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Chapter 1: Problem 75

In the United States, volume of irrigation water is usually expressed in acre- feet. One acre-foot is a volume of water sufficient to cover 1 acre of land to a depth of 1 ft \(\left(640 \text { acres }=1 \mathrm{mi}^{2} ; 1 \mathrm{mi}=5280 \mathrm{ft}\right)\) The principal lake in the California Water Project is Lake Oroville, whose water storage capacity is listed as \(3.54 \times 10^{6}\) acre-feet. Express the volume of Lake Oroville in (a) cubic feet; (b) cubic meters; (c) U.S. gallons.

Short Answer

Expert verified
The volume of Lake Oroville is approximately \( 1.54 \times 10^{11} \) cubic feet, \( 4.36 \times 10^{9} \) cubic meters, and \( 1.15 \times 10^{12} \) U.S. gallons.

Step by step solution

01

Determine the conversion factor for 1 acre-foot to cubic feet

Use the given information to calculate how many cubic feet are in an acre-foot. Since 1 acre is \( \frac{1}{640} \) of a square mile and 1 square mile is \( 5280^2 \) square feet, an area of 1 acre is equivalent to \( \frac{1}{640} \times \(5280^2) \) square feet, which equals to 43560 square feet. Since 1 acre-foot is a volume covering this area to a depth of 1 foot, therefore, 1 acre-foot corresponds to a volume of 43560 cubic feet.
02

Convert Lake Oroville volume to cubic feet

Use the conversion factor obtained in step 1 to convert the volume of Lake Oroville from acre-feet to cubic feet. Since Lake Oroville has a volume of \( 3.54 \times 10^6 \) acre-feet, its volume in cubic feet is \( 3.54 \times 10^6 \times 43560 \) cubic feet.
03

Convert the volume to cubic meters

To convert cubic feet to cubic meters, use the conversion factor of 1 cubic meter equals 35.31 cubic feet. Therefore, the volume of Lake Oroville in cubic meters is \( \frac{3.54 \times 10^6 \times 43560}{35.31} \) cubic meters.
04

Convert the volume to U.S. gallons

To convert cubic feet to U.S. gallons, use the conversion factor of 1 cubic foot equals 7.48 U.S. gallons. Therefore, the volume of Lake Oroville in U.S. gallons is \( 3.54 \times 10^6 \times 43560 \times 7.48 \) U.S. gallons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acre-Foot
In the realm of measuring water, especially for agricultural purposes, the term "acre-foot" often comes into play. An acre-foot represents the volume of water required to cover one acre of land to a depth of one foot.
This unit is especially practical because it ties directly to the land's area being irrigated, simplifying calculations for farmers and engineers.
  • An acre is defined as 43,560 square feet.
  • Thus, 1 acre-foot translates to 43,560 cubic feet of water.
This basic definition is vital when considering large bodies of water, such as reservoirs or lakes, where volumes can be expressed in millions of acre-feet.
This makes it easier to grasp the scale of water storage or usage.
Cubic Feet
Cubic feet are another common measure of volume in various contexts, from domestic water use to large-scale engineering projects. A cubic foot is defined as a volume with the dimensions of one foot by one foot by one foot.
To understand this unit better, consider that:
  • 1 cubic foot equals approximately 7.48 U.S. gallons.
  • This measure is three-dimensional, accounting for length, width, and height.
When converting from acre-feet to cubic feet, as in the case of Lake Oroville, the total volume in cubic feet is calculated by multiplying the number of acre-feet by 43,560.
Cubic feet provide a more granular measure than acre-feet, useful for detailed engineering calculations.
Cubic Meters
Cubic meters serve as the international standard unit for volume in the metric system.
This measure is part of the System International (SI), ensuring uniformity in scientific and engineering calculations worldwide.
  • 1 cubic meter is equivalent to 1,000 liters or 1,000,000 cubic centimeters.
  • In terms of conversion, 1 cubic meter equals approximately 35.31 cubic feet.
      • For projects that cross country boundaries or involve international partners, cubic meters are crucial as they enable clear communication and understanding.
      This is why, during the conversion of Lake Oroville's volume, converting to cubic meters aligned with global standards, making the information versatile across different systems.
U.S. Gallons
The U.S. gallon is a traditional volume measure often used in the United States to indicate the capacity of liquid containers, such as those for fuel or water.
Unlike the imperial gallon used in some other countries, the U.S. gallon is defined as exactly 231 cubic inches.
  • 1 cubic foot equals 7.48 U.S. gallons.
  • This conversion rate is helpful when converting from cubic feet, as many real-world applications involve understanding volumes in gallons.
In practical contexts, knowing the volume of water in gallons provides a tangible sense of scale.
For instance, understanding that Lake Oroville holds over tens of billions of gallons brings home the massive scale of this reservoir in terms ordinary people can understand.

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Most popular questions from this chapter

A class of 84 students had a final grade distribution of 18\(\%\) A's, 25\(\%\) B's, 32\(\%\) C's, 13\(\%\) D's, 12\(\%\) F's. How many students received each grade?

A tabulation of data lists the following equation for calculating the densities \((d)\) of solutions of naphthalene in benzene at \(30^{\circ} \mathrm{C}\) as a function of the mass percent of naphthalene. $$d\left(\mathrm{g} / \mathrm{cm}^{3}\right)=\frac{1}{1.153-1.82 \times 10^{-3}(\% \mathrm{N})+1.08 \times 10^{-6}(\% \mathrm{N})^{2}}$$ Use the equation above to calculate (a) the density of pure benzene at \(30^{\circ} \mathrm{C} ;\) (b) the density of pure naphthalene at \(30^{\circ} \mathrm{C} ;\) (c) the density of solution at \(30^{\circ} \mathrm{C}\) that is 1.15\% naphthalene; (d) the mass percent of naphthalene in a solution that has a density of \(0.952 \mathrm{g} / \mathrm{cm}^{3}\) at \(30^{\circ} \mathrm{C} .[\text { Hint: For }(\mathrm{d}),\) you need to use the quadratic formula. See Section A-3 of Appendix A.]

A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at \(25^{\circ} \mathrm{C}\), and the density of water at \(25^{\circ} \mathrm{C}\) is \(0.9970 \mathrm{g} / \mathrm{mL}\) ). At the left, grains of the mineral calcite float on the surface of the liquid bromoform \((d=2.890 \mathrm{g} / \mathrm{mL})\) At the right, the grains sink to the bottom of liquid chloroform \((d=1.444 \mathrm{g} / \mathrm{mL}) .\) By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined (Exercise 62).

The highest temperature of the following group is (a) \(217 \mathrm{K} ;\) (b) \(273 \mathrm{K} ;\) (c) \(217^{\circ} \mathrm{F} ;\) (d) \(105^{\circ} \mathrm{C} ;\) (e) \(373 \mathrm{K}\).

A lump of pure copper weighs \(25.305 \mathrm{g}\) in air and 22.486 g when submerged in water \((d=0.9982 \mathrm{g} / \mathrm{mL})\) at \(20.0^{\circ} \mathrm{C} .\) Suppose the copper is then rolled into a \(248 \mathrm{cm}^{2}\) foil of uniform thickness. What will this thickness be, in millimeters?
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