* From your hint, I arrived at an answer of 3.46 x 10^4 but the answer key says 3.11 x 10^5 Ib.
A backyard swimming pool holds 185 cubic yards (yd^3) of water. What is the mass of the water in pounds?
DrBob222:
185 yd^3 x (3 ft/yd) x (3 ft/yd) x *3 ft/yd) = ? cubic feet.
Density H2O is 62.3 lbs/ft^3 @ 70 F.
mass = volume x density
mass in lbs = volume in ft^3 x density (lbs/ft^3)
The problem with your problem is that it doesn't list a temperature so I had to assume a T of 70 F. The density of water changes with T. The density of H2O at 70 F is 62.31.
Here is where I obtained the density of 62.31.
http://www.engineeringtoolbox.com/water-specific-volume-weight-d_661.html
You must have punched the wrong numbers into your calculator.
185 x 27 x 62.3 = 3.11E5.
To find the mass of the water in pounds, you need to convert the volume of the water in cubic yards to cubic feet, and then multiply it by the density of water.
1. Start by using the conversion factor for yards to feet. Since there are 3 feet in 1 yard, you need to multiply the volume of the water (185 cubic yards) by the conversion factor (3 feet/yard):
185 yd^3 x (3 ft/yd) = 555 ft^3 (cubic feet)
2. According to the given information, the density of water is 62.3 lbs/ft^3 at 70°F.
3. Now, apply the formula for mass (in pounds) as mass = volume x density:
mass in lbs = volume in ft^3 x density (lbs/ft^3)
mass in lbs = 555 ft^3 x 62.3 lbs/ft^3
4. Calculate the multiplication:
mass in lbs = 34,591.5 lbs
So, the mass of the water in pounds is approximately 34,591.5 lbs.
If you arrived at an answer of 3.46 x 10^4 lbs, it is possible that you made an error during the calculation or rounding.