Answer:
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.
