Answer:
1.99 pounds per gallon of salt in t=2 in the tank.
Step-by-step explanation:
First we consider the matter balance equation that contemplates the input and output; the generation and consumption equal to the acomulation.
[tex] Acomulation = Input - Output + Generation - Consumption [/tex]
In this case we have no Generation neither do Consumption so, if we consider Acomulation = A(t), the rate of change of A(t) in time is given by:
[tex]\frac{dA}{dt}+R_{out}A(t)= CR_{s}[/tex] ---(1)
where C is th concentration, with the initial value statement that A(t=0) = 0 because there is no salt in the time cero in the tank, only water.
Given the integral factor -> [tex]u(t)= exp[R_{out}] [/tex] and multipying the entire (1) by it, we have:
[tex]\int \frac{d}{dt}[A(t) \exp[R_{out}t]] \, dt = CR_{in} \int \exp[R_{out}t] \, dt[/tex]
Solving this integrals we obtain:
[tex]A(t)=\frac{CR_{in}}{R_{out}}+Cte*\exp[-R_{out}t][/tex]
So given the initial value condition A(t=0)=0 we have:
[tex]Cte=- \frac{CR_{in}}{R_{out}}[/tex],
and the solution is,
[tex]A(t)=\frac{CR_{in}}{R_{out}}-\frac{CR_{in}}{R_{out}}\exp[-R_{out}t][/tex].
If we give the actual values we obtain then,
[tex]A(t)=2\frac{pounds}{gallon}-2\frac{pounds}{gallon} \exp[-3t][/tex].
So in t= 2 we have [tex]A(t)=2\frac{pounds}{gallon}[/tex].
