Answer and Explanation:
The rate of change of the volume of the brine is given by:
\(\Delta V = (\text{rate in}) - (\text{rate out})\)
\(\Delta V = (4 \text{ L/min}) - (3\text{ L/min})\)
\(\Delta V = +1\text{ L/m}\)
Since \(\Delta V\) is the same thing as \(V'(t)\), which is the derivative of \(V(t)\), we can integrate \(\Delta V\) to get \(V(t)\) (the total volume in the tank), with the constant of integration being the initial volume:
\(\displaystyle \int \Delta V\ dt = \int (1\text{ L/min}) \ dt\)
\(V(t) = \left(1{\rm\frac{L}{m}}\right) t + C\)
\(V(t) = t + 100\)
We can model the volume of brine by taking out the initial volume:
\(B(t) = t\)
The mass of brine there will be is:
\(\boxed{B(1) \cdot \dfrac{0.5\text{ kg}}{L} = 1 \cdot 0.5 = 0.5\text{ kg}}\)
We can solve for when the concentration of the salt is 0.1 kg/L by relating our equations and the definition of concentration (mass per volume):
\(M = \dfrac{B(t) \cdot 0.5\text{ kg/L}}{V(t)}\)
\(0.1 = \dfrac{t}{t + 100}\)
\(0.1(t + 100) = t\)
\(0.1t + 10 = t\)
\(0.9t = 10\)
\(\boxed{t \approx 11.11\text{ min}}\)
