# Latent Heat of Vaporization /w Temperature

## Main Question or Discussion Point

Hi,

I would like to ask, why is it that the specific latent heat of vaporization of water at, say 10 degrees Celsius, is considerably higher that at 100 degrees Celsius?

It would be great if you could provide an explanation from the molecular view.

Thanks.

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I'm not sure that the molecular viewpoint is going to be all that informative for you, because the reason has much more to do with the gas that the water vaporizes into.

At 100 C, when water boils or evaporates, it turns into 100 C, 1 atm pressure water vapor.

At 10 C, when water boils or evaporates, it turns into 10 C, ~0.01 atm pressure water vapor.

In other words, when the phase change occurs, at lower temperatures the end result is a more dilute gas. You can see that there is a difference in the energy state between a hot dense gas such as the 100C 1 atm water vapor and the cold dilute 10 C ~0.01 atm water vapor. This difference in energy is reflected in the difference in latent enthalpy of evaporation.

I guess from a molecular viewpoint the extra energy is used to make enough space for the vapor molecules to be more dilute at lower temperatures. As you get closer to the critical temperature (374 C for water), the amount of energy needed for this drops closer and closer to zero.

Thanks for the answer krysith - it makes sense to me I suppose. My original thought was that water at 100 C had slightly larger inter-molecular spacing (and thus P.E.), therefore it takes less energy for it to become gas compared to water at 10 C.

actually it isn't so much about the difference in vapor pressure at those temperatures. You can work out a quick calculation to show that the PV term is a fairly small contributor to the enthlapy of evaporation, less than 10%.

The second point is what matters. At 100C you are getting closer to breaking the bonds, so you don't need as much energy to vaporize it.