Great analysis! How much would you have to tweak it, to account for the weird way water behaves between 32*F and 39.2*F? (Or whatever the corresponding figures are for sea water.)
Good question… The density of water at 32.2 degrees F is 8.3452 lb/gallon so the increase in ocean height if the entire ocean temperature dropped from 39.2 to 32.2 degrees F would be to 1.00013 x 12,080 which is 12,081.57 feet. In other words, if the temperature of the ocean dropped from its densest point to the point just before it began to freeze, the increase in height would be 1.57 feet (18 7/8”).
You did raise a good point though which is not something I took into consideration in my earlier posts (or this one for that matter)….. the data I used is based on regular good old water, not sea water. For the record, ocean water has a typical salinity of approximately 3.5% which means that it is approximately 2 ½% more dense than standard water. The other thing about seawater is that the salt lowers its typical freezing temperature into the 28 degree F range so its density at that point would actually be a bit lower than the aforementioned density at 32.2 degrees F (if just taking into consideration the impact of temperature alone.) Bottom line is that to do all this in a highly precise way would need to take into consideration that the density of ocean water depends on three things… temperature, salinity and depth (which affects the pressure it is under). Having said that, I don’t think that any of these three have much impact on what I suggested my earlier posts which was just a quick and dirty estimate. However, somebody wanting to do this in a detailed way will of course be taking into consideration the fact that these three things will make some difference to the calculation….but at a first glance, not an appreciable difference.
I haven’t verified this but here is an online calculator I found that dials in the above three factors when calculating water density…. http://www.csgnetwork.com/water_density_calculator.html