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A typical backyard swimming pool holds 156 yd\(^3\) of water. Using the density of water at 25°C, 0.999 g/cm\(^3\), calculate the mass in pounds of the water.

Answer :

Final answer:

The mass of water in a typical backyard swimming pool with volume 156 yd^3, given the water density at 25°C is 0.999 g/cm^3, would be around 26,275,410.025 pounds.

Explanation:

The first step in solving this problem is to convert any units we given into standard International System (SI) units. Our volume is in yard cubed (yd^3) and needs to be converted to cm^3 (since our density is given in g/cm^3), and our final answer is required to be in pounds.

First we convert the volume from yd^3 to cm^3. 1 yard equates 91.44 centimeters, hence 1 yard cubed equates 91.44^3, or 764,554.857984 cm^3. So, our volume of water is 156 yd^3 * 764,554.857984 = 11,931,057,406.704 cm^3.

We know that the density of water is the mass of water divided by its volume, so the mass equals density multiplied by volume. Therefore, the water mass is 0.999 g/cm^3 * 11,931,057,406.704 cm^3 = 11,919,137,349.293 g or 11,919,137.349 kg (dividing by 1,000 to convert g to kg).

Finally, to obtain the mass in pounds, we need to multiply the mass in kilograms by the conversion factor 2.20462 (since 1 kg = 2.20462 lbs). So, the mass of water in pounds is 11,919,137.349 kg * 2.20462 = 26,275,410.025 lbs.

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