Chapter 7.3, Problem 47E

To determine

To find : Percentage of alcohol in the mixture after an hour.

Answer to Problem 47E

The percentage of alcohol in the mixture after an hour is $4.9\%$

Explanation of Solution

Given information :

A vat with $500$ gallons of beer contains $4\%$ of alcohol (by volume). Beer with $6\%$ alcohol is pumped into the vat at the rate of $5gal/\min .$ and the mixture is pumped out at same rate.

Formula used :

  $\frac{dA}{dt}=\text{ rate in }-\text{ rate out}$

  ${\displaystyle \int \frac{dx}{a-x}=-\ln \left|a-x\right|+C}$

Calculation :

Let, $A\left(t\right)$ be the amount of alcohol after $t$ minutes.

We know volume $=500\text{ gal}$ then $A\left(0\right)=\frac{4}{100}\times 500=20\text{ gal}$

Thus, $\text{rate in }=\frac{6}{100}\times 5\text{gal}/\min =\frac{3}{10}\text{gal}/\min$

  $\text{rate out }=\frac{A\left(t\right)}{500}\times 5gal/\min =\frac{A\left(t\right)}{100}gal/\min$

Then we have the equation $\frac{dA}{dt}=\frac{3}{10}-\frac{A}{100}=\frac{30-A}{100}$

  $\begin{array}{l}\frac{dA}{dt}=\frac{30-A}{100}\\ \Rightarrow \frac{dA}{30-A}=\frac{dt}{100}\end{array}$

Integrating we get,

  $\begin{array}{l}-\ln \left|30-A\right|=\frac{1}{100}t-C\text{ [C is integrating constant]}\\ \Rightarrow \left|30-A\right|=e^{C-\frac{t}{100}}\\ \Rightarrow \left|30-A\right|=e^C\cdot e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}\\ \Rightarrow 30-A=K{e}^{\frac{-t}{100}}\left[K=e^C\right]\end{array}$

So, $A\left(t\right)=30-K{e}^{\frac{-t}{100}}$

At $t=0$ , $A=20$

  $\begin{array}{l}\Rightarrow 20=30-K{e}^0\\ \Rightarrow K=10\end{array}$

So, $A\left(t\right)=30-10e^{\frac{-t}{100}}$

Here, we have to find the percentage of alcohol after an hour. So, $t=60$

The percentage of alcohol in the vat after an hour will be

  $\frac{A\left(60\right)}{500}=\frac{30-10e^{\frac{-3}{5}}}{500}=\frac{24.5}{500}=4.9\%$