Question

A tank initially holds $100gal~$ of a brine solution containing $20lb$ of salt. At $t=0,$ fresh water is poured into the tank at the rate of $5~gal/min,$ while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time $t.$
(A) $20{{e}^{\dfrac{-t}{40}}}$
(B) $20{{e}^{\dfrac{t}{40}}}$
(C) $40{{e}^{\dfrac{-t}{20}}}$
(D) $20{{e}^{\dfrac{-t}{20}}}$

Answer 1

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

Complete Step By Step Answer:
Expression for mass of salt as function of time;
Let $A=$ mass of salt after t min, ${{r}_{i}}=$ rate of salt coming into the tank and ${{r}_{o}}=$ rate of salt coming into the tank.
#Set up an expression for the rate of change of salt concentration.
$\dfrac{dA}{dt}={{r}_{i}}-{{r}_{o}}$ and here we can write ${{r}_{i}}$ as
${{r}_{i}}=\dfrac{5gal}{1\min }\times \dfrac{20lb}{1gal}=100\dfrac{lb}{\min }$
Similarly, we can write ${{r}_{o}}$ as ${{r}_{o}}=\dfrac{5gal}{1\min }\times \dfrac{Alb}{100gal}=\dfrac{A}{20}\dfrac{lb}{\min }$
Thus, $\dfrac{dA}{dt}=100-\dfrac{A}{20}$
Now integrating the above expression.
$\dfrac{dA}{dt}=\dfrac{120-A}{20}$
Taking $dA$ on one side and $dt$ on other side,
$\dfrac{dA}{120-A}=\dfrac{dt}{20}$
Now taking integration on both the sides we get,
$\int{\dfrac{dA}{120-A}}=\int{\dfrac{dt}{20}}$
$\Rightarrow -\ln (120-A)=\dfrac{t}{20}+C$ ………………………equation $1$
To get the value of we need to find the value of C first, for that we will substitute $A=1$ and $t=0$ ,
$-\ln (120-1)=\dfrac{0}{20}+C$
$\Rightarrow C=-\ln 129$
Now that we got the value of constant $C$ we can substitute it in equation $1$ ;
$-\ln (120-A)=\dfrac{t}{20}-\ln 129$
Multiplying throughout by $-1$ we get
$\ln (120-A)=\ln 129-\dfrac{t}{20}$
$\Rightarrow 120-A=129{{e}^{\dfrac{-t}{20}}}$ ……………………..since $\ln \left( a-b \right)=a{{e}^{-n}}$
$\Rightarrow A=120-129{{e}^{\dfrac{-t}{20}}}$
Thus, it can be rewritten as $A=20{{e}^{\dfrac{-t}{20}}}$
Therefore, the correct answer is option D i.e. $20{{e}^{\dfrac{-t}{20}}}$ .

Note:
Remember the formula of mass balance of a salt for time $dt.$ We have Concentration of the salt, for different values, of time $t$ we get the concentration at respective times. Also know that in definite integral we do not have integration constant. While in indefinite integral we have integration constant.