The point is, the example of the log can just as well be applied to a block of ice. For simplicity, let's continue with the example of 1 million kg.

If at any moment the ice block would displace more (or less) than 1 million kg. of water, the corresponding buoyancy would also be more (or less) than the weight of the ice block. Therefore, it would not be in equilibrium, and it would rise (or sink) until it is in equilibrium - i.e., until it displaces exactly it own weight in water.

Note that this is under the assumption that the block of ice is floating freely. If there are other forces, the situation changes. For example, if something pushes a block of ice forcefully under water, it would displace its volume in water.

>I'm well aware of Archimedes' Principle, and as for P 299 - yadder-yadder-yadder. I'm not concerned with upthrust and apparent loss of weight - I'm only concerned with VOLUME when it melts. My thinking may be totally erroneous, Hilmar, but you haven't explained what's wrong with my last suggestion/argument.
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>I'm willing to learn. This exercise was about my wondering if the predictions were worse than the possible future reality.
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>Terry
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>>The following exracts from my college physics book ("Physics for Scientists and Engineers") might shed further light on the subject.
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>>p.298: "... Archimedes' principle: the buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by that object."
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>>p.299: "Archimedes' principle applies equally well to objects that float, such as wood. In general, an object floats on a fluid if its density is less than that of the fluid. For example, a log whose specific gravity is 0.60 and whose volume is 2.0 m3 will have a mass of 1200 kg. If the log is fully submerged, it will displace a mass of water m = (density)V = (1000 kg/m3)(2.0 m3) = 2000 kg. Hence the buoyant force on it will be greater than its weight, and it will float to the top. It will come the equilibrium when it displaces 1200 kg of water, which means that 1.2 m3 or 0.60 of its volume will be submerged. In general, the fraction of the object submerged is given by the ratio of the object's density to that of the fluid."