Chapter 4 · The Hydrosphere

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

Answer 1

The mass balance is given by δ¹⁸O_{River A}·X_a + δ¹⁸O_{River B}·(1 − X_a) = δ¹⁸O_outflow. Solving −4.5·X_a + (−12.7)(1 − X_a) = −9.2, we obtain X_a = 0.42. Thus, 42% comes from River A, 58% from River B. [The printed text reads 68%, a misprint. River B, with the lower δ¹⁸O, is sampling the higher-altitude waters (constructed).]
We have an input of −9.2‰. (Remember, all other parameters are constant from part (a), so a similar equation can be set up.) The two sinks are the outflow (−4.9‰, the value of the lake) and vapor (−4.9 − 9 = −13.9‰). Evaporation accounts for 48% of the loss.

[Worked grid (constructed). The flow and summer-outflow rows carry the printed percentages through to m³/day, which the printed answer leaves implicit.]

Answer 2

At low temperatures, the fractionation between liquid and vapor is large. Therefore the Rayleigh distillation effect becomes more important and overwhelms all other considerations.

Answer 3

The δ¹⁸O and δD values of the two reservoirs (ocean and cryosphere) are given by the equation δ_total = Σ_i x_iδ_i. For oxygen, this yields 0.972 × 0.0 + 0.0215 × (−30) = −0.65‰. For hydrogen, the value is −4.95‰.

[Worked grid (constructed). The ice δD of −230‰ is reconstructed from the printed −4.95‰; check vs print.]

Answer 4

The Rayleigh fractionation equation for this process is

δ¹⁸O_v = [δ¹⁸O_v,i + 1000]·F^(α−1) − 1000Eq 4.16

δ¹⁸O_l = α·(δ¹⁸O_v + 1000) − 1000

From the Horita and Wesolowski equation and the given constants, the α value at 5 °C is 1.01126. The vapor will be lighter than the coexisting liquid. The values are plotted in the figure. [Figure not reproduced; the chapter's interactive Rayleigh plot draws the same curves.]

[Worked grid (constructed). Full precision gives α = 1.01124 vs the printed 1.01126. The plot shows vapor and instantaneous liquid against F with the model curves drawn from the fitted parameter cells.]

Answer 5

From the definition of δ, for (¹⁸O/¹⁶O)_SMOW = 1.008·(¹⁸O/¹⁶O)_NBS-1, δ¹⁸O(NBS-1) = 1000·(1/1.008 − 1) = −7.94‰ vs SMOW, and −37.7‰ vs PDB. check vs print [The printed PDB value is garbled in the extraction (digits read “22.728”); the value here is converted with δ_PDB = 0.97002·δ_SMOW − 29.98 (constructed).] Likewise, the δD value of NBS-1 is 1000·(1/1.050 − 1) = −47.6‰ vs SMOW. reconstructed

If a sample lies on the meteoric water line, the following equation is valid: δD = 8·δ¹⁸O + 10. For a δ¹⁸O value of −7.94‰, the δD value for a sample falling on the meteoric water line is −53.5‰. Thus, NBS-1 plots well above the global meteoric water line. For the North Carolina meteoric water line, the calculated value is in excellent agreement with the measured value. check vs print [The North Carolina line parameters and the calculated value are lost in the extraction.]

[Worked grid (constructed). NBS-1 sits 5.9‰ above the global meteoric water line in δD.]

Answer 6

Not printed in the 1st ed.

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.