Chapter 14.1, Problem 7RQ

Interpretation Introduction

Interpretation: With thegiven information the heat energy required to change $1.00$ mol of ice at $-10.0{}^{\circ }\text{C}$ to water at $15{}^{\circ }\text{C}$ is to be calculated.

Concept introduction:In going from ice at $-{10.0}^{\circ }\text{C}$ to water at ${15}^{\circ }\text{C}$ , heat energy is required which can be calculated as follows:

Heat = $\begin{array}{l}\text{mass of ice × specific heat of ice × temperature change +}\\ \text{ number of moles of ice }\times \text{molar heat of fusion +}\\ \text{mass of water × specific heat of water × temperature change}\end{array}$

Answer to Problem 7RQ

The heat energy required to change $1.00$ mol of ice at $-{10.0}^{\circ }\text{C}$ to water at ${15}^{\circ }\text{C}$ is $7515.0\text{8 J}$ .

Explanation of Solution

In going from ice at $-{10.0}^{\circ }\text{C}$ to water at ${15}^{\circ }\text{C}$ , heat energy is required which can be calculated as follows:

Heat = $\begin{array}{l}\text{mass of ice × specific heat of ice × temperature change +}\\ \text{ number of moles of ice }\times \text{molar heat of fusion +}\\ \text{mass of water × specific heat of water × temperature change}\end{array}$

It is given that,

  $\begin{array}{l}\text{specific heat of ice}=2.03{\text{ J/g }}^{\text{o}}\text{C},\text{specific heat of water}=4.184{\text{ J/g }}^{\text{o}}\text{C and}\\ \text{molar heat of fusion}=6.02\text{ kJ/mol}\end{array}$

The molar mass of ice = $\text{18 g/mol}$ and Temperature change = Final temperature − Initial temperature.

Calculation:

  $\begin{array}{c}\text{Q=18 g×2}{\text{.03 J/g×}}^{\text{o}}{\text{C×(0 }}^{\text{o}}\text{C-(-10}{\text{.0 }}^{\text{o}}\text{C))+1 mol×6}\text{.02 kJ/mol+18 g×4}{\text{.184 J/g}}^{\text{o}}{\text{C×(15 }}^{\text{o}}{\text{C-0 }}^{\text{o}}\text{C)}\end{array}$

    $\begin{array}{c}\text{=365}\text{.4 J+6}\text{.02 kJ+1129}\text{.68 J}\end{array}$

    $\begin{array}{c}\text{=365}\text{.4 J+6020 J+1129}\text{.68 J}\end{array}$

    $\begin{array}{c}\text{=7515}\text{.08 J}\end{array}$

Conclusion

Heat energy required to change $1.00$ mol of ice at $-10.0{}^{\text{o}}\text{C}$ to water at $15{}^{\text{o}}\text{C}$ is $\text{7515}\text{.08 J}$ .