SPU25_Sp18_HW1_KEY.docx - SPU 25 Energy and Climate Vision for the Future Spring 2018 Homework Set#1 Climate Basics DUE Monday 2\/12(1:00 pm in lecture

# SPU25_Sp18_HW1_KEY.docx - SPU 25 Energy and Climate Vision...

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A = 1.28*10 14 m 2 I = 1.76*10 17 W We calculate the cross-sectional area of the earth the multiply by the rate at which the sun is radiating energy to earth. (1 pt) Cross-sectional area of earth = r π 2 = π * (6.38 * 10 3 * 10 3 m) 2 = 1.27877 * 10 14 m 2 (1 pt) Solar energy intercepted = Cross-sectional area * rate at which the sun is radiating energy to earth = 1.27877 * 10 14 m 2 * 1379 W/m 2 = 1.76342 * 10 17 W The total rate at which solar energy is intercepted by earth is 1.76 * 10 17 W. b. Assuming that the energy from the Sun is eventually distributed equally over the entire surface area of the Earth, use your answer from part (a) to calculate the current average rate of incoming solar radiation. Express your answer in Watts per meter squared. Recall that the surface area of a sphere is 4 r π 2 . (2 points) I 0 = 1379 W.m -2 Divide by 4 to get I S = 344.8 W.m -2 We take the answer from part a and divide it by the surface area of earth (a sphere) (1 pt) Surface area of earth = 4 r π 2 = 4 π (6.38 * 10 3 * 10 3 m) 2 = 5.11507 * 10 14 m 2

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• Fall '10
• BOB
• Energy, Greenhouse gas

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