You have a cube of ice floating in a glass of water. The question is what fraction of the ice will be above the water line? Assume that the ice is not bobbing.

Most of you have probably heard the answer to this before. But please provide a proof or solution, along with your assumptions.

At http://www.simetric.co.uk/si_liquids.htm there is a table of densities of various liquids.

At http://www.simetric.co.uk/si_water.htm there is a table of densities of pure and tap water at various temperatures.

Using those tables as means, we can easily establish an acceptable range of values for the amount of ice which floats above the water.

An extreme case where we use sea water at 77°F and an ice cube made of pure water we get 915/1021.98 = 89.53% of the cube submerged.  That exposes about 2% more of the cube than Charlie demonstrated in his answer.  Colder sea water would expose more of the pure water ice cube.

A cube made of see water suspended in pure water would probably expose about 2% less than what Charlie demonstrated so I would venture to guess about 93% of the cube would be submerged.  Warmer pure waterr would cause more of the sea water ice cube to become submerged.

Therefore, an acceptable (to me) range of submerged ice cube would be between 89-93%.