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College Physics for AP® Courses 2e

11.2 Density

College Physics for AP® Courses 2e11.2 Density

Learning Objectives

By the end of this section, you will be able to:

  • Define density.
  • Calculate the mass of a reservoir from its density.
  • Compare and contrast the densities of various substances.

Which weighs more, a ton of feathers or a ton of bricks? This old riddle plays with the distinction between mass and density. A ton is a ton, of course; but bricks have much greater density than feathers, and so we are tempted to think of them as heavier. (See Figure 11.3.)

Density, as you will see, is an important characteristic of substances. It is crucial, for example, in determining whether an object sinks or floats in a fluid. Density is the mass per unit volume of a substance or object. In equation form, density is defined as

ρ=mV,ρ=mV,
11.1

where the Greek letter ρρ (rho) is the symbol for density, mm is the mass, and VV is the volume occupied by the substance.

Density

Density is mass per unit volume.

ρ=mV,ρ=mV,
11.2

where ρρ is the symbol for density, mm is the mass, and VV is the volume occupied by the substance.

In the riddle regarding the feathers and bricks, the masses are the same, but the volume occupied by the feathers is much greater, since their density is much lower. The SI unit of density is kg/m3kg/m3, representative values are given in Table 11.1. The metric system was originally devised so that water would have a density of 1g/cm31g/cm3, equivalent to 103kg/m3103kg/m3. Thus the basic mass unit, the kilogram, was first devised to be the mass of 1000 mL of water, which has a volume of 1000 cm3.

Substance ρ (× 10 3 kg/m 3 ρ(× 10 3 kg/m 3 or g/mL ) g/mL) Substance ρ ( 10 3 kg/m 3 ρ( 10 3 kg/m 3 or g/mL ) g/mL) Substance ρ ( 10 3 kg/m 3 ρ( 10 3 kg/m 3 or g/mL ) g/mL)
Solids Liquids Gases
Aluminum 2.7 Water (4ºC) 1.000 Air 1 . 29 × 10 − 3 1 . 29 × 10 − 3
Brass 8.44 Blood 1.05 Carbon dioxide 1 . 98 × 10 − 3 1 . 98 × 10 − 3
Copper (average) 8.8 Sea water 1.025 Carbon monoxide 1 . 25 × 10 − 3 1 . 25 × 10 − 3
Gold 19.32 Mercury 13.6 Hydrogen 0 . 090 × 10 − 3 0 . 090 × 10 − 3
Iron or steel 7.8 Ethyl alcohol 0.79 Helium 0 . 18 × 10 − 3 0 . 18 × 10 − 3
Lead 11.3 Gasoline 0.68 Methane 0 . 72 × 10 − 3 0 . 72 × 10 − 3
Polystyrene 0.10 Glycerin 1.26 Nitrogen 1 . 25 × 10 − 3 1 . 25 × 10 − 3
Tungsten 19.30 Olive oil 0.92 Nitrous oxide 1 . 98 × 10 − 3 1 . 98 × 10 − 3
Uranium 18.70 Oxygen 1 . 43 × 10 − 3 1 . 43 × 10 − 3
Concrete 2.30–3.0 Steam 100º C100º C 0 . 60 × 10 − 3 0 . 60 × 10 − 3
Cork 0.24
Glass, common (average) 2.6
Granite 2.7
Earth’s crust 3.3
Wood 0.3–0.9
Ice (0°C) 0.917
Bone 1.7–2.0
Silver 10.49
Table 11.1 Densities of Various Substances
A pile of feathers measuring a ton and a ton of bricks are placed on either side of a plank that is balanced on a small support.
Figure 11.3 A ton of feathers and a ton of bricks have the same mass, but the feathers make a much bigger pile because they have a much lower density.

As you can see by examining Table 11.1, the density of an object may help identify its composition. The density of gold, for example, is about 2.5 times the density of iron, which is about 2.5 times the density of aluminum. Density also reveals something about the phase of the matter and its substructure. Notice that the densities of liquids and solids are roughly comparable, consistent with the fact that their atoms are in close contact. The densities of gases are much less than those of liquids and solids, because the atoms in gases are separated by large amounts of empty space.

Take-Home Experiment Sugar and Salt

A pile of sugar and a pile of salt look pretty similar, but which weighs more? If the volumes of both piles are the same, any difference in mass is due to their different densities (including the air space between crystals). Which do you think has the greater density? What values did you find? What method did you use to determine these values?

Example 11.1

Calculating the Mass of a Reservoir From Its Volume

A reservoir has a surface area of 50.0km250.0km2 and an average depth of 40.0 m. What mass of water is held behind the dam? (See Figure 11.4 for a view of a large reservoir—the Three Gorges Dam site on the Yangtze River in central China.)

Strategy

We can calculate the volume VV of the reservoir from its dimensions, and find the density of water ρρ in Table 11.1. Then the mass mm can be found from the definition of density

ρ=mV.ρ=mV.
11.3

Solution

Solving equation ρ=m/Vρ=m/V for mm gives m = ρ V m = ρ V .

The volume VV of the reservoir is its surface area AA times its average depth hh:

V = Ah = 50.0 km 2 40.0 m = 50.0 k m 2 10 3 m 1 km 2 40.0 m = 2 . 00 × 10 9 m 3 V = Ah = 50.0 km 2 40.0 m = 50.0 k m 2 10 3 m 1 km 2 40.0 m = 2 . 00 × 10 9 m 3
11.4

The density of water ρρ from Table 11.1 is 1.000×103kg/m31.000×103kg/m3. Substituting VV and ρρ into the expression for mass gives

m = 1 . 00 × 10 3 kg/m 3 2 . 00 × 10 9 m 3 = 2.00 × 10 12 kg. m = 1 . 00 × 10 3 kg/m 3 2 . 00 × 10 9 m 3 = 2.00 × 10 12 kg.
11.5

Discussion

A large reservoir contains a very large mass of water. In this example, the weight of the water in the reservoir is mg=1.96×1013Nmg=1.96×1013N, where gg is the acceleration due to the Earth’s gravity (about 9.80m/s29.80m/s2). It is reasonable to ask whether the dam must supply a force equal to this tremendous weight. The answer is no. As we shall see in the following sections, the force the dam must supply can be much smaller than the weight of the water it holds back.

Photograph of the Three Gorges Dam in central China.
Figure 11.4 Three Gorges Dam in central China. When completed in 2008, this became the world’s largest hydroelectric plant, generating power equivalent to that generated by 22 average-sized nuclear power plants. The concrete dam is 181 m high and 2.3 km across. The reservoir made by this dam is 660 km long. Over 1 million people were displaced by the creation of the reservoir. (credit: Le Grand Portage)
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