Solution in a container

Question

A container initially contains 10L of salt solution. Water flows in at a rate of 3L/min, and the well-stirred mixture flows out at a rate of 3L/min. After 5 minutes the concentration of salt in the container is 0.2g/L. Find: (i) the amount of salt initially in the container (ii) the time when the concentration in the container is 0.1 g/L

Answer 1

So first, let's define variables and come up with an equation for the concentration of the system:

time in min: t

concentration in g/L: C(t)

total salt in g: T(t)

rate of change of total salt in g/min: T'(t)=(-3L/min)*C(t) (flow rate out times the concentration)

total liquid: 10L (constant because flow out=flow in)

equation: C(t)=total salt/total liquid=T(t)/10L

We don't know T(t), but we do know its rate of change, T'(t), so we need to take the derivative of our equation to get a differential equation which we can solve:

C'(t)=d/dt(C(t))=T'(t)/10L=(-3L/min)*C(t)/10L

Simplifying and rewriting this, and leaving off units for convenience, we can see that it is a separable differential equation:

dC/dt=-0.3C

dC/C=-0.3dt

Integral(dC/C) from 0 to t=Integral(-0.3dt) from 0 to t

ln|C| from 0 to t=-0.3t from 0 to t

ln|C(t)|-ln|C(0)|=-0.3t-0.3*0

ln|C(t)/C(0)|=-0.3t

C(t)/C(0)=e^(-0.3t)

C(t)=C(0)*e^(-0.3t)

Now we can use the point given, 0.2g/L at 5min, to find C(0):

C(5)=0.2=C(0)*e^(-0.3*5) --> C(0)=0.2e^1.5≈0.896 (g/L) (The answer to part i!)

Next, we plug in the concentration given, 0.1g/L, and the C(0) value we found in part i, to find the time at which it occurs:

0.1=C(t)=C(0)*e^(-0.3t)=(0.2e^1.5)*e^(-0.3t)=0.2*e^(1.5-0.3t)

--> 0.5=e^(1.5-0.3t)

1.5-0.3t=ln(0.5)

0.3t=1.5-ln(0.5)

t=(1.5-ln(0.5))/0.3≈7.310 (min) (The answer to part ii!)

Answer 2

Hi Christopher, I can get you started. Keep track of your units as you go through and make sure they match up. Use A(t) to be the grams of salt in the container at time t min. Now write a differential equation for

dA/dt which equals the rate of salt coming in - the rate of salt going out (g/min)

The rate in of salt is 0 since it is water going in. The rate out will be found using (L/min)(g/L). The L/min is given in the problem. See if you can figure out the g/L at time t. Multiply them together. This should give you a separable differential equation to solve. You will end up with a + C. This is when you will use A(5) and plug in. Remember A(5) is in grams so you use the 0.2 g/L at time 5 and the constant volume 10 L to find it. After you find C, you will have a formula for A in terms of t and can use it to answer both question using algebra. Tom S.