# Thermodynamic quantities for the freezing of water

I am trying to do the following problem:

Freezing of water at $273~\mathrm K$ and $1~\mathrm{atm}$

Which of the following is true for the above thermodynamics process?

p) $q=0$

q) $w=0$

r) $\Delta S_\mathrm{sys}<0$

s) $\Delta U=0$

t) $\Delta G=0$

My answer was r, s, and t:

• Since the reaction happens at constant temperature, internal energy is constant (yes for s)
• Since the reaction is open, it is isobaric in nature; and since the volume changes due to the phase change the work done is not zero (no for q)
• Heat must be exchanged (first law of thermodynamics) (no for p)
• Randomness decreases so $\Delta S_\mathrm{sys} < 0$ (yes for r)

But, the answer given is q, r, and t. I don't understand why q is correct but s is not.

• I recommend against closure; IMO sufficient explanation was given of the effort made on the problem. – hBy2Py Apr 27 '16 at 14:55
• Eh, there is plenty of justification in the question. No need to close as homework. (@Brian you beat me by a full nineteen seconds..) – orthocresol Apr 27 '16 at 14:56
• @orthocresol <pumps fist in air> – hBy2Py Apr 27 '16 at 14:57
• For option s, see this answer: chemistry.stackexchange.com/a/39719 – orthocresol Apr 27 '16 at 15:22

s is not correct because U is not just a function of temperature but also a function of chemical composition. The internal energy of liquid water is higher than the internal energy of ice. If you want to think of it in terms of temperature, then just view the heat capacity as being infinite at the phase transition (Dirac delta function of T), such that the integral over the phase change is equal to the heat of fusion. As far as q is concerned, I guess they are assuming that the change in volume from solid to liquid is negligible.

• sir please explain your answer in simple terms iam just a high school student and i don't know what is dirac delta function.and i have read in my books that internal energy is dependent only on temperature.is it also depended on chemical composition?please explain it sir. – Ajay Sabarish Apr 28 '16 at 1:27

Strictly speaking, you are correct that (q) is false, since the densities of ice and water differ at $0~\mathrm{^\circ C}$:

The density of ice is $0.9167~\mathrm{g\over cm^3}$ at $0~\mathrm{^\circ C}$, whereas water has a density of $0.9998~\mathrm{g\over cm^3}$ at the same temperature.
(source)

As the water freezes, the volume of the system expands slightly and thus $w = -P\Delta V \neq 0$ in a strict sense. (I'm using here the convention that $w$ is work performed on the system, not by the system.) However, for the purposes of the problem, the answer was probably written considering that this expansion is small enough to be neglected $(w\approx 0)$.

As for (s), $\Delta U = q + w \neq 0$ even though the process is isothermal. As you indicated, $q\neq 0$ is absolutely true since heat is being transferred. Since $w$ is very small, it has to be the case that there is a change in internal energy of the water/ice. This change is closely related to the enthalpy of fusion of water.

• thank you very much sir for replying,thought the volume change is small,how can it be neglected if that is the only one which is non zero? though we can explain change in internal energy being non zero through first law of thermodynamics doesn't it violate the fact that internal energy is dependent only on temperature?isn't enthalpy of fusion not the heat transferred during the reaction,please explain how is internal energy related to it. – Ajay Sabarish Apr 28 '16 at 1:26
• @AjaySabarish You're interpreting the phrase internal energy is dependent only on temperature too strictly. There absolutely is a change in internal energy as a substance undergoes a phase change. – hBy2Py Apr 28 '16 at 2:48
• @AjaySabarish In practical application of the sciences, one often must decide whether a phenomenon is of sufficient magnitude to actually affect the system of interest. I would guess the $PV$ work done by freezing water to be at most a few Joules per kilogram of water, which is far less than the heat transfer required to freeze it. Thus, while rigorously it is the case that $w\neq 0$, practically it might as well be the case that $w=0$. – hBy2Py Apr 28 '16 at 2:51
• sir then when can we use the expression internal energy is dependent only on temperature? – Ajay Sabarish Apr 28 '16 at 10:51
• @AjaySabarish Let's continue this discussion in chat: chat.stackexchange.com/rooms/39018/… – hBy2Py Apr 28 '16 at 13:49