Alcohol compared to Water Revisited
Date: Fall 2009
In regards to your response in Alcohol vs Water
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
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
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
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
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.
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)
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Update: June 2012