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

MS 20 Laboratory Density, Specific Gravity, Archimedes and Isostasy 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 (ρ, the ratio of a body’s 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 body’s 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 ship’s 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 (ρ), depth (d) and gravitational acceleration (g)

When applied to the Earth’s 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 Earth’s 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 water’s 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 are square (L = W), L x W = L2

Remember that the blocks have a cylindrical hole in the center that reduces the total mass. To account for this measure the diameter of the hole and calculate the missing mass in the cylinder:

Volume of a cylinder: Vhole = π x R2 x T (2)

π = 3.1416; R = D/2

Volume of block without hole: Vb - h = T x (L2 – π x R2) (3)

8. Now calculate the density for each of the blocks:

Density of block: ρ = mass/volume (4)


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B. Modeling the Continental and Oceanic Lithosphere Geologists divide the upper Earth into a chemically distinct crust and mantle. The crust comes in two flavors: continental crust varies in thickness, averaging about 35 km, and has a heterogeneous composition which, on average, resembles granite (silicate-based rocks rich in sodium, potassium and aluminum); oceanic crust is much thinner, typically 8-10 km thick, and is composed of basalt (a silicate-based rock higher in iron, magnesium and calcium). The mantle is a homogeneous, mostly solid mass made of a rock called peridotite (a silicate rock dominated by iron and magnesium). Only rarely does this rock get hot enough to melt, but the region known as the asthenosphere is close enough to its melting temperature that it may be up to 1% molten. As a result the asthenosphere, while not a fluid, displays fluid-like behavior under most conditions (geophysicists refer to this silly putty-like behavior as “plastic”). The region of the mantle located above the asthenosphere is completely solid. The combination of the solid crust and this solid, upper mantle is known as the lithosphere. In the rare case when we are referring only to this upper solid layer of the mantle (separate from the crust), it is called the “mantle lithosphere” (Figure 3). In this exercise we will create a simple physical model of the upper layers of the Earth using our wood blocks and water in an aquarium to represent the various layers of rock:

Figure six. Aquarium model of the Earth’s lithosphere and asthenosphere.

1. Fill an aquarium with about 10 cm of water and place the ring stand in the

aquarium (Figure 6).


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2. Slide a small pine block down the ring stand. Place the large pine block with the rectangular cutout over the ring stand so that it fits around the small pine block. This is a model of the continental lithosphere with no topography. Float the thin oak block next to the pine block – this represents the oceanic lithosphere.

3. Measure the water depth beneath the pine block and beneath the oak block using

a metric ruler. Enter the data in the space provided on the answer sheet.

NOTE: Measuring water depth with a ruler here is much easier if you put the ruler in the water next to the wood block and put your head down so that you are looking at the ruler from “underwater.”

4. Calculate the volume and weight of the water column under each block, assuming

a wood block and a water column with horizontal dimensions of 1 cm x 1 cm and whose thickness is the water depth (Figure 7). We can assume the density of fresh water to be 1.0 g/cm3. Enter these on the answer sheet.

5. Assuming wood blocks of these same thicknesses but with horizontal dimensions

of 1 cm x 1 cm, calculate the mass of each of the wood blocks using the wood densities you determined in part A.

6. Add up the wood and water masses under each column and enter them on the

answer sheet. 7. Measure the depth of the water with no overlying block of wood and determine the

mass of this water column, assuming the same horizontal dimensions (Figure 7).

Figure seven. According to Archimedes, the total pressure

acting on a square centimeter (or any fixed area) at a fixed depth should be constant.


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8. Put another small pine block over the ring stand to represent a mountain (Figure 8 below). Note how it displaces part of the lower "continental crust" in the water.

9. Again assuming blocks of 1 cm x 1 cm, measure and record the depth of water

under this "mountain." Calculate and record the mass of this water column. 10. Calculate the volume and mass of the pair of pine blocks, using the measured

thickness and horizontal dimensions of 1 cm x 1 cm. Calculate the mass of the pair of blocks using the previously determined density. Add this mass to the mass of the underlying water column (Figure 9).

Figure eight. If the lithosphere becomes thicker it should

increase in depth as well as in elevation.


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Figure nine. If the lithosphere becomes thicker it should

increase in depth as well as in elevation.

C. Density and Specific Gravity of Rocks In creating our simple wood/water model of the upper layers of the Earth we have made a couple of key assumptions: that the asthenosphere does behave as a fluid over relatively short time scales; that the asthenosphere is denser than the lithosphere, otherwise the lithosphere would not float (since the mantle lithosphere is approximately the same density as the asthenosphere, it is the lighter crust that decreases the overall density of the lithosphere).

NOTE: Much of what we know about the rheology of the asthenosphere comes from seismologists: We know, for instance, that the asthenosphere can transmit seismic shear waves (S-waves), indicating that it is not a pure fluid; we know that no earthquakes originate in the asthenosphere, indicating that it is too soft (plastic) to break under stress; and we know that all seismic body waves slow down significantly in this region, indicating that the rock there is closer to its melting point than adjacent layers (asthenosphere means “weak sphere,” and it is also known as the “low velocity layer”).


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In this portion of the lab we will consider how well our model works in the real world of rocks. Unlike cut wooden blocks, which can be measured with a ruler, rocks have odd shapes that make volume measurements very difficult. We can get around this problem by using Archimedes’ Principle: “an immersed object displaces water equal to its own volume.” To determine density of a rock we must know its mass and its volume. We could determine volume using the method we used in part A: immerse the rock in a full beaker of water and measure the water that spills out. This is complicated and messy however. But let’s assume that we have a beaker of water full to the brim with 100 grams (100

mL) of water. If you drop in a rock that is 1 cm3 (= 1 mL) and a mass of 3 grams (ρ = 3

g/cm3) then 1cm3 of water will spill out. The net gain in mass is 2 grams (+3 grams of rock, -1 gram of water). We conclude that even though the rock will not float, the water is exerting a buoyant force on the rock that negates a weight equivalent to the mass of the water displaced (this is what Archimedes did with the King’s crown). Therefore, the difference between the weight of the rock in air and its weight underwater is the weight of the water displaced, which is equivalent to its volume (since 1.0 gram of water = 1.0 mL of water), which is equal to the volume of the rock. This is the numerical gram equivalent of the volume, not the volume itself, as the unit is grams rather than cm3 or mL. Dividing this equivalent volume by the density of water yields the true volume of the rock:

VolumeWeightWeight

rock

water

waterair

(5)

Alternatively we can simply divide the weight in air by the difference between weight in air and weight in water, yielding the specific gravity of the sample:

Gravity_SpecificWeightWeight

Weight

waterair

air

(6)

Since the units here are grams divided by grams, specific gravity is a unitless quantity whose value is the numerical equivalent of density. Specific gravity is sometimes called “relative density”, and can be thought of as the ratio of rock density to the density of water. Measuring the specific gravity of rocks Each student will be given three rocks: granite, representing the continental crustal rocks; basalt, representing oceanic crust; and peridotite, representing the mantle rocks in which they are "floating." 1. Using the experimental set up shown below in Figure 10, determine the mass of

each rock on the top tray of the scale (weight in air). Record the results on the answer sheet.


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2. Determine the mass of each rock sample on the hanging tray below the scale (weight in water). Depth does not matter so long as the rock is completely submerged. Be sure that no part of the tray is touching the sides of the beaker. Again, record the results on the answer sheet.

3. Determine the specific gravity of each rock using equation 6. Record the results on

the answer sheet.

Figure ten. Lab set-up for determining specific gravity of rocks, or any odd-shaped

substances denser than water.

D. Determination of total mass exerted at a fixed depth within the

asthenosphere Using the specific gravity calculations from the previous exercise and Figure 11 below, calculate the total mass a column of rock exerts at 150 km depth beneath the mountains, the continents and the ocean. Make the following assumptions: 1. Assume that the water in the ocean has a density of 1.03 g/cm3. 2. Assume that the continental crust has the density of granite, oceanic crust has the

density of basalt, and the lithosphere and the asthenosphere each have the density of peridotite.

3. Since both layers have the same density, we will treat the mantle lithosphere and

the asthenosphere as a single layer of mantle. 4. We are determining the masses of three columns with varying layer thicknesses

and densities, but we will assume each to have horizontal dimensions of 1 cm x 1 cm.


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5. Record your data and calculations on the answer sheet. Pay attention to units and convert where necessary.

Figure eleven. Simple three-layer model of the Earth’s upper layers.


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Names Date Group No.

MS20 Laboratory: Density, Specific Gravity, Archimedes and Isostasy Answer sheet: record all data in the appropriate metric units (centimeters, grams, etc.). Remember to use significant figure rules and to indicate appropriate units (if the scale reads 13.4 g, your answer is not 13.4, but 13.4 g (or 13.4 grams). A. Demonstrating Archimedes’ Principle

Pine Block Oak Block Mass of block <grams>

Weight of water in pan <grams>

Displaced volume <cm3 or mL> (volume of water in pan)

Block thickness <cm>

Length of side <cm>

Vblock (equation 1) <cm3>

Rhole <cm> (½ diameter)

Vhole (equation 2) <cm3>

Vb - h (equation 3) <cm3>

ρwood (equation 4) <g/cm3>


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Archimedes' Principle states that a floating body will displace its own weight of fluid. Are the masses of the blocks the same as the masses of water each displaces? If there is a difference, what are the likely sources of error?

B. Modeling the Continental and Oceanic Lithosphere

Pine Block Oak Block Depth of water <cm>

Volume of water <cm3 or mL>

Mass of water <g>

Mass of block <g>

Combined masses <g>

(water column + wood block)

Depth of water/no blocks <g>

Mass of water/no blocks <g>

Again, according to Archimedes' Principle the pressure (or the total weight) acting on the bottom of the tank (or at some depth in the asthenosphere) should not change as more floating masses are added; i.e., the combined masses of each of the wood blocks and the water columns beneath them should be the same as the total mass of the open water. Is this true here? Again, try to account for likely sources of error.


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Figure 6 shows our model of the earth's crust using two different types of wood of to represent the two different types of crust. Assume you shaved some wood off the top of the pine block (representing continental erosion). What would happen to the pine block? Explain.

Depth of water under two pine blocks <cm>

Volume of water column <cm3 or mL>

Mass of water column <g>

Total mass of two pine blocks <g>

Combined masses of blocks and water column <g>

Is the total mass acting on the bottom of the aquarium approximately the same as in the previous calculations? Again, try to account for likely sources of error?


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Gravity and time, aided by various processes of physical and chemical weathering, removes rock from higher continental elevations and transports it to lower elevations, ultimately to the ocean floor. For every meter of rock removed from a mountain range, would you expect the elevation to decrease by one meter? Explain.

What happens to the oceanic crust as water, ice and wind continuously deliver and deposit continental sediments? Explain.

C. Density and Specific Gravity of Rocks

GRANITE BASALT PERIDOTITE Mass in air

Mass in water

Massair - Masswater

Rock volume (equation 5)

Rock density (equation 4)


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Can you think of another way to measure the volume of the rock specimens?

D. Determination of total mass exerted at a fixed depth within the

asthenosphere weight under ocean crust

5.0 x 105 cm x

x

1.0 cm

x

1.0 cm

=

A. Depth of water Water density Length (L) Width (W)

10 x 105 cm x

x

1.0 cm

x

1.0 cm

=

B. Depth of ocean crust Basalt density Length (L) Width (W)

135 x 105 cm x

x

1.0 cm

x

1.0 cm

=

C. Depth of mantle Peridotite density Length (L) Width (W)

Total weight under ocean (A + B + C)

=

weight under continental crust

30 x 105 cm x

x

1.0 cm

x

1.0 cm

=

D. Depth of crust Granite density Length (L) Width (W)

120 x 105 cm x

x

1.0 cm

x

1.0 cm

=

E. Depth of mantle Peridotite density Length (L) Width (W)

Total weight under continent (D + E) =

weight under mountains

55 x 105 cm x

x

1.0 cm

x

1.0 cm

=

F. Depth of crust Granite density Length (L) Width (W)

100 x 105 cm x

x

1.0 cm

x

1.0 cm

=

G. Depth of mantle Peridotite density Length (L) Width (W)

Total weight under continental mountains (F + G) =


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During ice ages slightly cooler temperatures at mid-to-high latitudes cause less snow to melt in spring and summer than accumulates in winter. Over many years this ice piles up to form continental glaciers that can exceed 2 kilometers in thickness. Glacial ice has a density of about 0.9 g/cm3, just below that of water, and about 1/3 that of granite. What would be the effect of a 2000 meter thick ice sheet on the continent? What would happen when the ice melted?

The total weight under the continents should precisely equal the weight under the ocean basin. The calculations, although close, don't really match. What does this tell us about our simple three-rock model? Explain.

The last ice age ended in northern Europe and in North America approximately 10,000 years ago, yet there is excellent evidence that the Scandinavian peninsula (which includes Norway, Sweden and Finland) is still uplifting rapidly. From this information, what can we conclude about the physical properties of the asthenosphere?

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

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