Chptr8PrblmSolns - SOLUTIONS FOR CHAPTER 8 8.1 From (8.1),...

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SOLUTIONS FOR CHAPTER 8 8.1 From (8.1), δ 18 O o oo () = 18 O 16 O sample 18 O 16 O standard 1 x 10 3 = 0.0020150 0.0020052 1 x 10 3 = 4.9 Since the sample has more 18 O in it, there would be more glaciation since ice selectively accumulates 16 O, increasing the concentration of 18 O left behind in seawater. 8.2 Plotting delta D versus temperature gives . As shown, δ D changes by 6.19 per mil per o C. 8.3 An ice core with ( 2 H/ 1 H) = 8.100 x 10 -5 . a. Using VSMOW 0.00015575 for deuterium in (8.1) gives D o / oo = 2 H/ 1 H sample 2 H/ 1 H standard 1 x 10 3 = 0.00008100 0.00015575 1 x 10 3 =− 479.8 b. With δ D( 0 / 00 ) = –435 per mil, and 6 per mil change in δ D( 0 / 00 ) per o C, the rise in δ D from –479.8 to – 435 (44.8 per mil) gives a temperature rise of 44.8/6 = 7.5 o C. 8.4 From the equation given for the ice core, T o C = 1.5 18 O o oo + 20.4 = 1.5x 35 ( ) + 20.4 32.1 o C Pg. 8.1
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notice, by the way, that since this sample is for glacial ice, not ocean water or sediment, the negative sign on δ 18 O o oo ( ) means colder temperatures. 8.5 Plotting the ice core data for T( o C) and δ D So: T o C () = 0.1661 D o oo + 72.45 8.6 The flat earth! 1370W/m 2 R E absorbed = E radiated S π R 2 = σ T 4 A = T 4 2 R 2 T = S 2 1 4 = 1370W /m 2 2x5.67x10 8 W/m 2 K 4 1 4 = 331.5K - 273.1= 58.4 o C 8.7 The basic relationship is S = k d 2 . Using d and S for Earth from Table 8.2 lets us find k: k = S d 2 = 1370W /m 2 x 150x10 6 km x 10 3 m/km ( ) 2 = 3.083x10 25 W a. Mercury: S = k d 2 = 3.083x10 25 W 58x10 6 km x 10 3 2 = 9163W/ m 2 Pg. 8.2
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b. The effective temperature (8.7) of Mercury would be: T e = S1 α () 4 σ 1 4 = 9163W/ m 2 1 0.06 4x5.67x10 8 W/m 2 K 4 1 4 = 441K (168 o C) c. Peak wavelength: λ max = 2898 TK = 2898 441 = 6.6 μ m 8.8 Solar flux variation of ± 3.3%, gives a range of S S max = 1370 (1+0.033) = 1415.2 W/m 2 S min = 1370 (1 -0.033) = 1324.8 W/m 2 T e ,max = 4 1 4 = 1415.2W/m 2 1 0.31 4x5.67x10 8 2 K 4 1 4 = 256.2K (-17 o C) T e ,min = 4 1 4 = 1324.8W /m 2 1 0.31 4x5.67x10 8 2 K 4 1 4 = 252K (-21 o The variation from –17 o C to –21 o C is a difference of about 4 o C, or about ± 2 o C. 8.9 After a nuclear war: 2 a. Surface temperature, T s 4 = 240W /m 2 Pg. 8.3
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T s = 240W/ m 2 5.67x10 8 W/m 2 K 4 1 4 = 255K (-18 o C) b. X, Atmosphere to space: Balance Incoming from space = Outgoing to space 342 = 69 + X X = 273 W/m 2 c. Y, Absorbed by earth: Incoming solar has to go somewhere, 342 = 69 + 257 + Y Y = 16 W/m 2 d. Z, Radiation from atmosphere to surface: Balance earth's surface radiation, Y + Z = 240 = 16 + Z Z = 224 W/m 2 8.10 A 2-layer atmosphere: 342 107 X Y 168 24 78 40 W Z W 350 Z T1 T2 390 a. At the surface: 168 + Z = 24 + 78 + 390 Z = 324 W/m 2 b. Extraterrestrial: 342 = 107 + 40 + W W = 195 W/m 2 c. Lower atmosphere: Y + 24 + 78 + 350 + 195 = 2 x 324 Y = 1 W/m 2 d. Incoming: 342 = 107 + X + 1 + 168 X = 66 W/m 2 e. Temperatures T 1 and T 2 (assuming blackbody radiation) can be found from σ T 1 4 = W = 195 T 1 = 195W/m 2 5.67x10 8 2 K 4 14 = 242K (-31 o C) Pg. 8.4
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σ T 2 4 = Z = 324 T 2 = 324W/m 2 5.67x10 8 W/m 2 K 4 14 = 275K (2 o C) 8.11 Hydrologic cycle: evaporation = 78W/ m 2 x5.1x10 14 m 2 x 1J/s W x3600 s hr x24 hr d x365 d yr 2465kJ /kg x 10 3 kg/ m 3 x 10 3 J/kJ = 5.1x10
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This note was uploaded on 04/03/2008 for the course C E 370 taught by Professor Burgos during the Spring '08 term at Pennsylvania State University, University Park.

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Chptr8PrblmSolns - SOLUTIONS FOR CHAPTER 8 8.1 From (8.1),...

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