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[Darvi]

Is that like tiramisu but without the cheese?

[ToonyMan]

Is that like tiramisu but without the cheese?

Much much cheaper, junk food cheaper.

[Darvi]

Eh, either way I couldn't do it. Mostly because I don't like coffee.

[Earthquake Damage]

So does your challenge allow wrapping the coffee cake in something to contain the crumbs, so they don't fall?  Or perhaps finding a coffee cake that's moist and sticky enough not to crumble?

Note that I don't really know what constitutes coffee cake, so maybe the second question is impossible.

[Il Palazzo]

2.Now, if there is a stone in the ice cube.
Let's assume it weights just as much as 1 liter of water(1kg). In this case, the cube is fully submerged.

Which it isn't, as stated by the question.

If the cube has got an additional weight attached(the stone) then it is more submerged than it would be otherwise(i.e.just pure ice, as analyzed in 1.). It doesn't matter to what extent exactly it is submerged, as the effect on the system will be of the same type only less extreme. Assuming that the stone weights as much as half of the ice cube simply makes the qualitative calculations easier.
In other words, if it weights anything between infinitesimaly close to zero and 1kg, the additional weight will cause the ice to submerge more than it would otherwise, displacing more water than the submerged part's own weight.
You can do the calculations with whatever weight you like(as long as it's not too heavy), but the extreme case is much easier to understand and calculate.

[ToonyMan]

Eh, I was just wondering with eating it normally, I'm sure people would find tricks with the wrapper and such and I don't care much about that.

[G-Flex]

2.Now, if there is a stone in the ice cube.
Let's assume it weights just as much as 1 liter of water(1kg). In this case, the cube is fully submerged.

Which it isn't, as stated by the question.

If the cube has got an additional weight attached(the stone) then it is more submerged than it would be otherwise(i.e.just pure ice, as analyzed in 1.). It doesn't matter to what extent exactly it is submerged, as the effect on the system will be of the same type only less extreme. Assuming that the stone weights as much as half of the ice cube simply makes the qualitative calculations easier.
In other words, if it weights anything between infinitesimaly close to zero and 1kg, the additional weight will cause the ice to submerge more than it would otherwise, displacing more water than the submerged part's own weight.
You can do the calculations with whatever weight you like(as long as it's not too heavy), but the extreme case is much easier to understand and calculate.

It doesn't matter that the stone makes it "more submerged". At all. The answer to the question remains the same. Now, if the stone itself were melting, it would be a different story, but I already addressed the issue.

The object is buoyant. The object sheds X of its mass in the form of water. X mass of water is added to the fluid, which would cause the water level to rise, except that X mass of water is also no longer being displaced by the object, so nothing changes. This is perfectly generalizable and doesn't matter what the actual density or mass of the object is (unless it becomes denser than water, which is not the case).

[smigenboger]

I'm so not shoveling my driveway right now.

[Darvi]

Sure, the stone pulls the ice down. But when it gets released, the ice goes up, displaces less water, causing the water level to go down.

[Earthquake Damage]

Eh, I was just wondering with eating it normally, I'm sure people would find tricks with the wrapper and such and I don't care much about that.

Pfft.  You're no fun. 

[ToonyMan]

I was saying you could do whatever you want.

[Il Palazzo]

It doesn't matter that the stone makes it "more submerged". At all. The answer to the question remains the same. Now, if the stone itself were melting, it would be a different story, but I already addressed the issue.

The object is buoyant. The object sheds X of its mass in the form of water. X mass of water is added to the fluid, which would cause the water level to rise, except that X mass of water is also no longer being displaced by the object, so nothing changes. This is perfectly generalizable and doesn't matter what the actual density or mass of the object is (unless it becomes denser than water, which is not the case).

Of course it does matter.
X mass of water is added to the fluid volume, but at the cost of more than the volume of X of water - the ice is less dense than water, remember?
If not for the stone "pressing" the ice cube down, the above wouldn't matter, as the extra volume of ice would always stay above water.

[Darvi]

You know what, I'll reproduce this experiment and tell you the answer later, okay?

[G-Flex]

X mass of water is added to the fluid volume, but at the cost of more than the volume of X of water - the ice is less dense than water, remember?

Dude, did you not hear what I said. Look up Archimedes' Principle. How many times do I have to explain it?

A buoyant object displaces its weight in water. If the object weighs 100g, then 100mL of water (assuming 1g/mL is the density of water) is displaced by it. If the object sheds 100g (100mL) of water, then the object also displaces 100g (100mL) less water. It balances out. It does not matter if the object is slightly more dense than pure ice is, as long as it is floating.

You have, say, 1000mL of water. You have an object weighing 100g. This object will -- assuming it is buoyant, which in this case it is -- displace 100g (100mL) of water. The total volume of the water and the part of the object that is submerged is the total of the surrounding water (1000mL) and the volume displaced below the surface by the object (100mL), or 1100mL.

10g of the object melts into water equivalent to the water it is floating in. This adds 10mL of water to the surrounding water. However, the object now only weighs 90g. There is now less of the object underwater, as only 90g (90mL) of water is being displaced by it. The total volume of the water and the part of the object that is submerged is the total of the surrounding water (1110mL) and the volume displaced below the surface by the object (90mL), or 1100mL.

The fact that there is a stone buried in the ice is 100% irrelevant to what I am saying here, unless it causes the density of the object as a whole to go below that of water, which is not the case.

This really isn't even contestable. This is all just a basic Archimedes' Law question at this point, with the original question being a sort of logic/scientific understanding puzzle.

[Tellemurius]

listen to him, i understand the problem and know the answer and hes got it, im not posting mine because im lazy.