Is the amount of heat required to melt a $0^\circ C$ 10cm x 10cm x 10cm cubic ice equal to the amount of heat required to melt $10^{15}$ piece $0^\circ C$ 1$\mu$m x 1$\mu$m x 1$\mu$m cubic ice ?

I think it should be equal.

But something confuses me.

I have two same $0^\circ C$ piece of ice and I cut one of them into half without melting it (which uses energy). Then, I melt them all.

If the amount of heat required to melt two half pieces of ice is equal to the amount of heat required to melt a single piece of ice, why cut-before-melt process use more energy to turn ice into water?

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    $\begingroup$ I don't understand the last sentence. Why do you think you need more energy for melting after you cut the block? $\endgroup$ – Jasper Apr 26 '18 at 13:06
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    $\begingroup$ I think if you do it perfectly it'll be slightly cheaper to melt the two halves, and the amount of energy saved will be exactly the energy spent cutting them. $\endgroup$ – Ryan Thorngren Apr 26 '18 at 13:07
  • $\begingroup$ Because "If amount of heat to melt two half piece of ice equal to amount of heat to melt a piece of ice" amount of heat in cut-before-melt process is (amount of heat to melt)+(energy to cut them into half) while amount of heat to melt a piece of ice is only (amount of heat to melt) I think (amount of heat to melt two half) = (amount of heat to melt full piece) - (energy to cut them into half) but why? and why my high school teacher never talk about them? $\endgroup$ – Ro Theory Apr 26 '18 at 13:13
  • $\begingroup$ Don't forget surface tension. It's a small correction, but ... $\endgroup$ – Bert Barrois Apr 26 '18 at 13:13
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    $\begingroup$ I think the cut-before-melt process in real life is more messy than just breaking molecular bonds. If you split a block of ice with an axe, you will deliver mechanical energy by friction. If you want to grind your ice into dust, you will have to consider carefully how much power your grinding device requires, and how much of that goes into heating the ice dust. My guess is, pulverizing the ice into 1 μm particles will raise its temperature by at least 10 K. $\endgroup$ – user27542 Apr 26 '18 at 16:05

It's approximately the same energy to melt little chips of ice and one big one, but not exactly because there is energy involved in creating surfaces, called the "surface energy" (or surface tension). However, an introductory physics course usually wouldn't include this in its model of melting, so you might not have heard of it.

When you cut something, you break atomic bonds in it, and this requires energy input. The energy is now a bit higher, which would suggest that the energy to get to a melted state would be a bit less.

However, we have to account for the surface energy of the water as well. When the tiny chips of ice melt, they create a large surface area of water. Water has surface energy, so the many tiny drops of water would have more energy than the one big drop of water you'd get from melting a single chunk of ice. Whether it takes more energy to melt the little chips of ice or the big block of ice depends on whether the surface energy of the ice or the water is higher. If the water has higher surface energy, you'd be putting more energy into the water during melting, and it would take a bit more heat to melt the small chips. If the ice has higher surface energy, it would take a bit less energy to melt the small chips. (All of this ignores gravity. If the gravitational energy is changing as you melt/freeze, you'd have to account for that as well.)

This extra energy in the little pieces of ice/water is stored in the surfaces. If you took many tiny drops of water and let them all converge, they would heat up as they did so, and you'd wind up with hotter water than you started with because the surface energy would turn into thermal energy in the water - that's where all that extra energy you put in to make the little drops would show up.

The effect is over all pretty small because atoms are small, meaning only a small fraction of them are on the surface for any macroscopically-sized stuff. The surface energy of water is about 0.07 J/m^2, while the energy to melt water is about 334 kJ/kg. So even for tiny droplets of water with radius 1 micron, the surface energy is only about 0.06 percent of the energy to melt that much ice; a small effect.

Finally, it's not clear just from everyday experience that breaking ice apart requires energy, in the sense that the broken-apart ice has higher energy than the ice you started with. This is true, but doesn't follow just from the fact that in real life, when you hack at ice with a chisel, you're using energy. Most of that energy is going to heat the ice (and surrounding environment).

  • $\begingroup$ "Finally, it's not clear just from everyday experience that breaking ice apart requires energy, in the sense that the broken-apart ice has higher energy than the ice you started with." So this is interesting because when ice freezes slowly (not supercooled) it forms a crystalline structure which must require some amount of energy. So whenever you break down a block of ice into particles, you are ending up with the same amount of mass without as much structure. Is it necessarily the case that these particles have higher energy? $\endgroup$ – JimmyJames Apr 26 '18 at 19:37
  • $\begingroup$ 3rd paragraph: "and it would take a bit more heat" Which "it" is it? You're discussing 4 separate cases, but have only disambiguated 2 of them: A) many pieces + liquid has higher tension, B) one piece + liquid has higher tension, many pieces + solid has higher tension, one piece + solid... $\endgroup$ – jpaugh Apr 26 '18 at 19:37

I'm not totally sure I understand the question. But I think you may be confusing the amount of heat energy needed to melt the ice with the amount of time needed to melt the ice.

You're correct that the energy needed to melt the ice, given that the ice is already at the temperature of the phase change, depends only on the mass. However the heat responsible for the phase change must enter the ice through its surface. You give the example of splitting a large block of ice into $10^{12}$ pieces. An interesting exercise for you is to compute the surface areas of those two distributions. If the rate of heat flow is proportional to the surface area, it's clear that the micron-sized ice powder will melt more rapidly than the large block.

An example of this that you may have seen is liquid nitrogen, which boils at 77 kelvin. An open bucket of liquid nitrogen may be stable for many minutes. Take that same bucket and dump it on the ground, and the nitrogen vaporizes more or less instantly. The difference isn't the amount of heat that's needed: it's the the surface area through which that heat can enter the material.

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    $\begingroup$ This sounds more like what the OP is getting at... $\endgroup$ – Oscar Bravo Apr 26 '18 at 14:05
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    $\begingroup$ I'm not confusing. I know that more surface area will melt more rapidly. But, I think micron-sized ice powder need less amount of heat to melt than same mass of a block of ice. $\endgroup$ – Ro Theory Apr 26 '18 at 15:14
  • $\begingroup$ WRT the liquid nitrogen, I think it's not just the surface area, but also that the ground has a higher heat content than the air. $\endgroup$ – jamesqf Apr 26 '18 at 17:35
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    $\begingroup$ @RoTheory What is the reason you believe that? And to what magnitude do you think the difference would be? $\endgroup$ – Michael Richardson Apr 26 '18 at 18:26
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    $\begingroup$ @Michael Richardson Because I think, in ideal, amount of energy to do same thing is equal and doesn't depend on process. But I don't know the magnitude. I think it is very small but it is possible to calculate. $\endgroup$ – Ro Theory Apr 27 '18 at 6:20

For simplicity let's assume that every molecule of ice (water) is a little cube and the bonds are between adjacent faces.

Your 10cm cube of ice contains approx. $3.07*10^{25}$ molecules, and thus $9.21*10^{25}$ bonds (i.e three times the amount of molecules, since each cube has six faces and almost each face is shared by two cubes).

A 1um cube of ice is $3.07*10^{10}$ molecules, thus one face of it contains $\sqrt[3/2]{3.07*10^{10}} = 9.80*10^6$ molecules of ice per layer. By cutting out such a cube you have broken thrice that amount of bonds. So for the $10^{15}$ little cubes you have split the 10cm cubed into you have just broken $2.93*10^{22}$ bonds. This is 3000 times less than the total number of bonds initially, but you have a point:

By cutting a large block into 1um cubes you have reduced the number of bonds to break by $0.0316\%$, and thus, to some approximation, the energy required to melt the rest by the same amount.

Mark raises a good point in the comment, that surface energy of the resulting water will play a role. The surface energy was mentioned in his answer, with a figure of 0.06% relative to the energy needed to melt a one-micron piece of ice. However, my result is of the same order. I expect the role of surface energy to be highly variable depending on the initial configuration of our ice dust. If we place each tiny piece far from every other, surface energy might win over. If everything is put into a pile, it might turn out to be a negligible correction.

  • $\begingroup$ I like the calculation, but disagree with the conclusion that you've reduced the energy to melt by that percentage because we also have to account for the surface energy of the water droplets the ice is melting into. $\endgroup$ – Mark Eichenlaub Apr 26 '18 at 13:36
  • $\begingroup$ @MarkEichenlaub hence "... to some approximation...". $\endgroup$ – LLlAMnYP Apr 26 '18 at 13:38
  • $\begingroup$ @Mark I tried to address this in an edit, but I think, the answer is "it depends". $\endgroup$ – LLlAMnYP Apr 26 '18 at 13:45
  • $\begingroup$ If all the small ice particles are in a pile (a reasonable assumption, I guess), then water from melted particles will wet the yet-unmelted particles, and release its surface energy. So, in my opinion, surface tension of water should be ignored. $\endgroup$ – user27542 Apr 26 '18 at 15:57
  • $\begingroup$ @user27542 precisely what I meant by "negligible correction [to the correction from cutting the ice up into pieces]". $\endgroup$ – LLlAMnYP Apr 26 '18 at 16:04

protected by Qmechanic Apr 27 '18 at 6:27

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