Chapter 5 · The Oceans

Answers · Sharp, Principles of Stable Isotope Geochemistry, 1st ed. (2007) · ← Problems

Answer 1

Although there is significant evaporation, most of the water is returned to the ocean by precipitation, so that the entire evaporation–precipitation cycle is a closed system.

Answer 2

See Chapter 9 for further discussion.

Answer 3

Four trajectories are shown in Fig. 5.1: equatorial precipitation, trade winds evaporation, polar contribution, and freezing. (1) The equatorial precipitation will lower salinity, but only slightly, as the dilution is countered by precipitation. The curve is relatively flat because the source of water is locally derived evaporation of the ocean, so that the δ18O value of precipitation is similar to that of evaporation. (2) Trade wind evaporation will raise both δ18O and salinity values because the vapor that is removed has a lower δ18O value than its source (the ocean) and a portion of the vapor is removed by transport to continental settings. (3) Polar freshwater contribution has a strong slope, because the infiltrating freshwater has very low δ18O values. (4) Freezing has a large effect on the salinity because salt is not incorporated into ice. The δ18O changes little because the fractionation between ice and water is small. The negative slope in Fig. 5.1 is explained because Δ18O(ice–water) is positive.

Answer 4

The Pacific deep waters do not originate at the surface of the Pacific Ocean. Instead, they are a mixture of the North Atlantic Deep Waters and the Antarctic Bottom Waters. See Fig. 5.3 and accompanying discussion.

Answer 5

Antarctic waters freeze. Ice formation concentrates dissolved salt in the remaining water, driving up salinity. However, the isotopic fractionation between water and ice is small, so that the isotopic composition of the remaining water changes only slightly.

Answer 6

The δ18O values increase with depth because oxidation of organic matter preferentially removes the light isotope, and in doing so, the concentration of oxygen decreases with depth. In contrast, δ13C values decrease with depth because there is a removal of light isotopes at the surface by incorporation into organic material and a rerelease at depth by oxidation of the light organic material, the latter causing a decrease in δ13C values. The concentration of dissolved carbon increases with depth because of the biological pump, discussed further in Chapter 7. [The printed text reads “an increase” in the final clause about δ13C at depth, which contradicts its own opening sentence; the sense here follows the opening sentence. Check vs print.]

Answer 7

Not printed in the 1st ed. (essay question).

Transcribed from Z. Sharp, Principles of Stable Isotope Geochemistry, 1st ed. (Pearson Prentice Hall, 2007), end-of-chapter problem set (published source). Reference-scale names are written VSMOW, VPDB and VCDT where the original used SMOW, PDB and CDT, with the original numerical constants unchanged; historical usages (e.g. Craig's definition of SMOW) are kept as printed.

The source PDF's text layer scrambles mathematics, so equations and some values were reconstructed during transcription. Bracketed notes and anything tagged reconstructed or check vs print are editorial (constructed), not part of the original, and the warn-tagged items should be verified against the print copy.