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Of course. On a hot, humid day, if you have a glass of ice water you'll get water droplets condensing onto the glass. The same happens onto the surface of the water, you just can't tell the difference between newly condensed water and the pre-existing water.

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Imagine an adiabatic container (like a Dewar bottle): perfectly insulated and pressure resistant, filled with water. A heating element provides a heat flow into the container but no heat can escape from it. No work can be done by the system either: the volume is constant. See diagram: Now if we add some heat to the system by means of the heating element ...

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First derivation: the equation $Lm=H_2-H_1$ is only valid for a process at p=constant, which is not your case. Second derivation: the equation $Q=C_p (T_2-T_1)$ is only valid for a process at p=constant, which is not your case. Plus the rest of the argument doesn't make sense: both $\Delta T=0$ and $\Delta V=0$ in a closed container with boiling water.

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The heat required to vaporize water increases with ambient pressure, not decreases. You can think of it as some heat is required to vaporize the water and some energy to do work against the pressure because the volume of the water is increasing. If you boil your water in an infinite container but under a piston, you can extract this energy by expanding the ...

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