perplexus.info :: Science : Ice Floating in Water

Ice Floating in Water (Posted on 2005-04-21)

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.

See The Solution Submitted by np_rt

Rating: 3.0000 (8 votes)

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Solution Based on Densities

| Comment 9 of 31 |

The principle of physics involved is that a floating object displaces an amount of water equal to its own weight. Hence

(Weight of Ice) = (Density of Ice) × (Volume of Ice) = (Weight of Water Displaced) =

= (Density of Water) × (Volume of Water Displaced)

or

(Volume of Water Displaced) / (Volume of Ice) = (Density of Ice) / (Density of Water).

But

(Volume of Water Displaced) / (Volume of Ice) =

= (Submerged Depth of Ice Cube) / (Total Height of Ice Cube).

Hence

(Submerged Depth of Ice Cube) / (Total Height of Ice Cube) =

= (Density of Ice) / (Density of Water).

The fraction of the cube above water is thus 1 minus the ratio of the density of ice to the density of water -- normally a little less than 10%.

Posted by Richard on 2005-04-21 20:48:30

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