Answer
300g
Work Step by Step
To find the amount of the chemical in the tank, we obtain:
$$\frac{dA}{dt}=c_1r_1-c_2r_2$$
We are given:
$c_1=10$
$r_1=4$
$r_2=2$
and $V(t)=(r_1-r_2)t+V_0=20+2t$
Since $V=40 \rightarrow 40=20+2t \rightarrow t=10$
Putting the values:
$$\frac{dA}{dt}=10\times4-2\frac{A}{V}$$
$$\frac{dA}{dt}+\frac{A}{10+t}=40$$
Solve for $A$:
$$A=e^{-\int \frac{1}{10+t}}dt(c+\int 40e^{\int \frac{-1}{10+t }dt}dt)$$
where $c$ is a constant of integration.
$$A=e^{-\ln (10+t)}dt(c+\int 40e^{\int \ln (10+t)}dt)$$
$$A=\frac{1}{(10+t)^2}(c+\int40(10+t)^2dt)$$
$$A=\frac{c}{(10+t)^2}+20(10+t)$$
Since $A(0)=0 $
Find $c$:
$0=\frac{c}{10} +200\rightarrow c=-2000$
The amount of salt in the tank at any time:
$$A=\frac{-2000}{10+t}+20(10+t)$$
When $t=10$:
$$A=\frac{-2000}{10+10}+20(50+10)=\frac{-2000}{20}+20\times60=300$$g
