Answer:
Salt Concentration = 40g/L
Explanation:
let v(t) is the volume of tank which is written as v(t) = 200 + 40t, considering v(t) ≤ 1000
dy/dt = Rate in - Rate out
y'(t) = 50 x 80 - 40y(t)/v(t)
y'(t) = 4000 - 40y(t)/40t + 500
y'(t + 12.5) + y(t) = 4000(t + 12.5)
∫d/dt . [t + 12.5] . y(t) . dt = ∫4000(t + 12.5) . dt
(t + 12.5) . y(t) = 2000(t + 12.5) . dt
(t + 12.5) . y(t) = 2000(t + 12.5)² + c
y(t) = 2000(t + 12.5) + c/(t + 12.5)
at y(0) = 5000, as tank has 500L x 10g/L = 5000
c = (5000)(12.5) - 2000(12.5)(12.5) = 62500 - 312500 = -250000
y(t) = 2000(t + 12.5) - 250000/(t + 12.5)
v(t) = 500 + 40t, at tank full v(t) = 1000, solving for t, we get t = (1000 - 500)/40 = 12.5
Concentration at t = 12.5 is y(t)/v(t)
Concentration = [{2000(12.5 + 12.5)² - 250000}/12.5 + 12.5]/1000
Concentation (12.5) = 40g/L
