Since ethanol has a lower dielectric constant than water would the water heat up and boil before the ethanol? Would the water transfer heat to the ethanol and, since ethanol has a lower boiling point, bring it to a boil before the water?


4 Answers 4


In a microwave the EMW energy is transferred to the water molecules, but, since they are in immediate contact with other molecules (as in any food), the whole volume gets heated. You will not have a two-temperature mixture.


In a liquid mixture such as ethanol-water, both components vaporize to some extent. If the combined vapor pressure of the two equals the external pressure, say 1 atm, the mixture will boil. The components DO NOT boil separately. Further, the composition of the vapor and the composition of the liquid will be different from each other. This is the basic principle behind distillation.

To make matters worse, intermolecular interactions between the two different molecular species can cause "strange" things to happen. For example at some concentrations the mixture can boil at a temperature ABOVE the boiling point of either pure component. For other mixtures the boiling can take place at a temperature BELOW the boiling point of either component.

There is a diagram of the actual boiling system for alcohol-water at


The way this diagram works is this: pick a concentration (a mole fraction is the percentage of the mixture by number of molecules of one kind) and a temperature. Move up the diagram from the mole fraction to the temperature. That's the system point. If the system point is below the vapor line, the system is liquid at that temperature and concentration. If it is above the liquid line, the system is totally vapor at that temperature and concentration.

If the system point is in between the two, one can read off the liquid composition by going to the left until the line is hit and then reading down to the concentration. If the vapor composition is wanted, read the other way.

Note that the alcohol/water system has a composition, way over on the pure alcohol side of the diagram, where the boiling point of the system is below the boiling point of alcohol.


I don't see any reason to think anything but the following will happen:

  • Both alcohol and water will evaporate constantly;
  • The microwave will heat the vodka just like any other water based solution;
  • Once the temperature of the vodka reaches 78.2 °C it will boil.

On the other hand if you use a tightly corked bottle of gin... this happens!

  • $\begingroup$ Vodka is essentially a mixture of ethanol and water. Since water is more polar than ethanol the microwave would heat it easier. The hotter water would then transfer heat to the vodka (so it would have both the microwave and the hotter water surrounding it providing heat). So if vodka, with those two things happening, increases in temperature less than the water and the water can reach its boiling point before the vodka does then the water boils off and the vodka remains; if not the opposite happens. $\endgroup$
    – user18684
    Jan 9, 2013 at 23:09
  • $\begingroup$ I assume you mean alcohol and not vodka above. Considering it's a real mixture, and the two components are not separated, I do not think it's even possible that one boils but not the other. $\endgroup$
    – Sklivvz
    Jan 9, 2013 at 23:15
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    $\begingroup$ I hate being picky, but boiling at 78.2\deg C applies only to alcohol. Mixed with water the boiling point will depend on the concentration of alcohol and be somewhere between 78 and 100 \deg C. $\endgroup$ Jan 10, 2013 at 0:50
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    $\begingroup$ @PaulJ.Gans actually that's incorrect, ethanol boils at 78.37... if you click on the value it will take you to chemistry.SE where someone actually calculated that for us. $\endgroup$
    – Sklivvz
    Jan 10, 2013 at 6:38
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    $\begingroup$ This answer and the linked one at Chemistry.SE are wrong. Vodka is about 40% ethanol by volume and will boil at about 84 °C. This can be seen clearly at this plot. We can check that by noticing that 40% V/V is a molar fraction of 0.17 and using Paul's more academic resource :) $\endgroup$
    – mmc
    Jan 12, 2013 at 23:56

Vladimir Kalitvianski answers to half of your question. I will try to answer the second half and give a general explanation.

Assuming the sample is small enough so there's no significant gradient (or assume gradient, it just makes a single graph more inacurate), you start heating it. As answered, they will both heat at the same pace since they are homogeneously mixed. If you could put molecules apart, there will be a small difference, but completely negligible at macroscopic scale.

Then 78.37 degrees will be reached. The temperature will be stabilized for a moment, while the ethanol boils, as all the heat that is absorbed is spent in boiling the alcohol at steady temperature.

Then, when all the alcohol is gas (I assume closed but big enough recipient), the temperature will keep going up. 100 degrees are reached and again, the temperature stabilizes until all the water has boiled.

Here is the approximate representation of Temperature vs time (the change of state is smooth in any real system).

Vodka Temperature vs time

$k_1$: Vodka; $k_2$: only water; $k_3$: mixture of gases.

$k_3>k_2$, $k_3>k_1$, since I assume a closed adiabatic recipient being the heat capacity of gas lower than of liquid (for water, I didn't check for alcohol). This means, for the same amount of heat, the gas temperature increases faster. Also, $k_2>k_3$, since the specific heat capacity of the water is higher than that of the ethanol:

$$C_{eth} < C_{wat}$$ $$\implies xC_{eth} + (1-x)C_{wat} < C_{wat}$$ $x$ being the ethanol percentage in the mixture.

Note: the graphic is not very accurate because the second slope should be about 2 times steeper than the first one assuming a microwave with a constant power.

Note 2: apologize and correct any language mistakes please, I'm not native.


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