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The question is basically:

During the Cretaceous there were no continental ice sheets. The ratio of O-18/O-16 in sea water was:

a) Higher than today
b) Lower than today

The answer is stated to be b, but it does not make sense as glaciers have a lower 18O/16O ratio and so the ratio should be higher than today. Can someone justify why it's b?

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  • $\begingroup$ Welcome to Earth Science Stack Exchange! If you need any assistance with the site, please visit The Help Center. Also, please see this Meta post on homework questions. The best possible way to keep a homework question from being closed is to show that you have made a good-faith effort to work the problem before posting here. $\endgroup$
    – L.B.
    Commented Apr 27, 2015 at 2:08

3 Answers 3

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It does make sense. As you said - glaciers have lower 18O/16O. This is because they have less 18O overall. Therefore, the oceans have more 18O. But this is today.

At the Cretaceous there were no continental ice sheets, so what we said previously is not relevant. There are no ice sheets to take all of the 16O, so it's all in the ocean. This results in a lower overall 18O/16O ratio.

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There is a quite relevant paper in Science with a fine analysis and interpretation on $\delta$ 18O evolution during the Phanerozoic Eon, which quite summarily display how this ratio depend on the sea level among other things.

Full reference here : Miller, K. G.; Kominz, M. A.; Browning, J. V.; Wright, J. D.; Mountain, G. S.; Katz, M. E.; Sugarman, P. J.; Cramer, B. S.; Christie-Blick, N. & Pekar, S. F. The Phanerozoic Record of Global Sea-Level Change Science, 2005, 310, 1293-1298

Actually, following this paper, sea level was somewhat higher than today during the late Cretaceous, with a 100 $\pm$ 50 m peak during that period (in the abstract). They established the concept of a small and ephemeral ice sheet in Antarctica as well between 100 and 33 Million years. Some significant factors regulating the sea level and oxygen ratios would be, for any given time:

  1. Sea Level
  2. Average global temperature
  3. Tectonic activity and related tectonoeustasy > implying impact on water displacement

Briefly to answer the question, Figure 3 in the paper show the $\delta$ 18O in all time during the Cretaceous as lower than during recent times - so b) is the answer.

Re-plot of Miller et al. 2005 Figure 3 showing oxygen isotopes and sea level change since 100Ma

Redrawn and simplified after Miller et al. 2005

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  • $\begingroup$ I am not sure if it is permitted to show figures stored behind a paywall - if anyone could inform me on such policies and how SE deals with this - well this answer would be more 'graphic' with the actual figure 3... I assume for now that this is not permitted $\endgroup$
    – marsisalie
    Commented Apr 27, 2015 at 1:26
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    $\begingroup$ In the mean time I added a plot I made using the supplementary material of Miller et al. 2005. $\endgroup$
    – plannapus
    Commented Apr 27, 2015 at 6:41
  • $\begingroup$ @plannapus nice one that really does it and really complete the answer IMO $\endgroup$
    – marsisalie
    Commented Apr 27, 2015 at 10:50
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First of all, oxygen isotopes are generally given as 16O/18O so it is confusing the way the questions is asked from a geochemical perspective. My students typically get confused about this. But it is very simple to understand if you keep in mind that 16O is lighter than 18O. Evaporation of ocean water preferentially takes 16O relative to 18O thus decreasing the 16O/18O in ocean water (keep in mind we are talking about a ratio here). During glacial periods, more water is taken out of the oceans and collected in glaciers which have a high 16O/18O ratio. Thus the ocean water during high glacial periods will be low in 16O/18O. Therefore during a period of no glaciation the 16O/18O would be higher than today. Or, in the silly way the questions states 18O/16O would be lower than today

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    $\begingroup$ The first part of your answer seems odd, as most of the classical works (e.g., Epstein & Mayeda, dx.doi.org/10.1016/0016-7037(53)90051-9) use the 18O/16O nomenclature. In fact the definition of delta-18O (en.wikipedia.org/wiki/%CE%9418O) uses the ratio as stated in the question. $\endgroup$
    – arkaia
    Commented Jul 7, 2015 at 18:26

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