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
Page 3 of 11
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.
Page 4 of 11
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
Page 6 of 11
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
Page 8 of 11
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