Chapter 17.3, Problem 31LC

Interpretation Introduction

Interpretation: The amount of heat to be removed to freeze a tray of ice cubes at $\text{0}°\text{C}$ is to be calculated.

Concept introduction: The molar enthalpy of solidification $\Delta H_{solid}$ , is the amount of energy that must be removed from 1 mole of a liquid substance to transform into 1 mole of solid at constant pressure.

Answer to Problem 31LC

The heat removed to freeze a tray of ice cubes at $\text{0}°\text{C}$ if the water has a mass of $\text{225 g}$ is $75.06\text{ kJ}$ .

Explanation of Solution

Molar enthalpy of solidification for water is $\text{-6}\text{.01 kJ}$ .

This means when 1 mole of water solidifies into ice, $\text{6}\text{.01 kJ}$ of heat is removed.

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

  $\begin{array}{l}\text{moles of water}=\frac{\text{given amount of water}}{\text{Molar mass of water}}\\ \text{moles of water}=\frac{\text{225 g}}{\text{18}\text{.01 g/mol}}=12.49\text{ mol}\end{array}$

For $\text{1 mol}$ of water to solidify at $\text{0}°\text{C}$ , heat released is = $\text{6}\text{.01 kJ}$

Thus, for $12.49\text{ mol}$ of water to solidify at $\text{0}°\text{C}$ , heat released is = $\frac{6.01\text{ kJ}}{1\text{ mol}}\times 12.49\text{ mol}=75.06\text{ kJ}$