This is a strange question I would like to have some explanation accompanied with:

We have an ice block, with in it a lead block. This entire block is put into a glass of water, resulting in a rise of water level until the edge of the glass. The block stays floating on the surface. What will happen with the water level when the ice melts?

The answer is that it will drop, but how is this even true?? My thoughts are: - block floats on water, so rho(block) < rho(water) - Archimedes law, the Fg of block = Fg of water pushed away. - Since a smaller volume of water has equal mass (and so Fg) to block, and ice is less dense than water, when it melts, no mass is lost, so it will take in less space??? Please help me with this point of thinking.

  • $\begingroup$ The given answer is wrong - the water level will not drop. (I won't say what it actually does because I think this is a homework question and if so it's better for you to work it out for yourself. But the place to start is to learn what happens when the block consists only of ice, and then think about what's different when the lead is there.) $\endgroup$ – Nathaniel Oct 2 '13 at 7:01
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    $\begingroup$ the given answer is correct. $\endgroup$ – gregsan Oct 3 '13 at 11:17

Let's deal first with a pure ice cube.

A 100 gram ice-cube (pure water) is floating in a glass of pure water, filled to the very top. By buoyancy laws the ice-cube must be displacing exactly 100 grams of pure water from the glass. The volume of the ice-cube that is below the water level is the volume of 100 grams of water. But that's what the ice-cube will become when it melts. The water level will remain unchanged as the ice melts and exactly fills the hole it occupied. No more water will overflow, and the level will stay right at the top of the glass.

Now let's add the lead, but in a particular way:

You now have a largish ice-cube, floating as before in a full-to-the-top container of water.

What happens if you carefully put 100 grams of lead on the ice-cube, which settles a little, but remains floating. How much, if any, water will overflow?

Remove the 100 grams of lead from the ice-cube. What will happen to the water level?

Put the lead, without splashing, into the water beside the ice-cube and let it settle to the bottom. What will happen to the water level? BY how much?

Now let the ice-cube melt. How does the water level change? (See first part of answer)

  • $\begingroup$ Ok, what you said in the first part is very logical. However, since the lead cube is inside (!) the larger ice cube, things start to change. Your proposals for testing it out (I tried to do it in my mind) gives me the same result as the asnwerbook gave me: first the lead pushed out the water, then the ice melts and fills some water with it but not enough to get it full again. I still don't get the implications of your answer, please elaborate. And please give me a situation when the lead is in the ice. $\endgroup$ – user209347 Oct 4 '13 at 7:18
  • $\begingroup$ For the last part, I can't do it. The location of the lead in, on ore even around the ice is immaterial. The only relevant information is the volume of the floating object submerged, and the mass of the floating object. Moving the lead in the ice cube may make the ice cube flip over, but it won't make it change from floating to sinking... $\endgroup$ – DJohnM Oct 4 '13 at 16:18

Archimedes principle: upthrust equal to weight of displaced fluid.

$\rho_{ice}*V_{ice}*g$ = $\rho_{water}*V_{submerged~ice}*g$

so $\frac{V_{submerged~ice}}{V_{ice}} = \frac{\rho_{ice}}{\rho_{water}}$

i.e, the fraction a floating ice cube that is under water is equal to the relative density of ice to water--roughly 0.92.

this value is also the same ratio that frozen water shrinks to when it melts. e.g. 100cc of ice will melt into 92cc of water. essentially, an ice cube will melt to occupy the same volume it displaced underwater. so as long as no other forces act on the ice (the ice cube was not forcibly submerged while it was still solid), the water level does not change after the cube is placed in it.

this particular question the lead displaces in weight in water, which is much more than just its volume in water (if it was allowed to sink)--it does so by causing the ice to submerge lower than it normally would on its own. i.e, more than 92% of the ice is submerged because of the lead. once the lead sinks completely, the amount of water it displaces drops by an amount equal to of $({\rho_{lead}}-{\rho_{water}})*V_{lead}$

  • $\begingroup$ But the OP says the ice stays floating on the surface. It matters... $\endgroup$ – DJohnM Oct 3 '13 at 0:31

OK, please comment on my final judgement on this question:

The Ice with lead in it will sink deeper than normal, so when all the ice will melt, it will not replace all the lost water. The lead is there from the beginning on, taking in the volume of the ice, so it will principally stay there. When all the ice is molten, the water level will be below the maximum, and the lead will sink, but since it already influenced the beginning (it forced the ice to push more water out and took in volume of the ice itself, meaning there's less ice) it won't do more influence than that. So, the result is a lower water level.

Please help my thinking in here, since I doubt some parts. I used answers you gave me to form these steps.

  • $\begingroup$ Looks correct... Another take is that the ice cube is a diversion. You've got some lead on a raft, pure and simple. Balsa wood, oil drums, whatever. Because of the added lead, the raft must displace an added volume of water whose mass equals the mass of the lead. When the lead is dumped off the raft, the raft stops this added displacement. And the sunken lead only displaces a volume of water equal to the volume of the lead. The fact that the raft is itself water, and melts, is just clouding the issue... $\endgroup$ – DJohnM Oct 4 '13 at 16:26

The ice block with the lead inside it has displaced more than the lead (individually). Let me take two cases. Case 1 when ice block and lead are in combined state, and case 2, when the ice block have been melted. So now in case1 the water displaced in combined state of ice with lead is equal to the (volume of ice + volume of lead) × density of water. Now in cases 2 the water displaced is equal to the (volume of lead) × density of water. Now you can see that the water displaced in case 1 is more than the case 2. In the case 2 the water will have same water level has that in case 1, because even if ice block melt it will have the same water level has before.

  • $\begingroup$ Youpost had the worst spacing i'veever seen.Please pay attentionto your postquality. $\endgroup$ – Brandon Enright Jul 11 '14 at 17:47

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