# Acetone Enthalpy & Observations

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?

• WolframAlpha does not know apples from oranges when it comes to chemistry. Acetone certainly has positive enthalpy of evaporation, like any other liquid. – Ivan Neretin Dec 31 '17 at 18:19
• @iAdjunct Apparently, you are confusing enthalpy $H$ and enthalpy of vaporization $\Delta H_\mathrm{vap}$. – Loong Dec 31 '17 at 19:09
• @Loong - I believe you are right; see my comment below on your answer. – iAdjunct Dec 31 '17 at 19:12
• 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/… – iAdjunct Dec 31 '17 at 21:00

• 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? – iAdjunct Dec 31 '17 at 18:48
• 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?) – iAdjunct Dec 31 '17 at 19:10