# Water evaporation from calorimeter giving less exothermic enthalpy value?

To find the enthalpy change of combustion for an alcohol, you use a calorimeter filled with water and heat it by burning an alcohol.

I was told that usually experimental values for the enthalpy change of combustion are less exothermic than data book values, because of many reasons (eg- heat loss, incomplete combustion etc.) One of the reasons why the experimental value may be less exothermic is because water has evaporated from the calorimeter.

I don't understand how water evaporating from the calorimeter would make the enthalpy change less exothermic. If the water evaporates, then the mass of water left in the calorimeter is lower, so it would heat up faster, so wouldn't this give a more exothermic result?

• Heat of vaporization is quite high, so if in the local area of the combustion, you vaporize water, that heat won't be distributed to the rest of water to warm it up. – Zhe May 18 at 16:42
• When water evaporates the heat to vaporize the water comes from the water itself, so the water cools. // Apologies to the "young" chemists, but I still think in calories. It takes 540 cal/gram to evaporate water, but 1 cal/g to change water temperature by $\pu{1 ^\circ C}$. – MaxW May 18 at 19:04

The specific heat of water is $$\pu{4200 J kg-1 K-1}$$ and it means that it takes $$\pu{4200 kJ}$$ amount of heat to increase the temperature of $$\pu{1 kg}$$ water by $$\pu{1 K}.$$ On the other hand, the amount of heat needed to turn $$\pu{1 kg}$$ water at $$\pu{373 K}$$ into $$\pu{1 kg}$$ water vapor is $$\pu{2268000 J kg-1},$$ i.e. the latent heat of vaporization of water is $$\pu{2268000 J kg-1}.$$