Problem with derivation of formula for change in entropy for latent heat We know, $dS=\frac{Q}{T}$[when $Q\to 0$]
$\implies \Delta S=\int\frac{Q}{T}$
$\implies \Delta S=\int \frac{ms}{T}$[m= mass, s= latent heat]
$\implies \Delta S=\frac{ms}{T}\int$
This is where I get stuck, this is clearly due to my lack of knowledge in integration. Could you please help me?
 A: The proper equation is
$$\Delta S=\int\frac{\delta Q_{rev}}{T}$$
Where $Q_{rev}$ = a reversible transfer of heat = $mh$ where $h$ is the latent heat
A phase change occurs at constant temperature, so $T$ along with $mh$ comes out of the integral and the change in entropy is simply
$$\Delta S=\frac{mh}{T}$$
Where $mh$ is positive if heat enters the system (causing vaporization or melting) and negative if heat exits the system (causing condensation or freezing).
Hope this helps.
A: You have written the wrong formula as far I can see:
$$dS=\int \frac{\delta Q}{T}=\frac{1}{T}\int \delta Q=\frac{mL}{T}$$
A: You can take out dQ out of the integral when is constant with respect to T.
This happens for a T range near a central vale (say 99-101 C for latent heat of water condensation at around 1 atm).
In case you want a general formula, you could compute dQ as function of the heat capacities
For this, subtract the heat capacities for final state minus initial and compute the integral. As equated below:
$$ @P_{constant}: \int \frac{dQ_{condensation}} {T}= \int_{T_{condensation}}^{T_{condensation}} \frac{\Delta c_P(T) dT}{T} =  \int_{T_{condensation}}^{T_{condensation}} \frac{\Delta[A + B *T + \frac{C}{T}+....]dT}{T}$$
You can find expressions for the regressed expressions for the heat capacities on websites such as NIST:
https://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Units=SI&Mask=1#Thermo-Gas
All the best.
BTW, If you are happy with the answer, please illustrate me on how to format equations, so I can learn and improve it.
LATER EDIT: Thank you for the links! You can find the formatted equations above.
