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$dS=dH/T_{boil}$ for the increase in entropy by changing a phase at saturation ($T=373\text K$ for $p=1\text {atm}$ for water). However, water also obviously evaporates below boiling point when equilibrium is not reached (below 100% relative humidity).

dS can also be interpreted using chemical potentials ($u$). The difference in the chemical of $u(p,T,N)$ between the 2 phases not at equilibrium is what drives the phase change. However, the increase in entropy when computed comes to $$ds=(u_{liquid}-u_{gas})\times N/T$$ where $N$ is the number of moles converted. But should the temperature $T$ be the "average" temperature of the system (i.e. the ground level atmospheric conditions for say a lake that evaporates water), or should $T$ be the temperature at which the water boils?

I ask in multiple interests, but motivated to ask due to the temperature discrepancy. Other ideas that can quantify entropy produced by non-equilibrium thermodynamic state is also appreciated.

Thanks

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$T$ should be the actual temperature at which the water evaporates. That is, the temperature at the interface between the air and the water, not the boiling point.

This is simply because $dU = TdS + pdV - \sum_i \mu_i dN$ (where $T$ is most definitely the temperature of the system), or by rearranging, $$ dS = \frac{1}{T}dU -\frac{p}{T}dV + \sum_i\frac{\mu_i}{T}dN_i, $$ which means that $$ \frac{\mu_i}{T} = \frac{\partial S}{\partial N_i}, $$ where $T$ is still the temperature of the system; the boiling point doesn't come into it.

In fact this holds independently for the water and gas phases. In that case $T$ might have a different value for the two phases. So in the case where (for example) cold water is evaporating into warm air, the entropy produced per mole will be $$ \Delta S = \frac{\mu_{liquid}}{T_{liquid}} - \frac{\mu_{vapour}}{T_{air}}. $$

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Specific entropies ($\mathrm{kJ\,kg^{-1}\,K^{-1}}$): \begin{array}{lrll} & \mathrm{^\circ C} & \text{liquid} & \text{vapor} \\ \text{Triple point} & 0.01 & 0 & 9.155 \\ \text{Normal boiling point} & 100. & 1.307 & 7.355 \\ \text{Critical point} & 374.15 & 4.430 & 4.430 \\ \end{array}

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