Question

A container initially containing10 L of water in which there is 20 g of salt dissolved. A solution containing 4 g/L of salt is pumped into the container at a rate of 2 L/min, and the well-stilled mixture runs out at a rate of 1 L/min. How much salt is in the tank after 40 min

Answer 1

Answer:

$A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196$

Step-by-step explanation:

For this case the solution flows at a rate of 2L/min and leaves at 1L/min. So then we can conclude the volume is given by $V= 10 +t$

Since the initial volume is 10 L and the volume increase at a rate of 1L/min.

For this case we can define A as the concentration for the salt in the container. And for this case we can set up the following differential equation:

$\frac{dA}{dt}= 4 \frac{gr}{L} *2 \frac{L}{min} - \frac{A}{10+t}$

Because at the begin we have a concentration of 8 gr/L and would be decreasing at a rate of $\frac{A}{10+t}$

So then we can reorder the differential equation like this:

$\frac{dA}{dt} +\frac{A}{10+t} =8$

We find the solution using the integration factor:

$\mu = -\int \frac{1}{10+t} dt = -ln(10+t)$

And then the solution would be given by:

$A = e^{-ln (10+t)} (\int e^{\int \frac{1}{10+t} dt})$

And if we simplify this we got:

$A= \frac{1}{10+t} (c + \int (10 +t) 8 dt)$

And after do the integral we got:

$A= \frac{c}{10+t} +4 (10 +t)$

And using the initial condition t=0 A= 20 we have this:

$20 = \frac{c}{10} +40$

$c= -200$

So then we have this function for the solution of A:

$A= \frac{-200}{10+t} +4 (10 +t)$

And now replacinf t= 40 we got:

$A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196$