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) $\Delta S_\mathrm{sys}&lt;0$
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 $\Delta S_\mathrm{sys} &lt; 0$ (yes for r)

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