Abrine solution of salt flows at a constant rate of 7 l/min into a large tank that initially held 100 l of brine solution in which was dissolved 0.1 kg of salt. the solution inside the tank is kept well stirred and flows out of the tank at the same rate. if the concentration of salt in the brine entering the tank is 0.02 kg/l, determine the mass of salt in the tank after t min. when will the concentration of salt in the tank reach 0.01 kg/l? determine the mass of salt in the tank after t min.

Explanation:

Let x(t) represent the amount of salt (in kg) at any time, t (in minutes).

Step 1

Write down the input rate and output rate in terms of x (showing how you get units). What is the differential equation for x (simplified)?

Input rate: (7L/min) x (0.02kg/L) = 0.14  = 7/50

Output rate: (7L/min) (x/100 kg/L) =   7x / 100

Differential Equation: δx/δt =

=> 7/50 - 7x/100

step 2

Find the general solution for your differential equation

DE is linear with μ(t) = e^(7t/100), so solving we have

x = x(t) = e^(- 7t/100) ∫7/50 e^(- 7t/100) dt

  = 50 + C e^(- 7t/100)

The initial condition mass of salt given is 0.1kg, using this we find the constant of integration

so, when t = 0, x = 0.1, therefore

0.1 = 50 + C

C = - 49.9

Hence x(t) = 50 - 49.9e^(- 7t/100)

When will the concentration of salt in the tank reach 0.01 kg/L?

concentration = 0.01kg/L

This is the same as

x/100 = 0.01

=> x = 1

Substituting x = 1 into the solution yields

1 = 50 - 49.9e^(- 7t/100)

t = 9.2486

Therefore the concentration of salt in the tank reach 0.01kg /L after 9.2486 minutes.

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answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.