Lab #2 - Density and Isostasy Lab
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School
Cypress College *
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Course
A105
Subject
Geology
Date
Jan 9, 2024
Type
Pages
7
Uploaded by DukeWorld11631
GEOL A105L:
General Geology Lab
-1-
Name_________________________________
Lab ____________________________
Density and Isostasy lab
PART I: Density of different types of rocks
Although the earth has many different kinds of rocks, some are more common than others and can be
considered the main constituents of the crust. In order to get a feel for the range of densities that occur
within average rock types, you will measure several rock samples to get an idea of the variation in density
on the earth.
Weigh and measure 1 sample each of granite, basalt, and peridotite.
Fill in the table
below and write your data on the board so we can gather class measurements and find averages. Use the
electronic balance to weigh the samples, and use water displacement to measure the volume. (Remember
that
1 mL = 1 cm
3
). To get density, then, all you have to do is divide the sample mass by its volume.
Basalt
Granite
Peridotite
Sample mass (g)
Original water
volume (ml)
Volume of water
and rock (ml)
Volume of rock
(cm
3
)
Density of rock
(g/cm
3
)
Class average
density (g/cm
3
)
Number 1-3, from
least-most dense
Which component
of the earth?*
* your choices are continental crust, oceanic crust, and mantle
Anna Do
#2
8.2 g
48.3 ml
51.0 ml
2.7 ml
3.0 g/cm3
3.1 g/cm3
2
Ocean crust
7.4 g
63.0 ml
65.7 ml
2.7 ml
2.7 g/cm3
2.6 g/cm3
1
Continental crust
Mantle
3
3.5 g/cm3
3.3 g/cm3
3.4 ml
54.6 ml
51.2 ml
11.3 g
GEOL A105L:
General Geology Lab
-2-
PART II: Isostasy and the buoyant force
In this part of the lab, you will measure some of the properties of wood floating in water and then
compare it to processes happening in the earth’s crust and mantle. Choose two of the different types of
wood, measure their mass (using the balance) and volume (using a ruler), and write your measurements in
the
table below
and on the board. Then float your block of wood in a tub of water and measure the
amount of each block that is above and below the equilibrium water line.
Type of
wood^
Mass
(g)
Height
(cm)
H
total
Width
(cm)
Length
(cm)
Volume
(cm
3
)
Density
(
ρ
)
(g/cm
3
)
X*
ρ
wood
/
ρ
water
**
Height
above
water
H
above
(cm)
Height
below
water
H
below
(cm)
Y*
H
below
/
H
total
***
Pine
Walnut
Oak
Maple
Basswood
Balsa
Hickory
^.
* X and Y are the columns you will graph. The values in these columns should be expressed as decimals.
** The density of water is 1 g/cm
3
*** H
total
should come from the third column.
1.
Graph columns X and Y on the next page. Your axes are already labeled for you.
2.
Based on the graph, describe the relationship between the density ratios and the height ratios.
Hopefully, your graph resembled a straight line.
Using this relationship, you can write an
equation that expresses how the density of the wood block relates to the height of the block below
the water line. The equation of this would be:
55 g
70 g
90 g
70 g
30 g
20 g
80 g
1 cm
1 cm
1 cm
1 cm
1 cm
1 cm
1 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm3
10 cm3
10 cm3
0.5 g/cm3
0.7 g/cm3
0.9 g/cm3
0.7 g/cm3
0.3 g/cm3
0.2 g/cm3
0.8 g/cm3
0.5
0.7
0.9
0.7
0.3
0.2
0.8
0.5 cm
0.3 cm
0.1 cm
0.3 cm
0.7 cm
0.8 cm
0.2 cm
0.5 cm
0.7 cm
0.9 cm
0.7 cm
0.3 cm
0.2 cm
0.8 cm
0.7
0.9
0.3
0.2
0.8
Based on the graph, there’s a linear relationship between the density
ratios and the height ratios.
100 cm3
100 cm3
100 cm3
100 cm3
100 cm3
100 cm3
100 cm3
0.4
0.6
0.6
0.6
GEOL A105L:
General Geology Lab
-3-
ρ
wood
ρ
water
=
H
below
H
total
We can rewrite the equation such that the only thing on the left-hand side of the equation is
H
below
:
H
below
=
ρ
wood
∗
H
total
ρ
water
(Remember: The wood block is floating in equilibrium when it has displaced a volume of water
that has the same mass as the entire wood block. If the wood block is 80% as dense as water, than
80% of the block will sit below the water line.)
Now, with some quick algebra (which we won’t go into for this course) we can rewrite the
equation again so that it is solved for the height of the wood block
above
the equilibrium line,
H
above
(remember that the height above the water line would be
H
total
–
H
below
).
H
above
= H
total
*
�
1
−
ρ
wood
ρ
water
�
3.
Describe how the wood block measurements would change if the block were floating in a tub of
honey instead of water (hint: the density of honey is about 1.42 g/cm
3
). What would you expect to
see and why?
The wood block would float higher in a tub of honey than in the water because
honey is about 1.42 g/cm3, which is denser than water (1 g/cm3). Therefore, the
ratio Pwood / P water would equal to zero which make the formula becomes to
Habove = Htotal
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