Lab 4 GEOL 1330 finished

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University of Texas *

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1330

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Geology

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Dec 6, 2023

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GEOL 1330 N ame: Brandon Perez Global Warming Lab #4 1. (20 points) Go to the online energy balance model available at: https:// climateresearchgroup.uta.edu/edu_outreach/studentProject/index.html . It calculates Earth’s average temperature in dependence of solar irradiance, Earth’s reflectivity (albedo, a), and Earth’s emissivity. With incoming energy being balanced by outgoing energy, a solar irradiance of 1365 Wm -2 , and the global albedo set to 30% (=0.3), the model calculates an average global temperature of 14.69°C (=”Year 0”, please disregard the other years). In the lower right, you can change the values for albedo, irradiance, and emissivity. Sunspot activity might change the irradiance by about 0.1 %. Assuming that the given irradiance of 1365 Wm -2 is a mean value, by how much does global temperature change if you increase the irradiance by 0.05 % or decrease it by 0.05 %? Please show which values you are using in the model. When it comes to the value of 1365 being decreased by 0.05 percent it leaves a value of 1364.3175. The original global temperature was 14.69 which is 0.03 points more then when using the decreased value as you end up with a global average of 14.66. N ow with the original value of 1365 being increased by 0.05 percent it leaves us with a new value of 1365.6825 which gives us a global temperature of 14.73 which is a a 0.04 point difference. By how much does global temperature change if the albedo decreases from 30% to 20% (using 1365 Wm -2 for solar irradiance)? The albedo change brings the original value of 14.69 up to 24.46 as the average temperature. 2. (10 points) Electromagnetic radiation (“light”) travels at a speed of ~3 x 10 8 m s -1 . The distance from Sun to Earth is ~150 x 10 6 km. How long does radiation from the Sun take to 1
reach Earth? Show your calculation. Speed of light (c) = 3 x 10^8 m/s Distance from Sun to Earth = 150 x 10^6 km Convert to meters to km and simplify Distance = 150 x 10^6 km = 150 x 10^6 km x 1000 m/km = 150 x 10^9 m Time = (150 x 10^9 m) / (3 x 10^8 m/s) Cancel out the units and divide to get 500 seconds or about 8minutes and 30ish seconds 2
3. (20 points) Draw schematically two curves into the figure below, one for the zonally averaged incoming visible short wave (Q SW ) radiation from the sun (solid curve) and one for the zonally averaged outgoing long-wave infrared (Q LW ) radiation (dashed curve) from the Earth’s surface depending on latitude. Label the two curves. Point out areas of radiative surplus (+) and deficit (-) on the Earth’s surface. 4. (20 points) Draw the major wind directions for the northern and southern hemisphere in the figure below and label the wind belts. 3 30°N 60°N 30°S 60°S Qlw Qsw Deficit (-) Deficit (-) Surplus (+) Sub polar low Subtropical high Intertropial convergence zone Subtropical high Sub polar low ~ nee x ** XXX + kX
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5. (10 points) Explain where and how deep-water forms in the ocean. 6. (20 points) The figure below shows average annual ocean salinity values. Explain why salinity in the polar regions is lower than that of the subtropical regions around 30 degrees latitude. How do you expect salinity values for polar regions to change during winter and during summer? 4 Deep water is formed when air temperatures is cold and the salinity is quite high. This combination makes the water denser and causes it to sink to the bottom. The polar regions have less salinity that those regions around 30 degrees latitude for several reasons. For one the melting ice; the melting ice releases freshwater which when “released” into surrounds bodies of water it lowers the salinity. A second reasons is the formation of ice, during winter the formation of ice expels salt away when melting back into freshwater. This relates to the changes of salinity during winter; the formation of ice expels the salinity present in the freshwater plus with the more ice on the surface it restricts some salinity reaching the atmosphere. During the summer salinity can be expected rise mainly due to rise in temperature causing increased evaporation.
Teamwork in the lab is encouraged. However, each student must provide his/her own answer. Copying of answers results in a grade of zero for all students involved. 5