Chapter 14.1, Problem 8RQ

Interpretation Introduction

Interpretation: The heat required to melt $100g$ of ice at $0^{\circ }C$ along with the mass of water that would vaporize with the same amount of heat needs to be determined.

Concept introduction: Heat required to melt ice at $0^{\circ }C$ is calculated as follows:

  $\text{ Heat required = number of moles of ice }\times \text{molar heat of fusion}$ .

Where number of moles of ice = $\frac{\text{Mass of ice}}{\text{molar mass of ice}}$

where Molar mass of ice =

With the same heat, number of moles of water that would vaporize is calculated as follows:

  $\text{Heat required = number of moles of water }\times \text{molar heat of vaporization }$

From the number of moles of water, mass of water can be calculated as follows:

  $\begin{array}{l}\text{Mass of water = Molar mass of water × Number of moles of water}\\ \text{where Molar mass of water}=18g/mol\end{array}$

Answer to Problem 8RQ

Heat required to melt $100g$ of ice at $0^{\circ }C$ is $\text{33}\text{.411}kJ$ and the mass of water that would vaporize with the same amount of heat is $\text{14}\text{.79 }g$ .

Explanation of Solution

It is given that molar heat of fusion of ice is $6.02kJ/mol$ and molar heat of vaporization for water is $40.6kJ/mol$ .

Heat required to melt ice at $0^{\circ }C$ is calculated as follows: $\text{ Heat required = number of moles of ice }\times \text{molar heat of fusion}$

Molar mass of ice Mass of ice =

Number of moles of ice

  $\begin{array}{l}=\frac{\text{Mass of ice}}{\text{Molar mass of ice}}\\ \text{= }\frac{100g}{18g/mol}\\ =5.55mol\end{array}$

  $\begin{array}{l}\text{Heat required}=5.55mol\times 6.02kJ/mol\\ \text{ = 33}\text{.411}kJ\end{array}$

With the same heat number of moles of water that would vaporize is calculated as follows:

  $\text{Heat required = number of moles of water }\times \text{molar heat of vaporization }$

  $\begin{array}{l}\text{Number of moles of water}=\frac{\text{33}\text{.411}kJ}{40.6kJ/mol}\\ \text{ = 0}\text{.822 mol}\end{array}$

From the number of moles of water, mass of water can be calculated as follows:

  $\begin{array}{l}\text{Mass of water = Molar mass of water × Number of moles of water}\\ \text{where Molar mass of water}=18g/mol\end{array}$

  $\begin{array}{l}\text{Mass of water = Molar mass × Number of moles of water}\\ \text{ = 18 }g/mol\times \text{ 0}\text{.822 }mol\\ \text{ = 14}\text{.79 }g\end{array}$

Conclusion

Heat required to melt $100g$ of ice at $0^{\circ }C$ is $\text{33}\text{.411}kJ$ and the mass of water that would vaporize with the same amount of heat is $\text{14}\text{.79 }g$ .