I know it's the most silly question one could ever ask but why is that so?

I am aware of what latent heat means and I also know it has something to do with freezing too.

Say I have with me, a mug of hot water and I put it to freeze. And so the temperature reduces first to 273 K and then remains constant to convert into ice and decreases once again. But why does the temperature remain constant though? Does this has something to do in molecular level?

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    $\begingroup$ In a broader sense, everything that happens to us and around us has something to do in molecular level, but I'd rather not put it this way. The temperature stays constant because it just can't go down. How would it go down? You can't have liquid water at below-freezing temperature (except in some special cases, but those are out of scope for now). $\endgroup$ Nov 9, 2018 at 9:57

1 Answer 1


Creating an ordered solid from a disordered liquid releases energy

At the molecular level this is what is happening when you cool something like water.

At first the cooling reduces the average kinetic energy of the water molecules so they slow down. Basically, temperature is the average kinetic energy so removing energy lowers the temperature.

At some point, though, water starts to form ice where the molecules are not wandering about randomly and get rigidly stuck together in ice crystals. But this process also releases energy: the ordered ice crystal has a significantly lower energy than water at the same temperature. Converting liquid water to ice releases that energy.

What this means is that when extracting energy from warm water the result is simply a lower temperature. But one you get to 0°C the extra energy extracted is used to convert the liquid to the solid rather than reducing the temperature. Once all the water has solidified, further removing energy reduces the temperature again.

The key is that the process of freezing itself involves energy changes so instead of reducing the temperature, the energy comes from the conversion of liquid to solid.

  • $\begingroup$ When you write "But this process releases energy" when talking about melting, it seems as if you are using this to distinguish it from simple cooling. But note that both cooling and melting release energy. Thus it's not the release of energy that makes melting qualitatively different from cooling. $\endgroup$
    – theorist
    Aug 30, 2021 at 17:33
  • $\begingroup$ @theorist Assume that you are removing energy at a steady rate. The energy comes from cooling the water. At the crystallisation point you are still removing energy at that rate. But a new process (crystallisation) starts in the liquid which, from the previous point of view, releases extra energy that stops the temperature falling steadily. Alternatively, you switch between two different energy releasing processes (cooling and crystallisation) at that point. $\endgroup$
    – matt_black
    Aug 30, 2021 at 17:46
  • $\begingroup$ I agree with what you wrote in your comment. I don't doubt you understand the process. My point is that the way you've written the portion of your answer that discusses melting ("But this process releases energy") makes it sound like what's distinct about the melting (vs. the cooling) is that the melting releases energy. (because you start with "but" and emphasize "releases"). So I'd recommend you instead write something like "As with the cooling, this process also releases energy. But what's different is...." $\endgroup$
    – theorist
    Aug 31, 2021 at 2:24
  • $\begingroup$ To give an analogy, suppose you were talking about the difference between two different versions of some car (let's call them the LX and SX models) and, in talking about the SX, wrote "But the SX uses gasoline..." That would make perfect sense if the LX didn't use gasoline (say, it was a diesel). But if both used gasoline, you would not want to write "But the SX uses gasoline...", because it implies the LX somehow does not. $\endgroup$
    – theorist
    Aug 31, 2021 at 2:28
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    $\begingroup$ @theorist OK good point. I've made a minor edit. $\endgroup$
    – matt_black
    Aug 31, 2021 at 11:25

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