Ice cube in water question

[missdoctor]

This is a question I came across in TBR Gen Chem 2 and was wondering why the answer is what it is..

If an ice cube floats in a glass of water in such a way that the surface of the water is flush with the brim, and the ice cube sticks out above the level of the water, what will occur as the ice cube melts?

A. The water level will drop below the top of the glass.
B. The water will overflow the top of the glass.
C. The water level will remain flush with the top of the glass.
D. The water level will rise above the top of the glass, but will not overflow it.

Answer: C

Explanation: For this question, it is important to remember that when ice melts, it becomes water. The mass of the water displaced by the floating ice cube is equal to the mass of the water generated by melting the ice cube. When the ice cube melts, the water that is formed has exactly the same mass and density as the water displaced. This means that it also has the same volume, so it fills the volume occupied by the submerged portion of the ice cube. The net result is that the level of the water remains constant at the top of the glass. As more and more ice melts, the ice cube drops lower into the water, but the top of the waters remain the same.


I was thinking that because ice is less dense than water, if you put an ice cube in water it will displace more mass than if it was the same amount of water being added because when water freezes, it expands.. and then when that ice melts, it compresses again so that original mass that was displaced won't be equal to the mass that actually melts. Though there must be something wrong with my logic because A isn't the right answer.

Edit: Okay now I realize that my logic is right but that assumes no part of the ice cube is sticking out.. which it is. So I guess the part that is sticking out, when it melts, will compensate for the decreased water level after the submerged part of the ice melted? How are you supposed to know that there was just enough ice cube sticking out to compensate for that? Or is my logic still not right? lol

Thanks for any help!

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[Isoprop]

This is a pain to answer online b/c the math requires a lot of subscripts. First, let's define some variables.
Mice = mass of ice
pw = density of water; Vw = volume of water; Mw = mass of water
Vt = volume level that is held by the cup
Mt = total mass
Vsub = volume of ice submerged in water

The buoyant force is equal to the weight of the water since it's in equilibrium.
(Mice)g = (pw)(Vsub)g
Mice = (pw)(Vsub)

We also know that the total volume that is held by the cup is:
Vt = Vsub + Vw

Finally, we know that the total mass of the ice and the water must be constant.
Mice + Mw = Mt = constant

Plug in the first equation into the second equation and rewrite Vw in terms of mass:
Vt = (Mice/pw) + Vw
Vt = (Mice/pw) + (Mw/pw)
Vt = (Mice + Mw)/pw

Look at the final equation. We know that Mice + Mw is constant, and pw is also constant. So Vt is constant. Therefore, the volume level held by the cup is constant.

 

[dextor2003]

i just thought of it like this: ice is less dense than liquid water, so that means, conserving mass, the ice cube must take up a larger volume than it will as liquid...so if the ice melts, the liquid will actually take up less volume than it will as a solid...so if less volume is taken up, shouldn't the level of the water decrease ??

 

[krnpanda]

i just thought of it like this: ice is less dense than liquid water, so that means, conserving mass, the ice cube must take up a larger volume than it will as liquid...so if the ice melts, the liquid will actually take up less volume than it will as a solid...so if less volume is taken up, shouldn't the level of the water decrease ??

i honestly thought this was the reasoning too

 

[Moosecakes]

This is a pain to answer online b/c the math requires a lot of subscripts. First, let's define some variables.
Mice = mass of ice
pw = density of water; Vw = volume of water; Mw = mass of water
Vt = volume level that is held by the cup
Mt = total mass
Vsub = volume of ice submerged in water

The buoyant force is equal to the weight of the water since it's in equilibrium.
(Mice)g = (pw)(Vsub)g
Mice = (pw)(Vsub)

We also know that the total volume that is held by the cup is:
Vt = Vsub + Vw

Finally, we know that the total mass of the ice and the water must be constant.
Mice + Mw = Mt = constant

Plug in the first equation into the second equation and rewrite Vw in terms of mass:
Vt = (Mice/pw) + Vw
Vt = (Mice/pw) + (Mw/pw)
Vt = (Mice + Mw)/pw

Look at the final equation. We know that Mice + Mw is constant, and pw is also constant. So Vt is constant. Therefore, the volume level held by the cup is constant.

I love reading your posts, just thought I'd let you know. Thanks for taking time out of your day to help everyone.