Answer:
(a) 0.288 kg/liter
(b) 0.061408 kg/liter
Step-by-step explanation:
(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute
The mass of salt exiting the tank per minute = 5 × (5 + x)/500
The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;
Δ/dt = x - 5 × (5 + x)/500
The increase, in mass, Δ, after an increase in time, dt, is therefore;
Δ = (x - 5 × (5 + x)/500)·dt
Integrating with a graphing calculator, with limits 0, 10, gives;
Δ = (99·x - 5)/10
Substituting x = 1 gives
(99 × 1 - 5)/10 = 9.4 kg
The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)
∴ The concentration of the salt and water in the tank after 10 minutes = (5 + 9.4)/500 = (14.4)/500 = 0.288
The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter
(b) With the added leak, we now have;
Δ/dt = x - 6 × (14.4 + x)/500
Δ = x - 6 × (14.4 + x)/500·dt
Integrating with a graphing calculator, with limits 0, 20, gives;
Δ = 19.76·x -3.456 = 16.304
Where x = 1
The increase in mass after an increase in = 16.304 kg
The total mass = 16.304 + 14.4 = 30.704 kg
The concentration of the salt in the tank then becomes;
Concentration = 30.704/500 = 0.061408 kg/liter.
as stated.</h3>