A Math Problem About Pools

Question

A pool is 32.3 m long and 23 m wide. If the average depth of water is 7.5 ft, what is the mass of water (in kg) in the pool?

Answer 1

Volume = length x width x depth. We have length and width in meters, so convert the depth to meters.

7.5 ft x 0.3048 meters/ft = 2.286 meters

Volume = 32.3 m x 23 m x 2.286 m = 1698 m³

Now we can convert m³ to liters: 1698 m³ x 1x10³ L/m³ = 1698x10³ L = 1.698x10⁶ L

Given a density of water of 1.0 g/ml = 1.0 kg/L, we have a mass of water equal to 1.7x10⁶ kg

This would be the answer to 2 significant figures since the pool measurements are limited to two sig. figs. by the value of the depth of 7.5 ft (2 s.f.)

Answer 2

This problem requires two pieces of knowledge for unit conversion. The first is the conversion from feet to meters and the conversion from cubic cm to grams. A foot is 0.3048 meters or 30.48 cm. A cubic centimetr of water has the mass of 1 gram. One needs to convert each dimension to centimeters before finding the pool volume. 32.3 meters is 3230 cm and 23 meters is 2300 cm. Multiply 3230 times 2300 times 7 times 30.48 to find the volume in cubic centimeters or 1585051440 cubic centimeters. Divide this amount be 1000 to have kilograms or 1585051.440 kg. But the answer can not have more significant figures that the data with the least. Seven feet has only one significant figure thus the answer can only have one figure. The answer must be in scientific notation to show this degree of accuracy. Thus the answer is 2 x 10^6 kg because 1.585051 rounds up to 2.