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

Jun 14, 2017

Here's what I got.

#### Explanation:

The idea here is that you're dealing with $\text{46.0 g}$ of liquid water at its normal boiling point of ${100}^{\circ} \text{C}$ and you're interested in calculating the amount of heat needed to convert this sample to vapor at ${100}^{\circ} \text{C}$.

The problem provides you with the heat of vaporization of water

$\Delta {H}_{\text{vap" = "2.0934 J g}}^{- 1}$

The heat of vaporization, also called the enthalpy of vaporization, tells you the energy needed in order to convert $\text{1 g}$ of water from liquid at its boiling point to vapor at its boiling point.

In this case, you know that $\text{1 g}$ of water requires $\text{2.0934 J}$ of heat in order to undergo a liquid $\to$ solid phase change.

Use this value as a conversion factor to calculate the energy needed for your sample

$46.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g H"_ 2"O"))) * overbrace("2.0934 J"/(1color(red)(cancel(color(black)("g H"_ 2"O")))))^(color(blue)(=DeltaH_ "vap")) = color(darkgreen)(ul(color(black)("96.3 J}}}}$

The answer is rounded to three sig figs.

SIDE NOTE It's interesting that the problem wants you to use that specific value for the enthalpy of vaporization of water because the actual value is

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

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

This means that the amount of heat that you would actually need is equal to

46.0color(red)(cancel(color(black)("g H"_ 2"O"))) * "2257 J"/(1color(red)(cancel(color(black)("g H"_ 2"O")))) = "104,000 J" ->  to three sig figs