Chapter 17.3, Problem 24SP

Interpretation Introduction

Interpretation: The heat absorbed by a certain quantity of liquid water changes into vapor at specified conditions.

Concept Introduction: The enthalpy of vaporization is the amount of energy that must be added to a liquid substance to transform a quantity of that substance into a gas.

Answer to Problem 24SP

The heat absorbed when $\text{63}\text{.7 g}$ of ${\text{H}}_2\text{O(l)}$ at $\text{100°C}$ and $\text{101}\text{.3 kPa}$ is changed to ${\text{H}}_2\text{O(g)}$ at $\text{100°C}$ is $\text{143}\text{.8 kJ}$

Explanation of Solution

The enthalpy of the vaporization of water at $\text{100°C}$ is $40.63\text{ kJ/mol}$

This means that to convert 1 mole of water at $\text{100°C}$ to $1$ mole of steam at $\text{100°C}$ , $\text{40}\text{.63 kJ}$ of heat must be absorbed by the water.

Moles of a substance is found by dividing the given amount of substance by the molecular mass of the substance.

Amount of water $$ given = $\text{63}\text{.7 g}$

Moles of water = $\frac{\text{63}\text{.7 g}}{\text{18}\text{.01 g/mol}}=3.54\text{ mol}$

For the conversion of 1 mole of water at $\text{100°C}$ to $1$ mole of steam at $\text{100°C}$ , heat absorbed is = $40.63\text{ kJ}$

Thus, for the conversion of $\text{3}\text{.54 mol}$ of water at $\text{100°C}$ to $1$ mole of steam at $\text{100°C}$ , heat absorbed is = $\frac{\text{40}\text{.63 kJ}}{\text{1}\text{.0 mol}}\text{×3}\text{.54 mol=143}\text{.8 kJ}$