I don't know how the liquid can drain and leave the substance behind, if it's dissolved. Maybe it's just in suspension, and the liquid drains through a filter.
Anyway, the volume of water at time t hours is
v = 60-4t
The concentration is
c(t) g/L = 100/v(t) = 100/(60-4t) = 25/(15-t)
dc/dt = 25/(15-t)^2
so, at t=5,
dc/dt = 25/(15-5)^2 = +0.25 g/L-hr
A tank contains 100 grams of a substance dissolved in a large amount of water. The tank is filtered in such a way that water drains from the tank, leaving the substance behind in the tank. Consider the volume of the dissolved substance to be negligible. At what rate is the concentration (grams/liter) of the substance changing with respect to time in each scenario?
(a) the rate after 5 hours, if the tank contains 60 L of water initially, and drains at a constant rate of 4 L/hr?
1 answer
1 answer