# Question 2ca27

Jun 14, 2017

$\text{2257 J}$

#### Explanation:

In order to be able to answer this question, you need to know the value of the enthalpy of vaporization of water

$\Delta {H}_{\text{vap" = "2257 J g}}^{- 1}$

https://en.wikipedia.org/wiki/Enthalpy_of_vaporization

The enthalpy of vaporization tells you the energy needed to convert '1 g" of a given substance from liquid at its boiling point to vapor at its boiling point.

In this case, you know that in order to convert $\text{1 g}$ of water at its normal boiling point of ${100}^{\circ} \text{C}$ to $\text{1 g}$ of vapor at ${100}^{\circ} \text{C}$, you need to provide it with $\text{2257 J}$.

You can use the molar mass of water to determine the molar enthalpy of vaporization, which represents the energy needed to convert $1$ mole of water from liquid at ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$.

18.015color(white)(.)color(red)(cancel(color(black)("g")))/"mol" * "2257 J"/(1color(red)(cancel(color(black)("g")))) = "40,660 J mol"^(-1)#

This means that in order to convert $\text{18.015 g}$ of water, the equivalent of $1$ mole of water, at its normal boiling point to vapor at ${100}^{\circ} \text{C}$, you need to provide

$\text{40,660 J " = " 40.66 kJ}$