Lab #2 - Density and Isostasy Lab

.pdf

School

Cypress College *

*We aren’t endorsed by this school

Course

A105

Subject

Geology

Date

Jan 9, 2024

Type

pdf

Pages

7

Uploaded by DukeWorld11631

Report
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
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help