Water with Ice

You have a glass of water and an ice cube floating in it. When the ice cube melts, will the water level increase, decrease or remain the same?

It will remain the same. The amount of water that the ice cube displaces is equal to its mass. Since the mass does not change and the density of water is equal to 1, the extra water after melting will be the same amount as the displaced water before that.

Unknown Author

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Responses

1. So if all the icebergs melt, should we all feel safe since the ocean levels won’t rise?

1. Hey K P, sea water is denser than fresh water and that’s why melted icebergs will cause sea levels to rise. This does not apply to the puzzle with the ice cube, since the densities are the same.

2. volume displaces the water, not mass. a block of gold the same size as the ice cube, will displace the same amount of water. therefore when the ice melts, its volume will decrease, therefore the level of the water should decrease……. I think.

1. The only caveat is that when the ice cube melts, it turns liquid and becomes part of the water in the glass:)

1. Daniel is right i think. According to the Archimedes principle, volume displaces the water, not weight. Since water is one of the few molecules that increse in volume when going into solid form (because of the forming of crystal structures). The volume will decrease when the ice cube melts.

Your argument about the weight however has some sort of truth to it. The weight of water does stay the same. But the density increases when going from solid to liquid. This also shows by the following formula: density = mass / volume

1. Hey David, please check the principle of floatation here:
https://en.m.wikipedia.org/wiki/Archimedes%27_principle#Principle_of_flotation

The weight of the floating cube is equal to the weight of the displaced water. The weight of the cube remains the same after it melts, and since the density of water is 1, its volume ends up the same as the volume of the displaced water.

Please note that the argument above wouldn’t work if the ice cube was submerged; in that case you are correct that the volumes would be the same, not the weights.