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

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Which weighs more, a ton of feathers or a ton of bricks? This old riddle plays v the distinction between mass and also density. A ton is a ton, that course; yet bricks have actually much greater density than feathers, and so we room tempted come think that them together heavier. (See figure 1.)

Figure 1. A ton the feathers and also a ton that bricks have actually the very same mass, but the feathers make a much bigger pile because they have actually a much reduced density.

Density, together you will see, is an essential characteristic that substances. It is crucial, for example, in determining whether an item sinks or floats in a fluid. Thickness is the mass per unit volume that a problem or object. In equation form, thickness is identified as

\rho =\fracmV\\,

where the Greek letter ρ (rho) is the symbol for density, m is the mass, and also V is the volume inhabited by the substance.

Density is mass every unit volume.

\rho =\fracmV\\,

where ρ is the symbol because that density, m is the mass, and also V is the volume lived in by the substance.

In the riddle regarding the feathers and bricks, the masses space the same, but the volume lived in by the feathers is much greater, because their thickness is much lower. The SI unit of density is kg/m3, representative worths are given in Table 1. The metric mechanism was originally devised so that water would have actually a thickness of 1 g/cm3, identical to 103 kg/m3. Thus the basic mass unit, the kilogram, was an initial devised to be the massive of 1000 mL the water, which has actually a volume the 1000 cm3.

Table 1. Densities of various SubstancesSubstance\rho \left(\text10^3\textkg/m^3\textor\textg/mL\right)\\Substance\rho \left(\text10^3\textkg/m^3\textor\textg/mL\right)\\Substance\rho \left(\text10^3\textkg/m^3\textor\textg/mL\right)\\
SolidsLiquidsGases
Aluminum2.7Water (4ºC)1.000Air1.29 × 10−3
Brass8.44Blood1.05Carbon dioxide1.98 × 10−3
Copper (average)8.8Sea water1.025Carbon monoxide1.25 × 10−3
Gold19.32Mercury13.6Hydrogen0.090 × 10−3
Iron or steel7.8Ethyl alcohol0.79Helium0.18 × 10−3
Polystyrene0.10Glycerin1.26Nitrogen1.25 × 10−3
Tungsten19.30Olive oil0.92Nitrous oxide1.98 × 10−3
Uranium18.70Oxygen1.43 × 10−3
Concrete2.30–3.0Steam (100º C)0.60 × 10−3
Cork0.24
Glass, typical (average)2.6
Granite2.7
Earth’s crust3.3
Wood0.3–0.9
Ice (0°C)0.917
Bone1.7–2.0

As you have the right to see by examining Table 1, the thickness of an object may help identify that composition. The thickness of gold, for example, is about 2.5 time the thickness of iron, i beg your pardon is about 2.5 times the thickness of aluminum. Density also reveals something about the phase of the matter and its substructure. Notification that the densities of liquids and also solids are roughly comparable, regular with the fact that your atoms room in nearby contact. The densities that gases are much much less than those of liquids and solids, since the atoms in gases room separated by huge amounts of north space.

### Take-Home Experiment Sugar and also Salt

A pile of sugar and also a heap of salt look at pretty similar, yet which weighs more? If the volumes of both piles space the same, any difference in mass is as result of their various densities (including the air room between crystals). Which perform you think has the greater density? What values did girlfriend find? What technique did you usage to determine these values?

### Example 1. Calculating the fixed of a Reservoir indigenous Its Volume

A reservoir has a surface ar area of 50.0 km2 and an typical depth the 40.0 m. What massive of water is organized behind the dam? (See number 2 for a view of a large reservoir—the 3 Gorges Dam site on the Yangtze flow in main China.)

Strategy

We have the right to calculate the volume V that the reservoir indigenous its dimensions, and find the density of water ρ in Table 1. Then the fixed m can be found from the meaning of density

\rho =\fracmV\\.

Solution

Solving equation ρ m/for m gives m = ρV. The volume V of the reservoir is its surface area A time its typical depth h:

\beginarraylllV& =& Ah=\left(\text50.0\textkm^2\right)\left(\text40.0\textm\right)\\ & =& \left<\left(\text50.0 k\textm^2\right)\left(\frac\text10^3\textm1\textkm\right)^2\right>\left(\text40.0 m\right)=2\text.\text00\times \text10^9\textm^3\endarray\\

The density of water ρ native Table 1 is 1.000 × 103. Substituting V and also ρ right into the expression because that mass gives

\beginarraylllm& =& \left(1\text.\text00\times \text10^3\text kg/m^3\right)\left(2\text.\text00\times \text10^9\textm^3\right)\\ & =& 2.00\times \text10^\text12\text kg\endarray\\.

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Discussion

A big reservoir contains a very large mass the water. In this example, the load of the water in the reservoir is mg = 1.96 × 1013 N, wherein g is the acceleration due to the Earth’s gravity (about 9.80 m/s2). That is reasonable come ask even if it is the dam must supply a pressure equal come this significant weight. The prize is no. As we shall watch in the adhering to sections, the force the dam have to supply deserve to be much smaller than the weight of the water it holds back.