Chapter 15, Problem 93E

Interpretation Introduction

Interpretation:

The amount of energy gained by the sample when the temperature is changed from $-11°\text{C}$ to $21°\text{C}$ is to be calculated.

Concept introduction:

The amount of energy required to change the state of a substance is known as enthalpy. It is the difference in the energy of the final and initial state of a substance. The negative and positive sign of enthalpy indicates the energy released and energy absorbed, respectively, during the phase change.

Answer to Problem 93E

The amount of energy gained by the sample when the temperature is changed from $-11°\text{C}$ to $21°\text{C}$ is $56.32\text{kJ}$.

Explanation of Solution

Three stages are involved when the temperature of ice sample is raised from $-11°\text{C}$ to $21°\text{C}$. The three states are heating of solid, melting of solid and heating of liquid.

The amount of energy required to raise the temperature of ice from $-11°\text{C}$ to $0°\text{C}$ is calculated by the formula shown below.

${\text{q}}_{\text{1}}=\text{mc}\left({\text{T}}_2-{\text{T}}_1\right)$…(1)

Where,

• $\text{m}$ is the mass of the sample.

• $\text{c}$ is the specific heat of ice.

• ${\text{T}}_{\text{2}}$ is the final temperature $\left(0°\text{C}\right)$.

• ${\text{T}}_{\text{1}}$ is the initial temperature $\left(-11°\text{C}\right)$.

The specific heat of ice is $2.06\text{J}/\text{g}\cdot °\text{C}$.

Substitute the mass, final, initial temperature and specific heat of ice in equation (1).

$\begin{array}{rll}{\text{q}}_{\text{1}}& =& \text{127}\text{g}\times 2.06\text{J}/\text{g}\cdot °\text{C}\left(0°\text{C}-\left(-11°\text{C}\right)\right)\\ & =& 2877.82\text{J}\end{array}$

The amount of energy required to melt a solid is calculated by the formula shown below.

${\text{q}}_{\text{2}}=\text{m}\Delta {\text{H}}_{\text{fus}}$…(2)

Where,

• is the heat of fusion.

The heat of fusion of water is $333\text{J}/\text{g}$.

Substitute the mass and heat of fusion in equation (2).

$\begin{array}{rll}{\text{q}}_{\text{2}}& =& \text{127}\text{g}\times 333\text{J}/\text{g}\\ & =& 42291\text{J}\end{array}$

The amount of energy required to raise the temperature of water from $0°\text{C}$ to $21°\text{C}$ is calculated by the formula shown below.

${\text{q}}_{\text{3}}=\text{mc}\left({\text{T}}_2-{\text{T}}_1\right)$…(3)

Where,

• $\text{m}$ is the mass of the sample.

• $\text{c}$ is the specific heat of ice.

• ${\text{T}}_{\text{2}}$ is the final temperature $\left(21°\text{C}\right)$.

• ${\text{T}}_{\text{1}}$ is the initial temperature $\left(0°\text{C}\right)$.

The specific heat of water is $4.18\text{J}/\text{g}\cdot °\text{C}$.

Substitute the mass, final, initial temperature and specific heat of water in equation (3).

$\begin{array}{rll}{\text{q}}_{\text{3}}& =& \text{127}\text{g}\times 4.18\text{J}/\text{g}\cdot °\text{C}\left(21°\text{C}-\left(0°\text{C}\right)\right)\\ & =& 11148.06\text{J}\end{array}$

The total amount of energy sample gained when temperature is changed from $-11°\text{C}$ to $21°\text{C}$ is shown below.

$\begin{array}{rll}\text{q}& =& {\text{q}}_1+{\text{q}}_2+{\text{q}}_3\\ & =& 2877.82\text{J}+42291\text{J}+11148.06\text{J}\\ & =& 56316.88\text{J}\end{array}$

Convert $56316.88\text{J}$ to $\text{kJ}$.

$\begin{array}{rll}56316.88\text{J}& =& \frac{56316.88}{1000}\text{kJ}\\ & =& 56.32\text{kJ}\end{array}$

Conclusion

The amount of energy sample gained when temperature is changed from $-11°\text{C}$ to $21°\text{C}$ is $56.32\text{kJ}$.