Latent heat of vaporization/evaporation

Question

It's the amount of heat required for liquid ---> gas phase change of a substance at it's Boiling point.

But my question is, does providing energy equivalent to latent heat of vaporization to a liquid at a temperature below it's Boiling point (if you need, also assume above it's Freezing point) 'Evaporate' it?

eg. When water is kept inside a clay pot out in the Sun, we know a very small amount of water somehow seeps out through it's pores and 'evaporates' (if not vaporizes). Obviously the temperatures aren't even close to 373K(boiling point of water), but the water is spontaneously turning into steam. Because it's not boiling, but is simply evaporating which can take place any where below it's boiling point. But the point here is, 'how much' of the energy from the rest of water-pot system is entrapped in order to convert into a form that has got more energy than its predecessor. Is it equivalent to only the amount of latent heat of vaporization or is it plus the amount of energy required to raise it's temperature upto it's boiling point.

Answer 1

Latent heat of vaporization/evaporation: It's the amount of heat required for liquid ---> gas phase change of a substance at it's Boiling point.

Not exactly. It's the amount of heat required for liquid ---> gas phase change of a substance at the particular temperature, there is usually and implicitly the substance boiling point. But generally, it is defined for any temperature.

Is it equivalent to only the amount of latent heat of vaporization or is it plus the amount of energy required to raise it's temperature upto it's boiling point.?

Neither, neither. It is between these 2 values.

The latent heat of water evaporation at temperature $T_2 \lt T_1$ (let we take $T_1$ as the boiling point ) is equal to the latent heat of vaporization at temperature $T_1$ plus the heat required to warm up water from $T_2$ to $T_1$ minus the heat released to cool down vapour from $T_1$ to $T_2$.

$$\Delta H_{\mathrm{evap},T_2} = \Delta H_{\mathrm{evap},T_1} + C_\mathrm{p,water}\cdot (T_1 - T_2) + C_\mathrm{p,vapour}\cdot (T_2 - T_1)$$

where $\Delta H$ are molar latent evaporation heats at constant pressure ( called molar enthalpies of evaporation) at respective temperatures,
$C_\mathrm{p}$ are respecive molar heat capacities of water and vapour ( for simplicity considered constant to avoid integrals).

It is application of the Hess' law, which is application of the energy conservation law. It says the energy change depends only on the initial and final state, not on the way how is the final state reached. If it were not so, we could easily construct perpetuum mobile by switching between 2 states by 2 different ways.

So getting A->B directly, or A->C->D->B is equivalent, where

• A is cold water
• B is cold vapour
• C is hot water
• D is hot vapour

Generally, the latent heat of evaporation increases with decreasing temperature, so you need more energy to evaporate water at lower temperature. At water critical temperature near $374 ^{\circ}\mathrm{C}$ the properties of liquid and gaseous water converge to the same values and the latent heat of evaporation converges to zero.

If water at $50^{\circ}\mathrm{C}$ is provided by latent heat of evaporation at $50^{\circ}\mathrm{C}$, then it will evaporate. If water at $50^{\circ}\mathrm{C}$ is provided by latent heat of evaporation at $100^{\circ}\mathrm{C}$, then some water will not evaporate. Because the latent heat of evaporation is bigger at $50^{\circ}\mathrm{C}$, compared to $100^{\circ}\mathrm{C}$.

Technically, it is different question, as evaporation at ambient temperature takes energy from the environment, while you have to supply energy for the boiling.

Note that physicists/chemists prefer to talk about vapour, a gaseous phase of substance, that is not considered as a "permanent" gas. Steam is rather a technical/engineering/everydays term, covering pure water vapour, mixtures of vapour with gases like air, or the same with an aerosol of water or ice particles. See water phase diagram. Water vapour exists even at at mesosphere at altitude near 80 km at temperatures almost $-100^{\circ}\mathrm{C}$. Also, ice can sublimate directly into vapour, without melting.

Answer 2

To answer your question, yes.

But my question is, does providing energy equivalent to latent heat of vaporization to a liquid at a temperature below it's Boiling point (if you need, also assume above it's Freezing point) 'Evaporate' it?

@Poutnik answered this fantastically, but with only theoretical examples. They did say though:

Generally, the latent heat of evaporation increases with decreasing temperature, so you need more energy to evaporate water at lower temperature.

…which is indeed what is stated in my College Physics textbook (Knight et. al, 2010, 2e). At standard atmospheric pressure, the (vaporisation) latent heat of water is:

• 24 × 10⁵ J/kg, at 30°C
• 22.6 × 10⁵ J/kg, at 100°C

Simply stated, you need 6% more energy to evaporate water at 30°C than you would at 100°C. Therefore, it can (and does) happen.