# A sample of H_2O with a mass of 46.0 grams has a temperature of 100 °C. How many joules are necessary to boil the water? (use 2.0934 J/g for the heat of vaporization of water)

Aug 25, 2017

I get $\text{104000 J}$ when I use the value given by many other textbooks.

(If I were to use your value, which is equal to $\text{0.0377 kJ/mol}$, I would only have gotten $\text{96.3 J}$, which is way too low.)

Sorry, but I refuse to use an incorrect enthalpy of vaporization. The correct value is $\text{40.67 kJ/mol}$, or about $\text{2257.6 J/g}$, over one thousand times greater than what you have cited.

This can be found

Boiling a substance would be done at constant pressure and temperature, so we can equate the heat required with the enthalpy of vaporization:

$q = n \Delta {\overline{H}}_{v a p}$

where $n$ is the mols of liquid, $\Delta {\overline{H}}_{v a p} = \text{40.67 kJ/mol}$, and $q$ is the heat absorbed.

Thus, the heat absorbed is:

$\textcolor{b l u e}{q} = 46.0 \cancel{\text{g" xx cancel"1 mol"/(18.015 cancel"g") xx "40.67 kJ"/cancel"mol}}$

$=$ $\textcolor{b l u e}{\text{104 kJ}}$

or $\text{104000 J}$.