Isostasy and Rock Density Equipment needed for each student group: 1 metric ruler 2 blocks of pine, 6.1 cm x 6.1 cm x 1.9 cm, with centered 1.4 cm hole 1 blocks of oak, 6.1 cm x 6.1 cm x 1.9 cm, with centered 1.4 cm hole aquarium three rock samples (granite, basalt and peridotite) squeeze bottle Funnel 600 ml glass beaker 100 ml graduate cylinder turkey baster 1000 ml beaker plastic collection dish availability of balance suitable for weighing objects suspended in water

Students should bring to all labs: 1 calculator 1 pencil

DISCUSSION:

Isostasy

In the 1920's, the German geophysicist, Alfred Wegner, hypothesized that the continents ‘float’ around on denser, subcrustal material. (Density is a term that refers to mass per unit volume of a substance. In the metric system, and in all of our labs, we will refer to grams/ cm3 or grams/ml as our unit of density.) Several decades ago, the scientific evidence became overwhelming that the outer, rigid portion of the earth's surface, the lithosphere, is actually composed of a number of plates that ‘float’ on denser, more plastic material beneath, the asthenosphere, which is often called the mantle. When objects of different density float on a more dense, non-rigid material, the depth to which they sink will vary with more dense objects sinking deeper than less dense objects. Think of an example, such as putting a beach ball, a handball and a basketball in a swimming pool. They will all reach an equilibrium position with the water level, but they will all be different. The beach ball will float highest in the water, the handball deepest and the basketball will be somewhere in between. In today's lab, we will be looking at the differences in density between continental crust (represented by pine blocks) and oceanic crust (represented by oak blocks) and how it affects the way they float on the denser astenosphere (represented by water). The process of reaching an equilibrium level as a function of relative densities is called isostasy. Isostasy literally translates from Greek to mean ‘equal standing’. In addition to using wooden blocks to look at isostasy, we will also determine the actual density of rocks samples which are representative of the continental and oceanic crust. In today's lab, we will be weighing objects to compare their masses. The mass of an object is a measure of its inertia. The weight of an object depends not only on the mass of the object, but on the local force of gravity as well. The mass of an object is constant and will be the same on earth as it is on the moon; however, since gravity on the moon is much less than on earth, the weight of the object will be less on the moon than on earth. Page 1 of 11

Isostasy, or ‘equal standing’, means that at some reference level well below the surface of the earth, there will be an equal mass of material above this reference level, whether it is continental crust, oceanic crust, or a combination of water and either of these crustal materials (Fig. 1). In this laboratory, we will be using different kinds of wood blocks in a water-filled aquarium to demonstrate isostasy. Think of the bottom of the aquarium as our reference level. If we put a more dense block of wood in the water, it will sink deeper relative to the water level, but since it displaces that water, the water level in the aquarium will also rise a little. The pressure at the bottom of the aquarium will be the same under the wood block as it is everywhere else.

FIGURE 1

In looking at Fig. 1, we might wonder how could the weight at the reference level be the same below the mountains as below the ocean when the mountains rise high above sea level. What would happen if the land were covered by glaciers during an ice age?

EXERCISE 1: Archimedes’ Principle Part of Archimedes' Principle states that a floating body will displace its own mass of fluid. Many of us have seen ocean tankers, either at sea or in harbors. When they are empty, they sit high in the water and you can see a lot of dull red paint above the actual waterline. When they are fully loaded, they sit much lower in the water and you can usually only see the company colors above the waterline. In this lab we will be using pine and oak blocks to demonstrate this concept. We will use the set up shown in Fig. 2 to ultimately determine their respective densities.

3 In these exercises we will assume that 1 ml of water weighs 1 gram and occupies a volume of 1 cm . Page 2 of 11

1. Weigh one pine block and one oak block on the digital balance to the nearest hundredth of a gram (0.01 g). Enter the weights in Table 1 on your form. 2. Carefully fill a 1000 ml beaker until the water is just at the lip/spout, but not quite overflowing (you may want to use a little soapy water to rub the lip of the beaker to reduce the effects of surface tension). You can top off the beaker with a squeeze bottle. Put the beaker in the center of the collection dish. 3. Carefully place the pine block in the water and catch the ‘runoff’ in the dish below. To prevent spilling, take a little water out of the beaker before lifting the beaker out of the dish. 4. Using a funnel, transfer the ‘runoff’ to a 100 ml graduate cylinder. According to Archimedes, this volume should be equal to the mass of the floating block. Repeat this exercise for the oak block. Enter the data in Table 1 on your forms and answer the questions. 3

In these exercises we will assume that 1 ml of water weighs 1 gram and occupies a volume of 1 cm .

FIGURE 2

TABLE 1 PINE weight (gms) displaced volume (water caught in dish)

OAK gms

gms

3

cm3

cm

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Answer the following questions: 3

A. If 1 cm or 1 ml of water weighs 1 gram, how do the displaced volumes compare with the actual weight? (Remember: Archimedes’ Principle states that a floating body will displace its own weight.) What causes the discrepancy?

B. How could you compute the actual volume of the pine and oak blocks? (hint: consider the dimensions of the solid block and the hole)

3

C. Assume the volume of the each block is approximately 69 cm . Given this volume, what would be the density of the pine and the oak blocks? (hint: density = mass/volume, use the weights from Table 1) density of pine =

3

g/cm

density of oak =

3

g/cm

DISCUSSION: Physical Model of the Earth’s Crust In the model of the earth’s crust most commonly accepted, continental crust is usually considered to be 30 to 40 km thick, consisting of rocks rich in silica and aluminum. (Some geologists use the term “sial” to denote this material.) In contrast, oceanic crust is thinner (5-10 km) and composed of basaltic type rocks rich in magnesium and iron (often called “sima”). The two main differences between the continental crust and the oceanic crust are the thickness and the rock type, which have varying densities. In our physical model, the continental crust can be represented by the pine block with lower density, while the oak slab represents the oceanic crust.

EXERCISE 2:

Physical Model of the Continental and Oceanic Crust with Continental Mountains

1. Fill the aquarium to a depth of 10 cm. Be sure to measure on the inside of the aquarium.Float the pine block on top of the water. This model represents the continental mass without mountains. 2. Float the oak block next to the pine ‘continental mass’. The oak block represents the oceanic crust. 3. Measure the water depth below the pine block and below the oak block with a metric rule as shown in Fig. 3, and enter the data in Table 2 on your forms and answer the questions.

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FIGURE 3

TABLE 2 4. Enter water depth under pine block

cm

under oak block

cm

5. Calculate the weight of the WATER under both wood blocks UNDER PINE (density)

x 3

(1.0 g/cm )

x

UNDER OAK (volume)

= mass

(1 cm x (1 cm x depth)

weight of water under pine =

(density) 3

=

g

x

(1.0 g/cm )

(volume)

x

= mass

(1 cm x 1 cm x depth) =

weight of water under oak =

g

2

6. Calculate the weight of 1 cm of pine and oak floating on top of the water. (hint: use the density of pine and oak as calculated in the previous exercise) WEIGHT OF PINE pine density 3

g/cm

x x

lxwxh 1 cm x 1 cm x 1.9 cm

WEIGHT OF OAK = mass =

weight of pine =

oak density 3

g/cm

weight of oak = Page 5 of 11

x

lxwxh

= mass

x 1 cm x 1 cm x 1.9 cm =

7. Now add up the wood and water weights under each column combined weights (water plus wood) under pine =

g

combined weights (water plus wood) under oak =

g

Answer the following question: Are these two combined weights approximately equal? Is this to be expected?

8. Put another pine block on top of the first pine block to represent a mountain (Fig. 4 below). Note how it displaces part of the lower ‘continental crust’ in the water. Measure the new depth under the ‘mountain’, enter the depth in Table 3, and calculate the new weight acting on the bottom.

FIGURE 4

TABLE 3

9. Enter new water depth under the two pine blocks

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cm

10. Calculate the NEW WATER weight under the two pine blocks UNDER TWO PINE BLOCKS

density x 3

1.0g/cm

volume x

=

1 cm x 1 cm x new depth =

mass of water g (weight of water)

11. Enter weight of two blocks. (hint: just double the pine weight in #6 above) =

g (weight of two pine blocks)

combined weight =

g (water plus two pine blocks)

Answer the following question: Is the weight acting on the bottom of the aquarium approximately the same or different from the previous calculations? Explain!

DISCUSSION:

Measuring the Specific Gravity of Rocks

Up to now we have been considering the model of the earth’s crust to be represented by wood of varying densities. Our next experiment will now consider actual rocks - the types that make up continental crust and oceanic crust. The experiment will make use of the second part of Archimedes’ principle which states: “when an object is immersed, it displaces water equal to its own volume.” We sometimes interchange the terms ‘density’ and specific gravity. Density as strictly defined 3 is mass per unit volume (g/cm ). Specific gravity or relative density is the ratio of the density of a substance to the density of water. The term has no units, since the units cancel out!

EXERCISE 3:

Measuring Rock Density

1. Each group 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.’ Weigh each rock type in air and enter the data in Table 4 on your forms. 2. Using the experimental ‘setup’ shown in Fig. 5 below, measure the weight of each rock immersed in water. In order to do this, you will need to learn how to read a vernier scale. An example is shown if Fig. 6. The innermost scale gives the units and tenths reading. Read the units first – in the case of Fig. 6 the innermost scale (ones) is more than four, but less than five. On the tenths between 0.7 and 0.8, so this scale gives a reading of 0.7. The Page 7 of 11

outermost scale gives you the hundredths reading. Look for the pair of lines that exactly line up between the innermost scale and the outermost scale. In the case of Fig. 3.6, this is at .05, so the total reading would be 4.75. Enter these data in Table 4 on your forms. The difference between the weight in air and the weight in water divided by the density of water is equivalent to the volume of the rock in the centimeter-gram-second (cgs) system. 3. Make the appropriate density calculation outlined in Table 4.

FIGURE 5

FIGURE 6

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FIGURE 7

GRANITE

PERIDOTITE

TABLE 4

PERIDOTITE

BASALT

GRANITE

BASALT

weight in air

g

g

g

weight in water

g

g

g

(wt in air - wt in water)

g

g

g

3 1.0 g/cm

)

volume = (wt in air - wt in water) = density of water volume of each rock =

density of each rock

3

(

3

cm

cm 3

g/cm

density = mass/volume 3 g/cm

3

cm

3

g/cm

Answer the following question: What is another way to measure the volume of the rock specimens (Remember the two parts of Archimedes Principle)?

Page 9 of 11

EXERCISE 4:

Calculation of the Overlying Crustal Mass at 50 km Depth

Using the calculations of density from the three rock types and Fig 7 below, calculate the mass acting at a 50 km depth under both the continent and ocean. Do the calculations in Table 5 on your forms.

FIGURE 8

TABLE 5 CALCULATING THE WEIGHT UNDER THE OCEAN CRUST Convert km to cm using the conversion factor of 105 cm/km 1. depth of water (cm) x seawater density x 1 cm x 1 cm = 5 3 (5 km x 10 cm/km) x 1.025 g/cm x 1 cm x 1 cm =

5

5.12 x 10 g

2. depth of ocean crust (cm) x basalt density x 1 cm x 1 cm = 5 x 1 cm x 1 cm = (10 km x 10 cm/km) x

g

3. depth of mantle (cm) x peridotite density x 1 cm x 1 cm = 5 (35 km x 10 cm/km) x x 1 cm x 1 cm =

g

Page 10 of 11

total weight under ocean = (1+2+3) =

g

CALCULATING THE WEIGHT UNDER THE CONTINENTAL CRUST WITHOUT MOUNTAINS 4. depth of continental crust (cm) x granite density x 1 cm x 1 cm = 5 3 (31 km x 10 cm/km) x g/cm x 1 cm x 1 cm =

g

5. depth of mantle (cm) x peridotite density x 1 cm x 1 cm = 5 3 (20 km x 10 cm/km) x g/cm x 1 cm x 1 cm =

g

total weight under continent = (4+5) =

g

CALCULATING THE WEIGHT UNDER THE CONTINENTAL CRUST WITH MOUNTAINS 6. depth of continental crust (cm) x granite density x 1 cm x 1 cm = 5 3 (42 km x 10 cm/km) x g/cm x 1 cm x 1 cm =

g

7. depth of mantle (cm) x peridotite density x 1 cm x 1 cm = 5 3 (10 km x 10 cm/km) x g/cm x 1 cm x 1 cm =

g

total weight under continent with mountains = (6+7) =

Page 11 of 11

g