Question

A tank with a capacity of 600 L is full of a mixture of water and chlorine with a concentration of 0.03 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 6 L/s. The mixture is kept stirred and is pumped out at a rate of 11 L/s. Find the amount of chlorine in the tank as a function of time.

Answer 1

i keep getting 18e^(-11t/600), but its wrong

Answer 2

if it is pumped out at the rate of 11L/s and pumped in at 6L/s
so net 5L/pumped out
so amount of chlorine is 18-0.15*t

Answer 3

ian important note is we are looking for the amount of chlorine, not the concentration. Also since more water is leaving than coming in the water level is decreasing at 5L/s so at t=120 the tank will be empty hence chlorine should be 0.
Reason it is not a linear function is because of difference in the concentration of water coming in and water leaving.
Function of water level with respect to t:
W(t) = 600-5t
chlorine = concentration * W(t)
so we need to find concentration in terms of t
C(t) = (initial chlorine - net change in chlorine)/W(t)
initial chlorine = .03*600 = 18
net chlorine is difference in whats going in and whats going out (Out-In)
IN: concentration is 0 and amount of water is 6t
In = 0
OUT: concentration is C(t) and water is 11t
Out = 11t*C(t)
Net chlorine = 11t*C(t) - 0
C(t) = (18 -11t*C(t))/W(t)
solve for C(t)
C(t) = 18/(600+6t)
Now to get function of chlorine multiply by W(t)
\[chlorine(t) = \frac{18(600-5t)}{600+6t}\]

Answer 4

what's wronge with my method

Answer 5

its assuming the concentration of chlorine stays the same
If they were just letting out 5L of water per sec without pumping in fresh water then you would be correct

Answer 6

oh........