Question

A tank has capacity 6 L and initially contains 11 mg of salt dissolved in 3 L of water. A solution containing 1 mg/L of salt enters the tank at a rate of 3 L/hour and the well-stirred mixture leaves the tank at a rate of 2 L/hour. (a) (4) Find the time when the tank is full. (b) (14) Find the amount of salt (in milligrams) in the tank at any time before the tank is full. What is the amount of salt (in mg) in the tank when it is full and what is the concentration of the mixture then (in mg/L)

Answer 1

Answer:

The answer is below

Step-by-step explanation:

a) The maximum capacity of he tank is 6 L and initially it contains 11 mg of salt dissolved in 3 L of water. Solution enters the tank at a rate of 3 L/hr, therefore in x hours, the amount of water that have entered the tank = 3x.

Solution also leaves the tank at a rate of 2L/hr, therefore in x hours, the amount of water that have left the tank = 2x

Hence the amount of water present in the tank at x hours is given as:

3 + 3x - 2x = 3 + x

The time taken to full the tank can be gotten from:

3 + x = 6

x = 6 - 3

x = 3 hr

b)

$\frac{dQ}{dx}=3-\frac{2Q}{3+x}\\ \\\frac{dQ}{dx}+\frac{2Q}{3+x}=3\\\\let\ u'=\frac{2u}{3+x}\\\\\frac{u'}{u}=\frac{2Q}{3+x}\\\\ln(u)=2ln(3+x)\\\\u=(3+x)^2\\\\(3+x)^2Q]'=3(3+x)^2\\\\(3+x)^2Q=(3+x)^3+c\\\\Q(0)=11\\\\(3+0)^2(11)=(3+0)^3+c\\\\x=72\\\\Q=x+3+\frac{72}{(x+3)^2}\\ \\Q(3)=3+3+\frac{72}{(3+3)^2}=8\ mg$

8 mg/ 6 L = 4/3 mg/L