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Name: Michael
Status: student
Grade: 9-12
Country: USA
Date: Fall 2009

In regards to your response in Alcohol vs Water Evaporative Cooling, :

The original question was "When rubbing alcohol evaporates, it 'feels' colder than water. Is this because it requires more energy for a phase change than water?" Your response was that rubbing alcohol requires more energy for changing from liquid to gas, due to both the phase change (heat of vaporization) as well as temperature change (specific heat capacity) while in the liquid phase. Where does the different volatilities of the two compounds come into this picture? I thought the increased volatility of rubbing alcohol (and higher tendency for evaporative cooling) was the main reason why it feels colder to the skin than water does.

I have read the original answer in the archives, and I won't argue with its facts and figures. However, I think it may be misleading. It quotes the molar enthalpies of vaporization of water and isopropyl alcohol, notes that the value is higher for isopropyl alcohol, and lets it go at that. What this means in English is that a molecule of isopropyl alcohol absorbs more energy (heat) when it evaporates than a molecule of water does.

However, the molecules aren't the same size. The molar mass of water is 18 g/mol, while isopropyl alcohol is 60 g/mol, more than triple water's. And note that their enthalpies of vaporization aren't very different--that means that when 1 gram of water evaporates, it absorbs nearly three times as much heat as 1 gram of isopropyl alcohol evaporating.

Now, isopropyl alcohol boils at a lower temperature than water, and it is more volatile than water at lower temperatures as well. This means that its equilibrium vapor pressure is higher at all temperatures. The effect of this is that if you have equal amounts of isopropyl alcohol and water on your skin, much more of the alcohol will evaporate over a short period of time. This means the evaporation process will consume more energy per second. Although evaporating equal masses of alcohol and water will require much more energy in the case of the water, it consumes energy faster in the case of the alcohol.

Richard Barrans, Ph.D., M.Ed.
Department of Physics and Astronomy
University of Wyoming

I think you're right, Michael. I can't compare numbers of latent heat of evaporation right now. I suspect that of water would be larger per unit volume or mass. Not sure about per mol or per volume vapor. Not sure which basis is most relevant for evaporative cooling.

However I'd start by considering latent heats of water and alcohol fairly similar. The most active variable is vapor pressure.

I'd see the evaporative cooling process like this: you have "wet" hands in nearly still air, which air is replaced occasionally by random breezes or hand movements. While the air is still, the vapor of alcohol or water builds up to steady-state concentrations in the air near the hands. ("Near" meaning on the order of one diffusion length of vapor in air, which varies as the square-root of the time that no motion occurs.) Once built up, the evaporation rate decreases to a low value and no longer feels very cool. Then some motion occurs, the vapor is dispersed and will build-up all over again, resulting in a new surge of cooling sensation. Your impression of the coolness of a liquid is proportional to the net energy removed by each surge of evaporation. And that will be very roughly proportional to the concentration of vapor in the air.

Isopropyl alcohol has a considerably higher vapor pressure at room temperature than water. Ethanol higher than that, and Methanol really high, being a largish fraction of an atmosphere.

So, perception of coolness will be in that sequence: water, isopropanol, ethanol, methanol. Presuming of course the latent heats are effectively similar, or at least less different than the vapor pressures.

One could create a steady-state scenario by imagining still air for a fixed distance from the hands, and a breeze continually exchanging all air beyond that. Then heat-removal rate would depend of vapor pressure and diffusivity in air.

Jim Swenson

Hi Michael,

Unfortunately "volatility" is a confusing term because it is used colloquially to mean two different things. It is often used to distinguish whether a substance has a lower boiling point, and it is also used to mean that a substance requires less energy in order to volatilize. Moreover, volatility is also a function of how strong the intermolecular attractive forces are in the liquid. As such, we should talk about boiling point versus evaporation, and the energy requirements for the process.

While it is true that alcohol has a lower boiling point than water (and will volatilize sooner given the same conditions), it is also true that alcohol requires more energy become vapor (as mentioned, because of its higher specific heat and enthalpy of vaporization).

Since we are essentially talking about a non-equilibrium condition of evaporating a liquid (rather than having it boil), then the issue is not whether alcohol reaches its boiling point sooner, but rather how much energy is required to convert a liquid to gas. In this case, alcohol requires more energy and therefore extracts more heat from our skin and so feels cooler.

Greg (Roberto Gregorius)
Canisius College

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