MS20 Laboratory Density, Specific Gravity, Specific Gravity, Archimedes and Isostasy Objectives

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  • MS 20 Laboratory Density, Specific Gravity, Archimedes and Isostasy Page 1 of 20

    Revised on 2/6/2018

    MS20 Laboratory Density, Specific Gravity, Archimedes and Isostasy Objectives

    To understand the definition of density

    To understand the relationship between density and specific gravity

    To learn several methods for determining density and specific gravity of solids and fluids

    To explain the relationship between isostasy, gravity and buoyancy

    To explain why continental crust is thicker and higher than oceanic crust

    To explain how erosion of mountains can lead to uplift and higher peak elevations

    Introduction Archimedes was a Greek mathematician and inventor who lived in the 3rd century BC. Famous in his time for various inspirations and inventions, he is probably best known today for determining the principle of buoyancy, known as Archimedes Principle. He was approached one day by Hiero, King of Syracuse, to solve a problem. The King had ordered a new gold crown and supplied the goldsmith the exact amount of gold he would need to complete the project. When the crown was delivered Hiero was concerned, though the total weight was correct, that the goldsmith had cheated him by substituting cheaper silver for some of his gold. Could Archimedes think of a simple way of determining whether the King was cheated? Archimedes had his inspiration one day when climbing into the bath. As he sat down he watched the water rise on the walls of the bathtub and realized that the volume of water being displaced weighed the same as he did, since the remaining water was supporting him buoyantly. He jumped out of the tub and ran through the streets, naked, shouting Eureka (I have found it)!

    What Archimedes realized was that, since the density (, the ratio of a bodys mass to its volume) of silver was less than that of gold, a crown of the same weight with silver substituted for some of the gold would have to have a greater volume than an all-gold crown. By weighing the crown under water (where it weighs slightly less because of the buoyant force) and comparing that to its weight in air Archimedes demonstrated that the goldsmith had indeed substituted silver for some of the gold (the goldsmith paid a large fine and was declared ineligible to bid on future government contracts).

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    NOTE: The terms weight and mass are not the same. The mass of a body is a quantitative measure of its inertia. The weight of a body is the force of gravity (g) acting upon the bodys mass. The mass is always constant regardless of gravity; however, the weight will change depending on changes in gravity (this is why an astronaut weighs less on the moon). At any fixed point on the Earth gravity is constant, so the weight of any body is directly proportional to its mass. In this laboratory we will be "weighing" objects on balances making use of gravity, but expressing their weight in "mass units such as grams.

    You can float in a body of water because your body is slightly less dense than water. A large oil tanker floats for the same reason: Though the ship may be constructed of materials, like iron, that are much denser than water, the overall density of the ship is its total mass divided by the total volume the ship occupies (the volume must include all of the empty space enclosed by the ships external walls, and the mass must include the mass of all of the air occupying that space).

    Figure one. The behavior of a boat with a single occupant, and with multiple

    occupants.

    Imagine you are sitting in a small boat when several friends decide to climb in with you (Figure 1). As more people climb into the boat it rides lower and lower in the water, because each person weighs much more than the volume of the air that they displace when the enter your boat. As the boat moves lower in the water column, the water is being displaced outward.

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    Figure two. The pressure at a fixed depth depends on the sum of the products of the

    weights and thicknesses of each overlying layer.

    Now imagine you are a small marine animal that swims beneath the boat (Figure 2). Pressure is the total weight of all the mass sitting above your head (at the surface, atmospheric pressure is the weight of the atmosphere above your head). As the fish swims beneath the boat, the pressure he feels does not change, because the weight of the open water above him is the same as the weight of the water plus the weight of the boat with one person, and that is the same as the weight of the water plus the weight of the boat with many people in it. This is true because of Archimedes Principle: as the weight of the boat increases, it always displaces a volume of water that is exactly equal to its own weight.

    NOTE: Pressure is weight per unit area, so the pressure is the

    product of density (), depth (d) and gravitational acceleration (g)

    When applied to the Earths lithospheric plates, the principle is exactly the same. The lithosphere floats on top of the slightly denser asthenosphere. If you were to go 200 kilometers below the Earths surface and measure the pressure in the asthenosphere, you would find it to be constant as the plates shift around on the surface (Figure 3). At equilibrium, the pressure at a particular depth (say, 200 km) in the asthenosphere is the sum of the products of the densities and the thicknesses of all of the layers above.

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    Figure three. The ability of the asthenosphere to flow under very low strain rates

    causes it to behave like a fluid in relation to the more rigid lithosphere.

    NOTE: The asthenosphere is not a fluid (like water) but a plastic (like silly putty, but thousands of times more viscous). That said, over time scales of thousands of years the asthenosphere functions as a fluid.

    A. Demonstrating Archimedes Principle Archimedes Principle states that a floating body will displace its own weight of liquid. In this first exercise we will test this idea by comparing the masses of different blocks of wood to the mass of water that each displaces. Figure 4 shows the experimental set-up. 1. First, determine the masses of one block of pine and one block of oak to the

    nearest 1/100 of a gram (0.01 g). The pine is the light-colored wood and the oak is the dark wood. Record the masses on the answer sheet.

    2. Weigh an aluminum pie plate. Set the 1000 mL beaker in the pie plate on top of

    several rubber stoppers. Fill the beaker until the water is just at the lip. Top it off with a squeeze bottle containing a small amount of Calgon (soap breaks the waters surface tension).

    NOTE: You want the beaker as close as possible to overflowing without spilling, and you want the pie plate to be absolutely dry. Line the plate with a couple of paper towels while filling the beaker in case of spills.

    3. Slowly lower the pine block into the water until it is just floating. The runoff should

    flow down the side of the beaker and into the pan. When it stops flowing, use a 25 mL plastic pipette and bulb to remove a bit of water from the beaker to prevent spillage when you lift the beaker.

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    4. Carefully remove the beaker and the rubber stoppers from the pie plate. Weigh the

    pie plate with the water. Subtract the weight of the pie plate from the total, and record the weight of the water on the answer sheet.

    5. Using a funnel, pour the runoff from the pie plate into a 25 mL graduated cylinder.

    There will be more than 25 mL. Repeat steps 2-4 with the pine block and record both of the water volumes on the answer sheet.

    NOTE: We are using a 25 mL graduated cylinder here because it has finer measurement divisions (greater precision). There will be more than 25 mL of water in the pan, so you will need to fill the cylinder almost to the top, record the volume, empty it, and then pour the rest of the water in, adding the second volume to the first. Do not fill the graduated cylinder higher than the 25 mL limit and estimate the volume or you are throwing out the precision this device provides.

    6. Since water has a density of 1.0 grams/cm3, and since 1 cm3 = 1 mL, the volume of

    water in mL should be equal to the mass of the water in grams. Also the volume of water in milliliters should be equal to the mass of the pine block, in grams. Is this true?

    Figure four. Set-up for determining the density/buoyancy relationship.

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    NOTE: The density of water can vary slightly with temperature (as you will learn in subsequent lab exercises). The small amount of soap used here can also affect the density, but there is enough error introduced here due to water sticking to the sides of the beaker, to the rubber stoppers and to the pie plate, that these slight density variations can be safely ignored. What other sources of error might be involved in this experiment?

    Figure five. Significant dimensions for determining the

    volume of a wood block with cylindrical hole.

    7. Measure the dimensions of each of the blocks and calculate the total volume of

    each of the blocks. Record your measurements on the answer sheet.

    Volume of a solid, square block: Vblock = T x L2 (1) *since the blocks a