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I was looking at the enthalpy of various liquids using WolframAlpha and noticed that at 30 °C, while water and ethanol have a positive enthalpy, acetone has a negative enthalpy.

If I'm understanding this correctly, this means that as acetone evaporates, the temperature of the acetone will increase (because the evaporation will release energy, thereby raising the temperature). I started searching up in temperatures to find the zero-crossing, and that appears to be around 56.074 °C.

This suggests that if a sufficient quantity of acetone is left around, it will eventually heat up to 56.074 °C (which interestingly happens to be when its vapor pressure is really close to the STP pressure).

However, if I get acetone on my hand, I get a distinctive "cool" feel — it, like other liquids, appears to be endothermic and cools down. I may be warm-blooded, but I'm not above 56 °C.

Is there something wrong with my understanding? If not, why does the theory seem to contradict my observations?

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    $\begingroup$ WolframAlpha does not know apples from oranges when it comes to chemistry. Acetone certainly has positive enthalpy of evaporation, like any other liquid. $\endgroup$ – Ivan Neretin Dec 31 '17 at 18:19
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    $\begingroup$ @iAdjunct Apparently, you are confusing enthalpy $H$ and enthalpy of vaporization $\Delta H_\mathrm{vap}$. $\endgroup$ – Faded Giant♦ Dec 31 '17 at 19:09
  • $\begingroup$ @Loong - I believe you are right; see my comment below on your answer. $\endgroup$ – iAdjunct Dec 31 '17 at 19:12
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    $\begingroup$ Congratulations: you put a question on hold after it was answered for being unclear what I'm asking, but kept this question open: chemistry.stackexchange.com/questions/73608/… $\endgroup$ – iAdjunct Dec 31 '17 at 21:00
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The absolute values of enthalpy H at a single state point are meaningless. It is only the difference between two different state points that matter. Thus, the value for a single state point can be set to any arbitrary value. Many handbooks set the arbitrary state point so that the values of these properties are positive for most liquid or gas states.

Apparently, WolframAlpha is setting enthalpy H to zero for the saturated liquid at the normal boiling point (NBP) for acetone but not for water and ethanol.

For ethanol, WolframAlpha is setting specific enthalpy to h = 200 kJ/kg for the saturated liquid at 0 °C (IIR reference state).

For water, the triple point is selected as the reference state, where the internal energy U of saturated liquid is assigned a zero value. Accordingly, the enthalpy H = U + pV of liquid water at the triple point is slightly larger than zero, at a specific enthalpy of about h = 0.6115 J/kg.

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  • $\begingroup$ I don't understand how a difference is used. My understanding was that when something says, for example, 30 J/kg, that means if you want to transition 2 kg of that substance from liquid to gas, you must add 60 J of energy to it. With a negative number, it seems to imply the removal of energy. How do you use a number with units of J/kg as references for differences? $\endgroup$ – iAdjunct Dec 31 '17 at 18:48
  • $\begingroup$ @iAdjunct The absolute value of enthalpy has no physical meaning. For example, the given value for the specific enthalpy of liquid water at equilibrium at 100 °C is 419.17 kJ/kg. This value alone doesn’t say anything. Since only the difference between two state points is relevant, you would need a second value; for example, the specific enthalpy of gaseous water at equilibrium at 100 °C, which is 2675.6 kJ/kg. Now you can calculate the specific enthalpy of evaporation at 100 °C as 2675.6 kJ/kg − 419.17 kJ/kg = 2256 kJ/kg. $\endgroup$ – Faded Giant♦ Dec 31 '17 at 19:02
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    $\begingroup$ Ohhhhhhhhhhh. I think that also answered why "Specific Heat of Vaporization" isn't just called "Enthalpy" - because the Specific Heat of Vaporization is ∆Hvap = Hgas - Hliquid at whatever temperature/pressure you specify? (therefor ∆Hvap = SpecificHeatOfVaporization?) $\endgroup$ – iAdjunct Dec 31 '17 at 19:10

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