Question

A tank contains 240 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Answer 1

`\boxed{240 - 230e^{-(t)/(40)}}`

Step-by-step explanation:

`\text{Let A = mass of salt after t min}\\\text{and }r_(i) = \text{rate of salt coming into talk}\\\text{and }r_(o)$ =\text{rate of salt going out of tank}`

1. Set up an expression for the rate of change of salt concentration.

`\frac{\text{d}A}{\text{d}t} = r_(i) - r_(o)\\\\r_(i) = \frac{\text{6 L}}{\text{1 min}} * \frac{\text{1 g}}{\text{1 L}} = \text{6 g/min}\\\\r_(o) = \frac{\text{6 L}}{\text{1 min}} * \frac {A\text{ g}}{\text{240 L}} =(x)/(40)\text{ g/min}\\\\\frac{\text{d}A}{\text{d}t} = 6 - (x)/(40)`

2. Integrate the expression

`\frac{\text{d}A}{\text{d}t} = (240 - x)/(40)\\\\\frac{\text{d}A}{240 - A} = \frac{\text{d}t}{40}\\\\\int \frac{\text{d}A}{240 - A} = \int \frac{\text{d}t}{40}\\\\-\ln |240 - A| = (t)/(40) + C`

3. Find the constant of integration

`-\ln |240 - A| = (t)/(40) + C\\\\\text{At $t$ = 0, $A$ = 10, so}\\\\-\ln |240 - 10| = (0)/(40) + C\\\\C = -\ln 230`

4. Solve for A as a function of time.

`\text{The integrated rate expression is}-\ln |240 - A| = (t)/(40) - \ln 230\\\\\text{Solve for } A\\\\\ln|240 - A| = \ln 230 - (t)/(40)\\\\|240 - A| = 230e^{-(t)/(40)}\\\\240 - A = \pm 230e^{-(t)/(40)}\\\\x = 240 \pm 230e^{-(t)/(40)}\\\\A(0) = 10 \text{ so we choose the negative sign}\\\\x = \boxed{\mathbf{240 - 230e^{-(t)/(40)}}}`

The diagram shows A as a function of time. The mass of salt in the tank starts at 10 g and increases asymptotically to 240 g.