**Question: **An alcohol rub can quickly reduce a fever victim’s temperature by taking away body heat, which is used to evaporate the alcohol from the skin. The evaporation is rapid, because air is normally devoid of alcohol vapor, and so the vapor pressure of alcohol is greater than its near zero partial pressure in air. How many degrees is the temperature of a 55.0 kg person reduced by the evaporation of 500 cm3 (half a liter) of ethyl alcohol from her skin?

**Interesting question**

Admittedly, I see some practical problems in covering yourself in a pint of ethanol, but let us assume that it can be done. Then we need some data:

– Specific heat capacity of water is 4.2 J/gK (Joules per gram times degree in Kelvin or Centigrade). Since roughly half your body mass is made up of water, a reasonable estimate for the specific heat capacity for your body would be 3.5 J/gK = 3500 J/kgK

– Density of ethanol is 0.789 g/dm^{3}.

– Enthalpy of evaporation of ethanol is 841 kJ/kg.

So the mass of .500 dm^{3} is 0.500dm^{3} x 0.789 g/dm^{3} = 0.3945 kg

Evaporating this requires 0.3945 kg x 851 kJ/kg = 331.7745 kJ

This will lower the temperature of your body by 332kJ/(55kg x 3.5kJ) = 1.7 K

So, in theory, you will lower your body temperature by almost two degrees.

Now, to the practical side of things:

– I think you would find it a bit difficult to cover yourself in half a liter of alcohol. It would drip off.

– Heat would be lost to the surroundings. (Heard that one before?)

– It would mainly affect your skin. It goes so fast that the core of your body would not have time to cool off.

Let us assume that you are able to affect the outer cm or so of your skin with an estimated weight of 10kg. One fifth the mass gives five times the drop in temperature i.e. 10 degrees! But again: Loss to the surroundings would be quite significant, so let’s assume 5 degrees. Interesting topic for a project if you can find a way of measuring it without peeling the skin off? Any volunteers?