Chapter 7.3, Problem 45E

(a)

To determine

To Find: The amount of slat in the tank after t minutes.

Answer to Problem 45E

The amount of salt in the tank is $15e^{\frac{-t}{100}}\text{kg}$ .

Explanation of Solution

Given:

The tank has 1000L of brine with 15 kg of dissolved salt.

The rate at which the pure water enters the tank is 10 L per minute.

Calculation:

Consider the amount of salt after t minutes is,

  $x\left(0\right)=15\text{kg}$

Then, the rate in and the rate out of the fresh water is,

  $\frac{dx}{dt}=ratein-rateout$

Then, from the given data the rate of pure water is 1000 litre per minute and the rate out is,

  $\begin{array}{l}\frac{dx}{dt}=\frac{x}{100}\\ \frac{-1dt}{100}=\frac{1}{x}dx\\ {\displaystyle \int \frac{1}{x}dx}={\displaystyle \int \frac{-1}{100}dt}\\ \ln x=\frac{-t}{100}+C\end{array}$

Consider that for $t=0$ the value of $x=15$ then,

  $\begin{array}{l}\ln x=\frac{-t}{100}+\ln 15\\ x=15e^{\frac{-t}{100}}\text{kg}\end{array}$

Thus, the amount of salt in the tank is $15e^{\frac{-t}{100}}\text{kg}$ .

(b)

To determine

To Find: The amount of slat after 20 minutes.

Answer to Problem 45E

The amount of salt in the tank is $12.3\text{kg}$ .

Explanation of Solution

Given:

The tank has 1000L of brine with 15 kg of dissolved salt.

The rate at which the pure water enters the tank is 10 L per minute.

Calculation:

Consider the given equation s,

  $x=15e^{\frac{-t}{100}}$

Then,

  $\begin{array}{c}x=15e^{\frac{-20}{100}}\\ =12.3\text{kg}\end{array}$