constant.
We know that at time \[t = 0\] and concentration will be zero \[c = 0\].
Now equation (1) becomes,
\[ \Rightarrow \ln (0.5) = k\]
\[ \Rightarrow k = \ln \left( {\dfrac{1}{2}} \right)\]
We know that \[\ln \left( {\dfrac{m}{n}} \right) = \ln m - \ln n\] applying we get,
\[ \Rightarrow k = \ln (1) - \ln (2)\]
But \[\ln (1) = 0\], we get
\[ \Rightarrow k = - \ln (2)\]
Now substituting in the equation (1)
\[ \Rightarrow \ln (0.5 - c) = - t - \ln 2\]
Put the base number \[e\] on both sides of the equation, we get:
\[ \Rightarrow {e^{\ln (0.5 - c)}} = {e^{ - t - \ln 2}}\]
Logarithmic and exponential cancels out, we have
\[ \Rightarrow (0.5 - c) = {e^{ - t - \ln 2}}\]
We know \[{e^{a + b}} = {e^a}.{e^b}\]
\[ \Rightarrow (0.5 - c) = {e^{ - t}}.{e^{ - 1 \times \ln 2}}\]
We have, \[\ln {a^b} = b\ln a\], we get,
\[ \Rightarrow (0.5 - c) = {e^{ - t}}{e^{\ln {{(2)}^{ - 1}}}}\]
Again logarithmic and exponential cancels out, we have
 \[ \Rightarrow (0.5 - c) = {e^{ - t}}.\dfrac{1}{2}\]
\[ \Rightarrow (0.5 - c) = {e^{ - t}} \times 0.5\]
\[ \Rightarrow c = 0.5 - {e^{ - t}} \times 0.5\]
Taking 0.5 has common term,
\[ \Rightarrow c = 0.5(1 - {e^{ - t}})\].
This is for one litres. Now for 10 litres \[10 \times c\]. That is
\[ \Rightarrow 10 \times c = 10 \times 0.5(1 - {e^{ - t}})\]
\[ \Rightarrow = 5(1 - {e^{ - t}})\]
\[ \Rightarrow = 5 - 5{e^{ - t}}\] or \[ - 5{e^{ - t}} + 5\]

Hence the required answer is option (A).

Note: Remember the formula of mass balance of a salt for time dt. We have Concentration of the salt \[ - 5{e^{ - t}} + 5\], for different values of time t we get the concentration at respective times. Also know that in definite integral we don’t have integration constant. While in indefinite integral we have integration constant.