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Chptr8PrblmSolns

# Chptr8PrblmSolns - SOLUTIONS FOR CHAPTER 8 8.1 From(8.1 18...

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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 δ 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 δ 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 δ D o oo ( ) + 72.45 8.6 The flat earth!