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Ice Floating in Water (Posted on 2005-04-21) Difficulty: 3 of 5

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 re: Uncertainty | Comment 2 of 31 |
(In reply to Uncertainty by Jonathan Chang)

"First, the weight of the ice.  The bigger the ice, the heavy that the ice is.  The bigger the ice, the less fraction of ice that floats above the water line.  Just imaging an ice weights 100 tons as compared to a pound of ice, which ice would have a bigger fraction of ice to be above the water line.  Of course!  The lighter ice."

Regardless of the size of the ice cube (it is specified as being a cube, rather than a boat shape for example), the buoyant force is equal to the amount of water displaced--the volume of the cube below the surface of the water.  So long as the two pieces of ice have the same density, the amount below the water will be proportional to the total amount of ice, and therefore the same fraction of the total.

Ice at 0 degrees Celsius has a density of .9150 g/cm^3.  Water at 0 deg C has a density of .9999 g/cm^3, increasing to 1.0000 at 4 degrees C--the maximum--and decreasing to .9982 at 20 deg C, room temperature.

So an ice cube in a glass of ice water will be 91.5% below the water line. And if placed into room-temperature water, the ice cube will be .9150/.9982 = 91.66% below the water level.  As it melts of course it will maintain 91.66% of its decreased volume at any given time below the water level.

  Posted by Charlie on 2005-04-21 14:23:08
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