College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 1 - Section 1.8 - Approximation - Practice Problem - Page 15: 1.10

Answer

Approximately: $2.4 \times 10^{11} \text{ liters/year}$ $2.4 \times 10^{11} \text{ quarts}/\text{year}$ $2.4 \times 10^{8} \text{ m}^3/\text{year}$

Work Step by Step

The population of the US is approximately $330$ million people $(3.3 \times 10^8)$. An average person drinks about 2 liters of water per day. $\text{Total liters per year per person }= 2 \text{ liters/day }\times \text{days/year}=730\text{ liters/year}$ Multiply by total population: $\text{Water consumed per year }=730\text{ L/person/year } \times 3.3 \times 10^8 \text{ people} \approx 2.4 \times 10^{11} \text{ liters/year}$ Convert to quart: $1\text{ liter} \approx 1\text{ quart}$: $2.4 \times 10^{11} \text{ liters/year} \approx 2.4 \times 10^{11} \text{ quarts}/\text{year}$ Convert to cubic meters: We know that $1\text{ L} = 10^{-3}\text{m}^3$, so: $2.4 \times 10^{11} \text{ liters/year} \times 10^{-3} (\text{m}^{3}/\text{liter})= 2.4 \times 10^{8} \text{ m}^3/\text{year}$
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