# Numerical related to crystallisation of water below its freezing point

Question:

There is $16\ \mathrm g$ of pure water in a container at temperature $-20\ \mathrm{^\circ C}$. A small piece of ice is added to start crystallization. Amount of water in container when temperature reaches to $0\ \mathrm{^\circ C}$ is: (assume specific heat of water below $0\ \mathrm{^\circ C}$ is $1\ \mathrm{cal\ g^{-1}\ ^\circ C^{-1}}$.)

My attempt:

I have read this: Liquid water below freezing temperature. So, I understand that the small piece of ice is only added to provide a nucleation site to promote the freezing of water.

But, all physical processes must follow the Law of Calorimetry, right? So, for water below $0$ degrees to freeze, it must first accept heat from somewhere, to reach zero degrees. But, the container is at an even lower temperature, so it won't provide water with any heat (negative temperature gradient). We have also not been given the mass of ice added, so we can't calculate the amount of heat the water will gain from the ice (the latter is at a high temperature than water). The system is isolated, so there is no other possible heat source...

I am apparently at a dead end in this problem. I hope I have put everything in detail and correctly. What is the logical mistake in my thinking? And what is the correct way to approach this problem? Thank you!

• sigh You start with water -20 end up with ice in higher temperature. Supercooled water is inherently metastable and no more activation energy then the nucleation site provides is needed, as this is kinda "chain reaction". – Mithoron Feb 1 '18 at 16:07

In the question Crystallization begins after you add a small piece of ice . So it has to begin that's been given in the question.

I don't know thermodynamically whether it's plausible or not, but it's a given in the question a logical alternative way to approach the question (according to me) is

I formatted it as a spoiler just in case you want to think a bit more or anybody else does

The temperature has increased because latent heat of fusion is being released i.e ice being formed from water is releasing heat.

PS: I may be wrong

• Water around the ice was initially entirely at $-20^\circ$ celsius. How would you account for the heat required to take water from $-20^\circ$ celsius to $0^\circ$ celsius? Will water start freezing at $-20^\circ$ celsius itself? – Gaurang Tandon Feb 1 '18 at 13:48
• Latent heat released by ice formed = increase in temperature of the system – Avnish Kabaj Feb 1 '18 at 13:53
• This equation is dimensionally incorrect... – Gaurang Tandon Feb 1 '18 at 13:54
• You did get what I'm trying to say. If not then multiply $\ce{[\delta]T}$ into the specific heat of the ice formed and water remaining – Avnish Kabaj Feb 1 '18 at 13:55
• Will water start freezing at $−20^∘$ celsius itself? – Gaurang Tandon Feb 1 '18 at 14:01