# Under what conditions does Latent Heat hold?

Let us say I have a liquid of mass $m$ at its boiling point and add heat to increase to cause turn it all into a vapour. Under what conditions will the heat I actually add be equal to: $$Q=ml$$ Where $l$ is the specific latent heat of vaporization of the liquid.

The equation $$Q=ml$$ Only holds in a system at constant pressure (isobaric) and constant temperature (isothermal) and generically it is not true.
If we look at the change in the internal energy (which is a state function) of the system: $$dU=dQ+dW$$ We see that for the above process we have: $$U=mL-p(V_2-V_1)$$ Where $V_1$ and $V_2$ are the initial and final volumes respectivly (with generally $V_2>V_1$)*. Now take a different process which has the same intial and final state but which does it in the following steps: 1. Free expansion to the final volume. 2. Isochoric (constant volume) vaporization of all of the liquid In this process no work is done but we still have the same change in internal energy (since it is a state function). In this case all that energy must come from the heat supplied and is thus simply not equal to $ml$ as one may naively expect. (i.e. for this process the heat added would be: $$\tilde Q=mL-p(V_2-V_1)$$ Which is actually less then that for the standard isobaric, isothermal phase change.
• @ChesterMiller Thanks for your comment, I wasn't really thinking about the details when I wrote it :). I believe the points you raise go way if we let step 2 becomes step 1 and vice versa (I will change it in my answer). Starting with a liquid and vapour in equilibrium we can perform a Joule expansion on the gas so its volume is $V_2$ (for an ideal gas this will keep the temperature constant). We then add heat to evaporate more liquid until we once again reach the equilibrium pressure. – Quantum spaghettification Jan 7 '16 at 13:17
• That makes much more sense. So the heat for the new version of the "different process" is just equal to the change in internal energy in going between the initial and final equilibrium states. I still don't get that $=p(V_1-V_1)$. I think it should just be $\bar{Q}=\Delta U$ – Chet Miller Jan 7 '16 at 14:40