Question

A tank initially contains 200 gallons of fresh water but then a salt solution of unknown concentration is poured into the tank at 2 gal/min. The well stirred mixture flows out of the tank at the same rate. After 120 min, the concentration of salt in the tank is 1.4 lb/gal. What is the concentration of the entering brine?

Answer 1

Let \(C(t)\) be the concentration of salt in the tank at time \(t\), then
\[\begin{align*}
\frac{dC}{dt}&=\text{(rate in)}-\text{(rate out)}\\\\
&=\left(2\frac{\text{gal}}{\text{min}}\right)\left(x\frac{\text{lb}}{\text{gal}}\right)-\left(2\frac{\text{gal}}{\text{min}}\right)\left(\frac{C(t)}{200}\frac{\text{lb}}{\text{gal}}\right)\\\\
&=2x-\frac{1}{100}C(t)
\end{align*}\]
We're given that \(C(0)=0\) and \(C(120)=1.4\).

You can solve the differential equation (hint: it's linear), then use the initial (and intermediate/final?) condition(s) to solve for \(x\).

Answer 2

**Treat \(x\) as a constant**

Answer 3

Is C the total weight of salt in pounds at time t instead of the concentration of salt in pounds per gallon at time t ?

If so, C(0) = 0, C(120) = 1.4 * 200

Answer 4

Yeah @aum you're right, I misspoke, I meant amount. Thanks.

Answer 5

You are welcome.