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The Question

There are three ice cubes floating in three different cups of water.

  • In the first cup, the ice cube has a pocket of air inside.
  • In the second cup, the ice cube has a pocket of liquid water inside.
  • In the third cup, the ice cube has a metal pellet inside.

How does the level of water in each cup change as the ice cubes melt?

(a) no change, no change, no change

(b) up, no change, down

(c) down, no change, up

(d) no change, no change, down (e) down, no change, down


My Work

So, I know that if you have just normal ice, the level of water in each cup will be the same as the ice will displace the same volume of water when melted (since the ice has a lower density). I do not really know how to apply that knowledge when there are different materials within the ice.

My first instinct is that the ice will still be displacing how much it weighs, but the pocket of air will not add to the level of the water. And, I am not sure if the metal pellet or water will act in the same way as ice because they are both denser than ice. Any help?

EDIT: Also, I think it is looking at the water level after it melts. Sorry for the confusion.

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closed as off-topic by Aaron Stevens, Emilio Pisanty, stafusa, John Rennie, Jon Custer Aug 30 at 18:00

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Sounds like a homework question so I will only give conceptual hints:

The ice can melt at both external and internal interfaces.

What determines the rate of melting at an interface?

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  • $\begingroup$ It is a homework question from a statics unit. The rate of melting is based on the temperature difference, the surface area of the interface, and specific heat. $\endgroup$ – Ojasw Upadhyay Aug 28 at 3:00
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    $\begingroup$ Why would the rate of melting matter at all in this problem? $\endgroup$ – BowlOfRed Aug 28 at 8:11
  • $\begingroup$ It is an inhomogeneous medium. The rate of melting will determine how much melting occurs where $\endgroup$ – Paul Childs Aug 29 at 0:51
  • $\begingroup$ Another constraint is that the air pocket is a compressible medium, so liquid melting there changes the air pressure. Internal melting in other cases creates stress which can lead to cracking. $\endgroup$ – Paul Childs Aug 29 at 0:52
  • $\begingroup$ Water that melts in the air pocket won't change the specific gravity of the ice block whereas water that melts at the outer interface will. $\endgroup$ – Paul Childs Aug 29 at 0:54
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Before the ice cube melts, the contents of the pocket are displacing a quantity of water that has the same mass. Imagine a fixed volume, and guess the mass of that size for air, for water, and for metal.

After the ice melts, the ice itself doesn't change the level (as you say). So the only thing that can change it is the substance inside.

If you put that air, that water, or that metal into the container and let the water come to equilbrium, does it displace the same amount of water or a different amount of water? If it displaces more, the water level will rise. If it displaces less, the water level will fall.

P.S. When I first looked at this, I thought the answer would basically be linear with the density of the inclusion (one goes one way, one goes the other, and the third in between). That's not the case. It was much more subtle and confused me for a bit.

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  • $\begingroup$ But the question does not ask for the water level after it melts but as it melts $\endgroup$ – Paul Childs Aug 29 at 0:58
  • $\begingroup$ @PaulChilds, You're quite right. I hadn't thought of it that way. I interpreted it as "the change you get when the pocket melts, releasing the contents." Otherwise, as it melts is undefined, since the pocket could be breached at any point during the melting. $\endgroup$ – BowlOfRed Aug 29 at 1:56

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