Reversible path for Entropy change on crystallisation

I am trying to re-learn thermodynamics but have come stuck with this question so am seeking help.


Two moles of supercooled water at $265$ K freeze to give ice at the same temperature. Using $\Delta H\_f = 6.008 $ kJ/mol and $C\_p(ice) =36.8$ J/mol/K at $273$ K, calculate $\Delta H$ and find a reversible route to calculate $\Delta S$ for this process.


(a) I tried to calculate the enthalpy change assuming that $C\_p$ is constant over the small temperature range; $\Delta H\_{265}=-(\Delta H\_f+\Delta T C\_p)$, where $\Delta T =-8$ which for two moles gives $-11.43$ kJ/mol and is the answer given.


(b) I'm not sure how to calculate the entropy change. $\Delta S=2\Delta H\_{265}/265 = -43.12$ J/K would seem to be the entropy change due to the phase change in crystallizing from the supercooled water but as this is a natural process it is presumably under irreversible conditions. The same point could be reached by calculating the entropy change in crystallizing at $273$ K and then the change from $273 \to 265$ as $\Delta S = -\Delta H\_f/273 + C\_P\ln(265/273)$ which for two moles gives $-41.8$ J/K which is close to the answer given ($-41.76$ J/K) but how could this be a reversible pathway? Additionally does this not ignore the heat used in initially taking the water from $265$ to $273$, so I think that my reasoning and answer is wrong.




Part (b) The supercooled liquid is not in equilibrium with the solid and therefore $\Delta G$ is not zero, so you cannot calculate the entropy for the phase change from $\Delta H(265)/265$. You can't make the supercooled liquid from the solid phase without melting it and allowing it to supercool again. (You lose the heat released on crystallisation to the lattice vibrations of the solid and then to the surroundings in the case of constant T.)


The only path remaining is then to start with the heat of fusion and correct for the change in temperature using the heat capacity of ice, as you show in your question. ( I presume the very slight difference in values is due to a typo some where in the calculation)


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The path that you have carried out is perfectly fine. If we add that both processes take place at constant pressure named $P\_1$, then we can also calculate the heat exchanged by $Q =n \Delta h$
\begin{align}
 \ce{H2O}(P\_1, T\_1, \pu{liquid}) &\rightarrow \ce{H2O}(P\_1, T\_2, \pu{liquid})
 \hspace{1 cm} (\pu{Step 1}) \\ 
 Q\_1 = nc\_\pu{p} (T\_2 - T\_1) = 588.8 \; \pu{J} &\hspace{0.5 cm}
 \Delta S\_1 = n c\_\pu{p} \ln\bigg(\frac{T\_2}{T\_1}\bigg) = 2.1890 \; \pu{J/K} \\
 \ce{H2O}(P\_1, T\_2, \pu{liquid}) &\rightarrow \ce{H2O}(P\_1, T\_2, \pu{solid}) 
 \hspace{1 cm} \pu{(Step 2)} \\
 Q\_2 = n(-\Delta\_\pu{fus} H) = -12016 \; \pu{J} &\hspace{0.5 cm}
 \Delta S\_1 = n\bigg(\dfrac{-\Delta\_\pu{fus} H}{T\_2}\bigg) = -44.0147 \; \pu{J/K} \\
\end{align}
and indeed you get $\Delta S\_\pu{sys} = \Delta S\_1 + \Delta S\_2 = -41.2681 \; \pu{J/K}$.


Since we are treating with changes in state functions, $\Delta H$ and $\Delta S$ will be the same for the process carried in a reversible manner, as well as irreversible. Now, one way of *stating a reversible path*, is for example:


1. Mechanically reversible process of heating of water from $T\_1$ to $T\_2$ at constant pressure.
2. Thermally reversible process of crystallization of water from the liquid state to the solid state.


To be honest, this doesn't say much, so we are going to do some additional calculations to understand what is happening. Indeed, the process is **irreversible** and it has to do with **how** the heat is exchanged between the surroundings


**Step 2** The second step is fine since the heat given **to** the surroundings is $Q\_\pu{2,surr} = -Q\_2 = 12016 \; \pu{J} $, and if we imagine it as a heat reservoir at $T\_2$
\begin{equation}
 \Delta S\_\pu{2,sys} + \Delta S\_\pu{2,surr} = -44.0147 \; \pu{\dfrac{J}{K}} +
 \pu{\dfrac{12016 \; J}{273 \; K}} = (-44.0147 + 44.0147)
 \; \pu{\dfrac{J}{K}} \rightarrow
 \boxed{\Delta S\_\pu{2,universe} = 0 \; \pu{\dfrac{J}{K}}}
\end{equation}


**Step 1** Here lies the problem. The heat given **by** the surroundings is $Q\_{2,surr} = -Q\_2 = -588.8 \; \pu{J}$. For example, if this one is a heat reservoir at $T\_2$, the 2nd law states
\begin{equation}
 \Delta S\_\pu{1,sys} + \Delta S\_\pu{1,surr} = 2.1890 \; \pu{\dfrac{J}{K}} - 
 \pu{\dfrac{588.8 \; J}{273 K}} \rightarrow
 \boxed{\Delta S\_\pu{1,universe} \; 0.032222 \; \pu{\dfrac{J}{K}} >0}
\end{equation}
We can do better if we split the heating in multiple steps, so that the heat reservoir can gradually accompany the system. In this case
\begin{equation}
 \Delta S\_\pu{1,universe} = \Delta S\_\pu{1,sys} + \Delta S\_\pu{1,surr} =
 2.1890 \; \pu{\dfrac{J}{K}} - \sum\_{k=1}^N \dfrac{Q\_\pu{k,surr}}{T\_\pu{k,surr}}
\end{equation}
E.g., two heat reservoirs that give $(588.8/2) \; \pu{J} = \pu{294.4 J}$ to the system, the first one at $\pu{265 K}$ and the second one at $\pu{273 K}$. The results are the following:


[![enter image description here](https://i.stack.imgur.com/dVJrv.png)](https://i.stack.imgur.com/dVJrv.png)


So now after all of this we can say something better: a reversible path for this process, is that of a heat exchange with the surroundings in the heating step, so that the surroundings is only infinitesimally at a higher temperature than the system. When $N\rightarrow \infty$ a reversible process may be approached.


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