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This question already has an answer here:

I have this:A piece of ice is gently placed on the surface of water filled in glass tumbler, so the water rises to the brim. What will happen to the level of water when the ice melts?

I still don't get it why the level of water remains the same. I tried to find out from other sources but they proved it using fancy physics formulas which I don't have and know currently. Please, can anyone give me a hand in this?

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marked as duplicate by John Rennie, sammy gerbil, Ben51, Kyle Kanos, ahemmetter Dec 3 '18 at 8:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ yena, this works when the ice is floating in fresh water. If the ice floats in salt water, the water level actually rises as the ice melts. $\endgroup$ – David White Dec 1 '18 at 16:39
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The rule of buoyancy (see Archimedes Principle) is that the buoyant force on an object is equal to the weight of the fluid it displaces.

When an ice cube with weight $W_{ice}$ is floating in water, the buoyant force (which again, is equal to the weight of the displaced water) must be equal to $W_\text{ice}$. Therefore, the amount of water being displaced has the same weight as the ice cube. If $\rho_\text{water}$ is the density of water and $V_\text{disp}$ is the displaced volume, then $\rho_\text{water}gV_\text{disp} = W_\text{ice}$, and so $V_\text{disp} = \frac{W_\text{ice}}{\rho_\text{water}g}$, where $g=9.8 \frac{\text{m}}{\text{s}^2}$ is the acceleration due to gravity.

When the cube melts, its volume decreases so its density becomes that of water. However, the volume of glass-water being displaced by the cube-water is the same as before, because the cube-water weighs the same as the original ice cube did (no weight is lost by melting), so $V_\text{disp} = \frac{W_\text{cube-water}}{\rho_\text{water}g}=\frac{W_\text{ice}}{\rho_\text{water}g}$.

If the volume of water being displaced is the same, then the water level stays the same.

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When the ice cube is floating the weight of the displaced water is equal to the weight of the ice.

Now think about it, you have a mass of displaced water and an equal mass of floating ice.
If that mass of ice melted it would have the same mass when it became water but the melt water would have a smaller volume than the ice but equal to the volume of the water which was displaced when the ice was floating.

So the melted ice exactly fills a volume exactly equal to the volume of the displaced water.

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