# Question b91ab

Apr 10, 2017

$\text{100. g H"_2"O}$

#### Explanation:

The problem essentially wants you to convert a given amount of energy to grams of steam at ${100}^{\circ} \text{C}$, so right from the start, you know that you're going to need a conversion factor.

This conversion factor is called the enthalpy of vaporization of water

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

http://www.engineeringtoolbox.com/fluids-evaporation-latent-heat-d_147.html

The enthalpy of vaporization tells you how much heat is needed in order to convert $\text{1 g}$ of liquid water at its normal boiling point of ${100}^{\circ} \text{C}$ to steam at ${100}^{\circ} \text{C}$.

In your case, you need $\text{2257 J}$ of heat in order to convert $\text{1 g}$ of liquid water at ${100}^{\circ} \text{C}$ to steam at ${100}^{\circ} \text{C}$.

This means that $\text{226,000 J}$ will allow you to convert

"226,000" color(red)(cancel(color(black)("J"))) * "1 g"/(2257color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)("100. g")))#

The answer is rounded to three sig figs, the number of sig figs you have for the amount of heat.