Question

Initially 5 grams of salt are dissolved in 20 liters of water. Brine with concentration of salt 2 grams of salt per liter is added at a rate of 3 liters a minute. The tank is mixed well and is drained at 3 liters a minute. How long does the process have to continue until there are 20 grams of salt in the tank?

Answer 1

Answer:

3.731 minutes

Explanation:

Let the amount of salt in the tank at any time be x(t)

Since x(0)=5 g is  dissolved in 20 liters of water

Brine with 2 grams per liter salt enters the tank at the rate of 3 liters/min

Salt entering per minute is 2* 3=6 grams/min

Volume of liquid leaving the tank is the same as the volume of liquid of tank entering, 3 liters/min

volume of liquid remains at 20 liters at all times

At any given points of time, the concentration of salt is $\frac{x(t)}{20}$ grams/liter

Amount of liquid leaving per minute is 3 liters/min so that the amount of salt leaving is $\frac{x(t)}{20}* 3=\frac{3x(t)}{20}$ grams/minute

Differential equation governing the salt amount in the tank is

$\frac{dx(t)}{dt}=6-\frac{3x(t)}{20}$

Therefore,  $\frac{dx(t)}{dt}+\frac{3x(t)}{20}=6$  

Integrating factor is  $\exp\left(\frac{3t}{20} \right )$ and so the equation becomes

$\frac{d}{dt}\left[\exp\left(\frac{3t}{20} \right )x(t) \right ]=6\exp\left(\frac{3t}{20} \right )$

Therefore,  $\left[\exp\left(\frac{3t}{20} \right )x(t) \right ]=\int 6\exp\left(\frac{3t}{20} \right )=40\exp\left(\frac{3t}{20} \right )+C$

 $x(t)=40+C\exp\left ( -\frac{3t}{20} \right )$

Using the initial condition  $x(0)=5\Rightarrow C=-35$

 $x(t)=40-35\exp\left ( -\frac{3t}{20} \right )$ is the amount of salt at any point of time

$x(t)=40-35\exp\left ( -\frac{3t}{20} \right )=20\Rightarrow 35\exp\left ( -\frac{3t}{20} \right )=20$

$\exp\left ( -\frac{3t}{20} \right )=\frac{20}{35}\Rightarrow -\frac{3t}{20}=\ln\left(\frac{20}{35}\right ) \approx -0.559616$

$t \approx 0.559616\times \frac{20}{3}\approx 3.730733$

After approximately 3.731 minutes, we have 20 grams of salt in the tank