Answer:
Step-by-step explanation:
Given that a tank contains 300 liters of fluid in which 20 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate
Water at any time t = 300 L +incoming rate - outgoing rate
= 300L (since both are the same)
Quantity of salt at any time t is A(t)
A(0) = 20 gm (initial salt level)
Rate of change of salt level = incoming rate - outgoing rate
Incoming rate =0
Outgoing rate is Q(t) *5/200 at time t (assuming mixed well)
So [tex]A'(t) = 0-\frac{-5A(t)}{300}[/tex]
We separate the variables and solve
[tex]ln A(t) = -t/60 +C\\A(t) = Be^{-t/60}[/tex]
Use the fact A(0) = 20
We get
[tex]A(t) = 20e^{-t/60}[/tex]
