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.