Chapter 17, Problem 98A

Interpretation Introduction

Interpretation: The amount of heat required for the conversion of 1 L of water to steam on the Summit of Mount Everest under given conditions should be calculated.

Concept Introduction: The energy required to change the amount of a substance from liquid to gaseous state is said to be the heat of vaporization.

Answer to Problem 98A

The amount of heat required for the conversion of 1 L of water to steam on the Summit of Mount Everest under given conditions is $\text{2}\text{.3}\times {\text{10}}^3\text{ kJ}$ .

Explanation of Solution

Given Information:

The molar heat of the vaporization of water at various temperatures is given in the following graph:

  

The boiling point on the Summit of Mount Everest is ${\text{70}}^{\text{o}}\text{ C}$ .

Converting 1 L to mL:

  $1\text{ L = 1000 mL}$

First, calculate the mass of water using the known value of density of water as 1 g/mL:

  $\begin{array}{l}\text{mass = density }\times \text{ volume}\\ \text{mass = 1}\text{.0 g/mL }\times \text{1000 mL}\\ \text{mass = 1}000.0\text{ g}\end{array}$

Now convert the mass of water to mol, as:

  $\begin{array}{l}\text{mol of water = }\frac{\text{mass of water}}{\text{Molar mass of water}}\\ \text{mol of water = }\frac{\text{1000}\text{.0 g}}{\text{18}\text{.0 g/mol}}\\ \text{mol of water = 55}\text{.6 mol}\end{array}$

The energy required to vaporize one mole of a liquid at its boiling point is known as the molar heat of vaporization, ${\text{ΔH}}_{\text{vaporization}}$ .

It is given that the boiling point on the Summit of Mount Everest is ${\text{70}}^{\text{o}}\text{ C}$ so, the value of heat of vaporization, ${\text{ΔH}}_{\text{vaporization}}$ at ${\text{70}}^{\text{o}}\text{ C}$ can be determined using the given graph as:

  

Thus, the value of molar heat vaporization ${\text{ΔH}}_{\text{vaporization}}$ at ${\text{70}}^{\text{o}}\text{ C}$ is $41.9\text{ kJ/mol}$ which means vaporizing one mol of water $41.9\text{ kJ}$ of heat is required. So, to vaporize 55.6 mol of water the required heat is:

  $\begin{array}{l}{\text{ΔH = ΔH}}_{\text{vaporization}}\times \text{mol of water}\\ \text{ΔH = 41}\text{.9 kJ/mol}\times 55.6\text{ mol}\\ \text{ΔH = 2329}\text{.6 kJ }\simeq \text{ 2}\text{.3}\times {\text{10}}^3\text{ kJ}\end{array}$

Hence, the amount of heat required for the conversion of 1 L of water to steam on the Summit of Mount Everest under given conditions is $\text{2}\text{.3}\times {\text{10}}^3\text{ kJ}$ .