Question

A tank initially contains 900 liters of pure water. Then water containing 6 mg of detergent per liter starts to enter the tank at the rate of 20 liters per hour. (a) How long until the average concentration of detergent in the tank is 2 mg per liter?

Answer 1

A tank initially contains 1000 liters of pure water. Then water containing 4 mg of detergent per liter starts to enter the tank at the rate of 30 liters per hour.  

Let; / ( if; )

Volume at any time t = V(t)

Concentration at any time t = C(t)

Amount at any time = Q(t)  

(then;)

C(t) = Q(t) / V(t)

dV/dt = 30

V(t) = 30t + 1000

dQ/dt =  (4 )(30)  

dQ = 120 dt

Q = 120t

(a) How long until the average concentration of detergent in the tank is 2 mg per liter?  

____hours  

C(t) = Q(t) / V(t)

C(t) = (120t) / ( 30t + 1000)

when C(t) = 2 mg/hr

(120t) / ( 30t + 1000) = 2

60t + 2000 = 120t

t = 100/3 hrs

t = 33 ⅓ hours

——————- ( 1 day 9 hours 20 minutes )

(b) How long until the average concentration of detergent in the tank is x mg per liter?  

(120t) / ( 30t + 1000) = x

30xt + 1000x = 120t

3xt + 100x = 12t

12t - 3xt = 100x

t = 100x/(12 - 3x)