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The Liquid Phase Help (page 3)

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By — McGraw-Hill Professional
Updated on Sep 5, 2011

Solution 2

All we need to do in this case is multiply our mass density answer by 9.81 m/s 2 . This gives us

d w = 1090 kg/m 3 × 9.81 m/s 2

= 10,700 N/m 3 = 1.07 × 10 4 N/m 3

Note the difference here between the nonitalicized uppercase N, which represents newtons, and the italicized uppercase N , which represents the number of moles of atoms in a sample.

Measuring Liquid Volume

The volume of a liquid sample is usually measured by means of a test tube or flask marked off in milliliters or liters. However, there’s another way to measure the volume of a liquid sample, provided we know its chemical composition and the weight density of the substance in question. This is to weigh the sample of liquid and then divide the weight by the weight density. We must, of course, pay careful attention to the units. In particular, the weight must be expressed in newtons, which is equal to the mass in kilograms times the acceleration of gravity (9.81 m/s 2 ).

Let’s do a mathematical exercise to show why we can measure volume in this way. Let d w be the known weight density of a huge sample of liquid too large for its volume to be measured using a flask or test tube. Suppose that this substance has a weight of w , in newtons. If V is the volume in meters cubed, we know from the preceding formula that

d w = w/V

because w = ma , where a is the acceleration of gravity. If we divide both sides of this equation by w , we get

d w / w = 1/ V

Then we can invert both sides of this equation and exchange the left-hand and the right-hand sides to obtain

V = w/d w

All this is based on the assumption that V , w , and d w are all nonzero quantities. This is always true in the real world; all materials occupy at least some volume, have at least some weight because of gravitation, and have some density because there is some “stuff in a finite amount of physical space.

Pressure In Liquids

Have you read or been told that liquid water can’t be compressed? In a simplistic sense, this is true, but it doesn’t mean liquid water never exerts pressure. Liquids can and do exert pressure, as anyone who has been in a flood or a hurricane or a submarine will tell you. You can experience “water pressure” for yourself by diving down several feet in a swimming pool and noting the sensation the water produces as it presses against your eardrums.

In a fluid, the pressure, which is defined in terms of force per unit area, is directly proportional to the depth. Pressure is also directly proportional to the weight density of the liquid. Let d w be the weight density of a liquid (in newtons per meter cubed), and let s be the depth below the surface (in meters). Then the pressure P (in newtons per meter squared) is given by

P = d w s

If we are given the mass density d m (in kilograms per meter cubed) rather than the weight density, the formula becomes

P = 9.8 l d m s

Pressure in Liquids Practice Problem

Problem

Liquid water generally has a mass density of 1000 kg/m 3. How much force is exerted on the outer surface of a cube measuring 10.000 cm on an edge that is submerged 1.00 m below the surface of a body of water?

Solution

First, figure out the total surface area of the cube. It measures 10.000 cm, or 0.10000 m, on an edge, so the surface area of one face is 0.10000 m × 0.10000 m = 0.010000 m 2 . There are six faces on a cube, so the total surface area of the object is 0.010000 m 2 × 6.0000 = 0.060000 m 2 . (Don’t be irritated by the “extra” zeroes here. They are important. They indicate that the length of the edge of the cube has been specified to five significant figures.)

Next, figure out the weight density of water (in newtons per meter cubed). This is 9.81 times the mass density, or 9,810 N/m 3 . This is best stated as 9.81 × 10 3 N/m 3 because we are given the acceleration of gravity to only three significant figures, and scientific notation makes this fact clear. From this point on let’s revert to power-of-10 notation so that we don’t fall into the trap of accidentally claiming more accuracy than that to which we’re entitled.

The cube is at a depth of 1.00 m, so the water pressure at that depth is 9.81 × 10 3 N/m 3 × 1.00 m = 9.81 × 10 3 N/m 2 . The force F (in newtons) on the cube is therefore equal to this number multiplied by the surface area of the cube:

F = 9.81 × 10 3 N/m 2 × 6.00000 × 10 −2 m 2

= 58.9 X 10 1 N = 589 N

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