Question

A recreational swimming lake is treated periodically to control harmful bacteria growth. Suppose that t days after a treatment, the concentration of bacteria per cubic centimeter is given by C(t) = 29.4ᵗ - 244 +500, where 0

Answer 1

To achieve a concentration of 3.0 ppm in a 5,000-gallon pool, 56.78 grams of chlorine must be added after converting the pool's volume to liters and then using the desired ppm concentration to find the mass of chlorine required.

Step-by-step explanation:

Calculating the Mass of Chlorine for Pool Disinfection

The question concerns the calculation of the mass of chlorine to be added to a swimming pool to reach a disinfectant concentration level of 3.0 ppm. To find the required mass of chlorine, we can use the formula:

Mass of Chlorine (g) = Volume of Water (L) × Desired Concentration (ppm)

First, we convert the pool's volume from gallons to liters, knowing that 1 gallon is approximately 3.78541 liters. For a 5,000-gallon pool, this is:

5,000 gal × 3.78541 L/gal = 18,927.05 L

With the volume in liters, we can now find the required mass of chlorine:

18,927.05 L × 3.0 ppm = 56,781.15 mg

Converting milligrams to grams, we have:

56,781.15 mg ÷ 1,000 = 56.78 g

To achieve a concentration of 3.0 ppm of chlorine, 56.78 grams of chlorine must be added to the pool.