HSmith_IsostasyForm

ISOSTASY FORM Name: Section: Hannah Smith OCNG-252-598 Lab 1: Isostasy EXERCISE 1: Archimedes’ Principle Table 1 Part 1: Floating Washers (1) Measured weight of washers (g) 41 g (2) Initial volume of beaker (ml) 500 ml (3) New volume of beaker (ml) 541 g (4) Volume displaced (ml) (3) – (2) 41 ml (5) Weight of washers from volume (4) ml x 1 g/cm 3 x 1 cm 3 /ml 164 g (6) Measured versus displaced difference. (5) – (1) 123 g How well does your measured (1) versus displaced mass (5) compare for the floating scenario? Are these the results what you expected? If your numbers are different, what could explain these differences? Based on Archimedes Principle, the volume of displaced water equals the mass of the floating object, which in this case, is the washers. Here, the volume of displaced water is 41 ml, which is equal to the measured (1) mass of the washers, which is 41 g. However, there is a difference than in the floating scenario because the displaced mass of the washers is 164g, which does not equal the volume of displaced water as it is greater. Table 2 Part 2: Submerged Washers (7) Calculated volume of washers (cm 3 ) 0.92 cm 3 x number of washers 11.04 cm3 (8) Initial volume of graduated cylinder (ml) 90 ml (9) New volume of graduated cylinder (ml) 101 ml (10) Volume displaced (ml) (9) – (8) 11 ml (11) Volume of washers in cm 3 (10) x 1 cm 3 /ml 11 cm3 (12) Measured versus displaced difference. 0.04 cm3 Page 1 CLWvSp17

ISOSTASY FORM Name: Section: (11) – (7) How well does your measured (7) versus displaced volume (11) compare for the submerged scenario? Are these the results that you expected? If your numbers are different, what could explain these differences? My results aligned with what I expected to happen based on the information I learned before, especially regarding Archimedes Principle and the submerged scenario. In the submerged scenario, this principle states that the volume of displaced water is equal to the volume of the submerged object. Here, my measured volume of the washers was 11.04 cm3 and my displaced volume was 11 cm3. Additionally, I found the volume of displaced water to be 11 ml, which is the same value that I found for the displaced volume of the washers. Therefore, my results compare well to the submerged scenario as the volume of displaced water that I found was equal to the volume of the submerged object. Page 2 CLWvSp17

ISOSTASY FORM Name: Section: The average density of the earth is about 5.5 g/cm 3 . How does this number compare to the density of your three rock samples? Why are they different? It makes sense that the density of the three rocks would be lower than that of the earth, with Peridotite and earth having a difference in density of 2.29 g/cm3 and Basalt and Granite having a 2.82 g/cm3 difference in density. It makes sense that peridotite is the closest in density to the earth, because in this lab it is meant to simulate the earth's mantle. The reasoning for the difference in density is that the earth is made up of several different layers, including the mantle, oceanic crust, and continental crust, so it is clearly going to have a greater density. These three rocks are meant to measure the density of each individual layer. With your understanding of the principle of isostasy, if your granite and basalt rock samples were of equal volume, which of these samples would float highest when resting on the mantle? Why? Not only did Granite and Basalt have equal volumes, but they also had equal densities. Based on the principle of Isostasy, these two rock samples would each float at the same level because how "high" or "low" they float is based on density. Which rock sample, granite, or basalt, would float lowest on the mantle? Why? The Peridotite rock sample would float the "lowest" on the mantle as it has the greatest density of the three samples. EXERCISE 3: Isostatic Adjustment Table 4 Pine Oak Weight (g) 33.52 g 47.96 g Density (g/cm 3 ) 0.486 g/cm3 0.695 g/cm3 Compare the density of the pine and oak wood blocks with the density of water (1 g/cm 3 ). Which is greater? Water has a greater density than both the pine and oak wood blocks in this scenario. The Oak wood block is closer to water as its density is 0.695 g/cm3 whereas the density of pine wood blocks is only 0.486 g/cm3. Page 4 CLWvSp17

ISOSTASY FORM Name: Section: Would you expect the wood blocks to float in water? Why? Yes, I would expect the wood blocks to float in water because they both have a lower density than that of the liquid that they are in. Now compare the density of water to the densities of the rock samples you determined in Exercise 2. Would you expect any of these rocks to float in water? Why or why not? I would not expect any of these rock samples to float in water because all three have a greater density than that of water. The density of Granite is greater than that of water by 1.69 g/cm3, Basalt is greater by 1.91 g/cm3, and Peridotite is greater by 3.21 g/cm3. But, if they were to float, Granite would float the "highest" because it has the lowest density and Peridotite would float the "lowest" because it has the greatest density. Table 5 1 Pine Block 2 Pine Blocks 1 Oak Block (1) Thickness of block above water (cm) 1.1 cm 2.1 cm 0.07 cm (2) Thickness of block below water (cm) 0.08 cm 1.7 cm 1.2 cm (3) Total thickness of block (cm) 1.9 cm 3.8 cm 1.9 cm (4) % of block above water = (1) / (3) x 100 57.89% 55.26% 3.68% (5) Thickness of water beneath block (cm) = 10 cm – (2) 9.92 cm 8.3 cm 8.8 cm (6) Mass of water under block (g) = (5) x 1 cm x 1cm x 1 g/cm 3 9.92 g 8.3 g 8.8 g (7) Mass of wooden block (g) density x 1 cm x 1 cm x (3) 0.9234 g 1.8468 g 1.3205 g (8) Mass of Water + wooden block (g) (6) + (7) 10.8434 g 10.1468 g 10.1205 g Page 5 CLWvSp17

ISOSTASY FORM Name: Section: Compare the 1 pine, 2 pine and 1 oak scenarios. Are the combined masses of wood and water similar? Is this what you expected? Explain your answer. (Hint: Use Table 5, Row 8) In this scenario, all the combined masses of wood and water are very similar. My recorded combined mass of wood and water for 1 pine block was 10.8434 g, the combined mass for 2 pine blocks was 10.1468 g, and the combined mass for 1 oak block was 10.1205 g. This is not what I had expected, aa I thought that the 2 pine blocks would have the greatest combined mass, but they had the second largest combined mass with the 1 pine block mass being 0.6966 g heavier. However, based on the previous calculations in table 5, I did expect that the 1 pine block and 2 pine blocks would each individually have a greater combined mass than the 1 oak block scenario, which was proved true as it had the lowest combined mass of all 3 scenarios. Page 7 CLWvSp17

ISOSTASY FORM Name: Section: SUMMARY QUESTION: What would happen to the continent of Antarctica if all the ice were removed? To answer this, assume that the ice did not melt into the ocean, e.g., no large volume of water was added to the surrounding ocean. Using what you learned in this lab, what happens to the continent? Where is sea level on the continent now without the ice compared to with the ice? What is this process called? (Hint: This is similar to removing the 2 nd pine block “mountain” used in Exercise 3.) This process would be called glacial isostatic adjustment. This is when continental crust sinks due to its weight increasing because of ice growth, sediment deposit, or lava. The weight of the mass on a continent causes the crust to subside, but when this weight from the ice is removed, the isostatic rebound takes place and the land would be uplifted, creating a valley that would be relatively flat. This is also because the sea level will lower due to the “rebound” and the area of land will increase. Page 8 CLWvSp17