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 [tex]\frac{x(t)}{20}[/tex] grams/liter
Amount of liquid leaving per minute is 3 liters/min so that the amount of salt leaving is [tex]\frac{x(t)}{20}* 3=\frac{3x(t)}{20}[/tex] grams/minute
Differential equation governing the salt amount in the tank is
[tex]\frac{dx(t)}{dt}=6-\frac{3x(t)}{20}[/tex]
Therefore, [tex]\frac{dx(t)}{dt}+\frac{3x(t)}{20}=6[/tex]
Integrating factor is [tex]\exp\left(\frac{3t}{20} \right )[/tex] and so the equation becomes
[tex]\frac{d}{dt}\left[\exp\left(\frac{3t}{20} \right )x(t) \right ]=6\exp\left(\frac{3t}{20} \right )[/tex]
Therefore, [tex] \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[/tex]
[tex]x(t)=40+C\exp\left ( -\frac{3t}{20} \right )[/tex]
Using the initial condition [tex]x(0)=5\Rightarrow C=-35[/tex]
[tex]x(t)=40-35\exp\left ( -\frac{3t}{20} \right )[/tex] is the amount of salt at any point of time
[tex]x(t)=40-35\exp\left ( -\frac{3t}{20} \right )=20\Rightarrow 35\exp\left ( -\frac{3t}{20} \right )=20[/tex]
[tex]\exp\left ( -\frac{3t}{20} \right )=\frac{20}{35}\Rightarrow -\frac{3t}{20}=\ln\left(\frac{20}{35}\right ) \approx -0.559616[/tex]
[tex]t \approx 0.559616\times \frac{20}{3}\approx 3.730733[/tex]
After approximately 3.731 minutes, we have 20 grams of salt in the tank
