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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!

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    $\begingroup$ 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". $\endgroup$ – Mithoron Feb 1 '18 at 16:07
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Another answer explains more generally how to interpret the problem. This completes that spoiler:

If one can assume that the enthalpy of fusion ($\pu{334 J/g},~$2) is independent of temperature between the normal melting point ($\pu{0^\circ C}$) down to $\pu{-20 ^\circ C}$, then the heat released during freezing equals that absorbed during heating of the liquid to $\pu{0^\circ C}$ so that $$\begin{align} \pu{334 J/g} \times m &= \pu{20 K \times 16 g \times 4.18 J/g*K} \\ \rightarrow m&= \pu{4.0 g} \end{align}$$ Note the important condition that the heat capacity of water remains constant and equal for liquid and solid over that temperature span.

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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

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  • $\begingroup$ 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? $\endgroup$ – Gaurang Tandon Feb 1 '18 at 13:48
  • $\begingroup$ Latent heat released by ice formed = increase in temperature of the system $\endgroup$ – Avnish Kabaj Feb 1 '18 at 13:53
  • $\begingroup$ This equation is dimensionally incorrect... $\endgroup$ – Gaurang Tandon Feb 1 '18 at 13:54
  • $\begingroup$ 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 $\endgroup$ – Avnish Kabaj Feb 1 '18 at 13:55
  • $\begingroup$ Will water start freezing at $−20^∘$ celsius itself? $\endgroup$ – Gaurang Tandon Feb 1 '18 at 14:01
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So, for water below 0 degrees to freeze, it must first accept heat from somewhere, to reach zero degrees.

No, water can freeze at lower temperatures, see e.g. https://www.youtube.com/watch?v=Fot3m7kyLn4.

As it freezes, the exothermic process warms up the ice and the surrounding water. The difference in enthalpy between liquid and solid water is larger than the difference between liquid water at the two temperatures, so not all the water freezes. Instead, you end up with liquid and solid water at equilibrium once the water and the ice reach the normal freezing point (not because they are warming up from the environment, but as ice continuously forms).

Instead of adding ice to start the process, you can also shake the container like in the video.

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The crystallization is an operation that delivers heat, and a good deal of heat. This is sufficient to heat the metastable water form -20°C to 0°C.

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The previous answers are good ones. I just want to add a (longwinded) metaphorical picture to help understanding the freezing of water at -20C.

Imagine a tower of bricks (2 x 4 x 8 inches each) about 20 high. It is metastable, but the tiniest push will topple it because gravity pulls each brick to the lowest level. This is a metaphor for the freezing of water at -20C, right? Yes, but there is a great dissimilarity, because the bricks fall down to a lower height, but the water temperature rises as it freezes. We all know that in reality, we can't reduce the temperature of water to -20C and have it be as stable as a pile of bricks 20 high. Then, too, freezing is a process that 99.9% of the time requires a reduction of temperature.

Now let's try another metaphor: take 20 bricks of low-density foamed polystyrene and push them down into a large tank of water. You may be able to hold them down perfectly straight, or even balance a heavy lead brick on top of them to hold them down while they are semi-floating, but if the slightest imbalance occurs (like the ice-seed), the polystyrene bricks will all float up to the surface (~ ~ supercooled water will freeze) and the heat released will equal the energy required to push the polystyrene into the tank. The heat released as the polystyrene bricks rise is really from the water settling in a gravity field (~ ~ the heat of fusion that was added to the water to keep it from freezing while it was being cooled to -20C.).

Now how can you cool water to -20C without removing the heat of fusion? In your imagination, separate the specific heat and the heat of fusion into different accounts. Take the heat out of the one and leave the other, but don't let the water know. This allows you to mentally reduce the temperature without allowing the solidification process to occur - it's like applying a "negative catalyst". Then when you remove the "negative catalyst" (i.e., add the ice-seed), some of the water freezes, releasing bottled-up heat of fusion to flow into some of the -20C water - the temperature rises. The result will be some ice and some water, and when the system comes to equilibrium, the temperature will have risen to 0C.

And the process can be stated much more succinctly by an equation, but sometimes equations are too succinct!

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