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In my chemistry course, the professor has discussed how a sample that contains both solid and liquid water would be at 0 degrees Celsius no matter what the proportion of solid to liquid water is. I find this concept to be confusing, because wouldn't the areas of the sample that had already changed from solid to liquid begin increasing in temperature even while the other parts of the sample continue to be in the process of fusion?

The other part of this idea that confuses me is that if the sample was at a uniform temperature, then why do bodies of water (ex. lakes that freeze in the wintertime) continue to have different temperatures of water within them?

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    $\begingroup$ And in addition to @DrMoishePippik's answer below on how, in equilibrium, the liquid and solid are at the same temperature, note that a lake tends not to be in equilibrium - the bottom will not be at the same temperature as the top. $\endgroup$ – Jon Custer Dec 21 '15 at 22:38
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What is confusing you is the thermodynamic concept of equilibrium. A system at equilibrium has no external source of energy flowing onto it and is perfectly mixed internally. These are not common circumstances in the real world.

In a system that is in equilibrium, a solid and its corresponding liquid can exist at one temperature. Any extra energy added will, when equilibrium is restored, serve to convert the solid to liquid (this takes energy). Since, at equilibrium, we have perfect mixing this will not result in a temperature change. For water it takes a lot of energy to convert ice to liquid water.

Of course, in the real world, nothing is at equilibrium and there is always a flow of energy in or out of the system. And for an isolated system at equilibrium, when energy is added there will be a period where parts of the system will be hotter than others. But, since being at equilibrium means the system will be perfectly mixed, this is temporary and the system will regain equilibrium with a constant temperature but a different mix of water and ice.

So you professor is right if he specifies that the system is in equilibrium but you are right in the real world for most systems. The idea of equilibrium is a useful idea for understanding the limiting behaviour systems if a little unrealistic as a description of the real world.

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    $\begingroup$ I'll throw in that many times in chemistry we'll assume an adiabatic process - in other words the system doesn't gain or lose heat to the rest of the universe. In reality such perfect isolation of a system is impossible. So it is a great theoretical concept even through it is impractical in an absolute sense. $\endgroup$ – MaxW Dec 23 '15 at 2:18
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For a system in equilibrium: Think of it as a closed system, with a certain amount of energy shared between the solid and liquid components (discounting the container). At any instant, some of the faster-moving molecules of the liquid could hit the ice, melting a bit, and thereby losing some of their own energy and freezing.

It takes ~80 cal/g to change ice at 0 °C to water at 0 °C. This enthalpy of fusion, aka latent heat of fusion , "absorbs" any additional heat applied until all the ice has melted. So a gram of water at 0 °C has ~80 calories more heat than an equal mass of ice at 0 °C. If you have a 1 gram chunk of ice and apply 40 calories as heat, then about half will melt.

BTW, pure water does not necessarily freeze at 0 °C... it may supercool, which will change the calculations a bit. See this video or this one for a supercool experiment!

As @MaxW states, in real life most systems are not in equilibrium (they need to be shaken or stirred, despite Ian Fleming's direction). To view a clearly non-equilibrium demo, where there is water in three states at once, see Red Hot Nickel Ball on Ice.

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    $\begingroup$ Ice/water at 0C assumes that the ice and water are at an equilibrium temperature. As the OP questioned, it is possible to have a lake where the bottom water is warmer than the surface water (with ice perhaps). $\endgroup$ – MaxW Dec 21 '15 at 22:39
  • $\begingroup$ True... I'm amending my answer, which had applied only to a system in equilibrium. Thanks for that correction. $\endgroup$ – DrMoishe Pippik Dec 23 '15 at 0:01

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