# Rising sea levels due to thermal expansion

According to NASA, one of the main reasons for the rising of sea levels is the increase in ocean temperature. The increase was of $$0.4^\circ \text{F}\sim 0.2^\circ C$$ for waters with depth $$\sim700\text{ m}$$. The observed sea level rise in that period was of around $$\sim 10\text{ mm}$$.

If the radius of earth is $$R$$, sea level heigth is $$h$$, and $$\beta$$ is the volumetric temperature coefficient at $$17^\circ C$$, a very simple model gives the volume change by $$\Delta V = 4\pi R^2h\beta\Delta T.$$ The volume of the thin spherical shell due to volume change is $$\Delta V = 4\pi R^2\Delta h.$$ Hence $$\Delta h = h\beta\Delta T$$. Considering that $$\beta = 1.7\times 10^{-4}/^\circ C$$, we find $$\Delta h \sim (700\times 10^3\text{ mm})\times 1.7\times 10^{-4}\times 0.2 =23.8\text{ mm}.$$

This is huge, much larger than the observed sea level rise. What is the greatest source of error of this calcultation? I want to make this calculation in class and then present the reasons why it is not precise.

• My guess is that the 0.2°C change is for the ocean surface, the change will be less than that at 700 m. – PM 2Ring Jun 24 '19 at 15:09
• The calculation is within the same order of magnitude. The proper statement is not that this is a huge and much larger value. Also, when you drop the mean temperature to 10 $^o$C, you will cut your relative expansion in half. In summary, the very fact that you can do this type of simple calculation and end up with a comparable order of magnitude (e.g. mm rather than km) is in itself informative both in terms of the geometry approximation (a thin shell) let alone the application of properties of sea water. – Jeffrey J Weimer Jun 24 '19 at 15:31