Question

1) A tank contains 10 kg of salt and 2000 L of water. A solution of concentration 0.025 kg of salt per liter enters a tank at the rate 7 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) Find the amount of salt in the tank after 1.5 hours. amount = ____ (kg) c) Find the concentration of salt in the solution in the tank as time approaches infinity. concentration = ____ (kg/L)

Answer 1

if x= amount of salt then x'=(rate in)-(rate out)

x'(t)= [.0125Kg/L * (7L/min)] - [7L/min * (x(t)Kg/2000L)]

dx/dt = .088 - (7x/2000)

multiply terms by 2000 and by dt:

2000 dx = (175 - 7x) dt

devide by (175 - 7x):

(2000 dx)/(175 - 7x) = dt

integrate both sides:

(-2000 * ln(abs(x - 25)))/7 = t + c

multiply both sides by -7/2000:

ln(abs(x - 25)) = (-7t/2000) + c

take e^ of both sides and add 25 to both sides:

x(t) = ce^(-7t/2000) +25

use initial condition x(0) = 50Kg to solve for c:

50 = ce^0 +25

c=25

x(t) = 25e^(-7t/2000) +25

at 3.5 hrs t is 3.5*60min = 210 min

so x(210) = 25 * e^((-7*210)/2000) +25 = 36.988 Kg of salt



For part B, as time goes to infinity, the salt concentration will equal the concentration that is being added to the tank which is 0.0125 Kg/L