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A backyard swimming pool holds 185 cubic yards of water. What is the mass of water in pounds? Assume the density of water is 1.00 g/mL. (1 yard = 3 feet, 1 foot = 12 inches, 1 inch = 2.54 cm, 1 lb = 453.59 g).

A) 1,051,456 pounds
B) 2,301 pounds
C) 8,230 pounds
D) 7,634 pounds

1 Answer

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Final answer:

To find the mass of water in pounds for a pool holding 185 cubic yards, multiple conversions between cubic yards, cubic feet, cubic inches, cubic centimeters, grams, and pounds are performed. Ultimately, approximately 934,572 pounds of water are contained in the pool given its volume and the density of water.

Step-by-step explanation:

The question concerns the conversion of units to determine the mass of water in pounds contained in a swimming pool of a known volume in cubic yards, assuming the density of water is 1.00 g/mL. To calculate this, several unit conversions are necessary: cubic yards to cubic centimeters to determine the volume, and grams to pounds to find the mass.

First, convert cubic yards to cubic feet: 185 cubic yards × (3 feet/1 yard)^3 = 14,985 cubic feet. Then, convert cubic feet to cubic inches: 14,985 cubic feet × (12 inches/1 foot)^3 = 25,848,240 cubic inches. Next, convert cubic inches to cubic centimeters: 25,848,240 cubic inches × (2.54 cm/1 inch)^3 = 423,776,000 cubic centimeters.

Since the density of water is given as 1 g/cm^3, the mass of the water in grams is equal to the volume in cubic centimeters: 423,776,000 g. Finally, convert grams to pounds: 423,776,000 g ÷ 453.59 g/lb = approximately 934,572 pounds.

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