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Currently, I am looking at a system that is composed of a metal cylinder (lets say Iron) that is being heated by a laser inside of a vacuum chamber. As the laser strikes the surface of the iron, part of the incident radiation is absorbed and an energy balance (and thus a steady state temperature, $T$) is reached when the incoming heat flux from the laser is balanced by the outgoing radiation (which follows the SB law which scales as $T^4$). As the laser power is increased, a point will be reached where the iron begins to melt and evaporate, which also carries away a portion of the incoming laser energy, thus cooling the liquid down.

This scales according to the Hertz-Knudsen Equation:

\begin{equation} \frac{1}{A}\frac{\mathrm{d}N}{\mathrm{d}t} \equiv \varphi = \frac{p}{\sqrt{2\pi m k_\text{B} T}} \end{equation}

Where $p$ is the vapor pressure, $m$ is the mass of a particle (in kg), $A$ is the surface area of the evaporating surface and $N$ are the number of particles leaving the evaporating surface. Therefor the energy, $E$ leaving an infinitesimal portion of the surface, $dA$ is.

\begin{equation} \frac{dE}{dA} = \Delta H \varphi \end{equation}

Where $\Delta H$ is the enthalpy of vaporisation (in kJ/mol). However, I am wondering whether or not the evaporated particles collect kinetic energy during evaporation and does this kinetic energy provide an extra source of energy loss from the evaporating liquid?

If it does account for another source of energy loss, how can this be described via a $T$ dependence?

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The enthalpy of vaporization incorporates all the energy needed to turn condensed matter into a gas at the same temperature. You don't need to add any additional energy to account for the fact that the gas is a gas (i.e., the fact that the free particles have relatively large kinetic energy).

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