I read that the heat capacity of water when it is at 100 °C or 0 °C is infinite because all the heat given to it will not be used to increase temperature but to change the state.

I am unable to understand this. Say, we have water at 100 °C. If we continuously provide heat, water will change to steam and then it's temperature will increase, right?
Sure, we'll have to provide more heat than before to break the bonds. But it will definitely become a 101 °C steam after a while.

So why is heat capacity infinite at state change?

  • $\begingroup$ The temperature of the steam stays at 100°c unless it is superheated. The extra heat/energy being added (latent heat) is consumed by the water changing phase. So the more heat you add the faster all the water will convert to steam when it reaches 100°C. $\endgroup$ – Technetium Nov 26 '15 at 10:52
  • $\begingroup$ Say, you have 1 kilogram of water at 20°C. You add 42kJ of heat, and your water gets warmer by 10°C. You divide 42000 by 10, and that's the heat capacity. Now you take 1 kilogram of water at 100°C. You add the same amount of heat, and... what is the change of temperature? $\endgroup$ – Ivan Neretin Nov 26 '15 at 12:38
  • $\begingroup$ Both the comments are pointing at the fact that we need to add a lot of heat to change temperature. So I'm guessing infinite heat here refers to a lot of heat, not to the fact that it is impossible to increase temperature. $\endgroup$ – Mahathi Vempati Nov 26 '15 at 16:00
  • $\begingroup$ You got it wrong. A lot of heat does not matter that much (also, it is not such an awful lot, if we are talking about the boiling of liquid helium, for example; but the infinite heat capacity would still be there even in this extreme case). What matters is that $\rm{c={Q\over\Delta T}}$, and $\Delta\rm T$ is zero. Not just very small, but zero. What happens when you divide by zero? $\endgroup$ – Ivan Neretin Nov 26 '15 at 16:19
  • $\begingroup$ Oh! OK. See if this is right: when you heat water at 100°c by a certain amount (not a VERY large amount) and divide by the temperature change which is zero, you get infinity, hence the heat capacity. Is that correct? $\endgroup$ – Mahathi Vempati Nov 26 '15 at 16:27

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