The amount of heat required to vapourize one mole of a liquid at constant temperature and under standard pressure (1 bar) is called its standard enthalpy of vapourization.

The enthalpy of vaporization is often quoted for the normal boiling temperature of the substance.

Suppose, some amount of water is kept inside a perfect thermally conducting container at 1 bar pressure and $\mathrm 50~^\circ \mathrm{C}$ within a thermal reservoir of $\mathrm 50~^\circ \mathrm{C}$ also. Due to evaporation, slowly liquid will convert to vapours.

After one mole of water evaporates, if we measure the heat energy received by the container from the reservoir, would it be equal to standard enthalpy of vapourization of water at $\mathrm 50~^\circ \mathrm{C}$?


1 Answer 1


Assuming that the reservoir does not absorb any heat and there is no loss of heat from the setup by:-

  1. Radiation of heat from the reservoir,
  2. Convection currents in the air/surrounding fluid (if any),
  3. The entire water is at the same temperature so that heat absorbed is not used for increase in temperature,
  4. Conduction loss from the reservoir to the ground/support it is on and air(negligible but thought I should mention)

Fulfilling these conditions ensures that the energy is only used by the water for vaporization. So, yes, these conditions are fulfilled. But, that is never the case when experiments are conducted in labs.

Edit: The above conditions are more than just an isolated system.

  • 1
    $\begingroup$ Only evaporation can take place at 50 C which will be very slow... So, how enthalpy of vapourization can be defined at any other temperature than its boiling point? $\endgroup$
    – Apurvium
    Jan 8, 2021 at 17:48

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