Answer
$196$ gm
Work Step by Step
To find the amount of the chemical in the container, we obtain:
$$\frac{dA}{dt}=c_1r_1-c_2r_2$$
We are given:
$c_1=4$
$r_1=2$
$r_2=1$
$t=0$
$A=20$
and $V(t)=(r_1-r_2)t+V_0=600+3t$
Putting the values:
$$\frac{dA}{dt}=4\times2-\frac{A}{10+t}$$
$$\frac{dA}{dt}+\frac{A}{10+t}=8$$
Solve for $A$:
$$A=e^{-\int \frac{1}{10+t}}dt(c+\int e^{\int \frac{1}{10+t }dt}8dt)$$
where $c$ is a constant of integration.
$$A=e^{-\ln (10+t)}dt(c+\int e^{\int \ln (10+t)}8dt)$$
$$A=\frac{1}{10+t}(c+\int(10+t)8dt)$$
$$A=\frac{c}{10+t}+\frac{1}{10+t}\times \frac{8}{2}(10+t)^2$$
$$A=\frac{c}{10+t}+4(10+t)$$
Find $c$:
$20=\frac{c}{10} +40\rightarrow c=-200$
Find $A$:
$$A=\frac{300000}{200+t}+\frac{30(200t+\frac{t^2}{2})}{200+t}$$
The amount of salt at everytime is:
$$A(60)=80-60e^{-\frac{1}{10}}$$
After 40 minutes, the amount of salt is:
$$A=\frac{-200}{10+40}+4\times50=196$$ gm
