Answer
$3.52$ g/L
Work Step by Step
To find the amount of the chemical in the container, we obtain:
$$\frac{dA}{dt}=c_1r_1$$
We are given:
$c_1=5$
$r_1=6$
$r_2=3$
$t=0$
$A=1500$
and $V(t)=(r_1-r_2)t+V_0=600+3t$
Putting the values:
$$\frac{dA}{dt}=5\times6-3\times\frac{A}{600+3t}$$
$$\frac{dA}{dt}+\times\frac{3A}{600+3t}=30$$
$$\frac{dA}{dt}+\times\frac{A}{200+t}=30$$
Solve for $A$:
$$A=e^{-\int \frac{1}{200+t}}dt(c+\int e^{\int \frac{1}{200+t }dt}30dt)$$
where $c$ is a constant of integration.
$$A=e^{-\ln (200+t)}dt(c+\int 30e^{\int \ln (200+t)}dt)$$
$$A=\frac{1}{200+t}(c+30(200t+\frac{t^2}{2}))$$
Find $c$:
$1500=\frac{c}{200} \rightarrow c=300000$
Find $A$:
$$A=\frac{300000}{200+t}+\frac{30(200t+\frac{t^2}{2})}{200+t}$$
After one hour (60 minutes), the amount of salt is:
$$A(60)=\frac{300000}{200+t}+\frac{30(200\times60+\frac{60^2}{2})}{200+60}=\frac{35700}{13}$$
The concentration of chemical in the tank after one hour is:
$$c_2=\frac{A}{V}=\frac{\frac{35700}{13}}{600+3\times60}=3.52$$ g/L
