Atank contains 50 kg of salt and 1000 l of water. a solution of a concentration 0.025 kg of salt per liter enters a tank at the rate 9 l/min. the solution is mixed and drains from the tank at the same rate.

(a) what is the concentration of our solution in the tank initially?
concentration = (kg/l)

(b) write down the differential equation which models the amount y of salt in the tank:
dydt=

(c) find the amount of salt in the tank after 2.5 hours.
amount = (kg)

(d) find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg/l)

A. 0.05kg/l

B. dy/dt = 9/1000(25 - y)

C. 20.05 kg of salt

D. 0.0025kg/l

Step-by-step explanation:

A. Concentration of salt in the tank initially,

Concentration (kg/l) = mass of salt in kg/ volume of water in liter

= 50kg/1000l

= 0.05kg/l

B. dy/dt = rate of salt in - rate of salt out

Rate of salt in = 0.025kg/l * 9l/min

= 0.225kg/min

Rate of salt out = 9y/1000

dy/dt = 0.225 - 9y/1000

dy/dt = 9/1000(25 - y)

C. Collecting like terms from the above equation,

dy/25 - y = 9/1000dt

Integrating,

-Ln(25 - y) = 9/1000t + C

Taking the exponential of both sides,

25 - y = Ce^(-9t/1000)

Calculating for c, at y = 0, t = 0;

C = 25

y(t) = 25 - 25e^(-9t/1000)

At 2.5 hours,

2.5 hours * 60 mins = 180 mins

y(180 mins) = 25 - 25e^(-9*180/1000)

= 25 - 25*(0.1979)

= 20.05kg of salt

D. As time approaches infinity, e^(Infinity) = 0,

y(t) = 25 - 25*0

Concentration (kg/l) = 25/1000

= 0.0025kg/l