Chapter 15, Problem 8PE

Interpretation Introduction

Interpretation:

The amount of energy released to lower the temperature of $50.0\text{-g}$ of sample of steam from $133°\text{C}$ to ice at $-22°\text{C}$ 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 different in the energy of 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 8PE

The amount of energy released to lower the temperature of $50.0\text{-g}$ of sample of steam from $133°\text{C}$ to ice at $-22°\text{C}$ is $-156\text{kJ}$.

Explanation of Solution

The mass of sample of steam is $50.0\text{g}$.

The boiling temperature of the water is $100°\text{C}$.

The melting temperature of the ice is $0°\text{C}$.

The required initial temperature is $133°\text{C}$.

The required final temperature is $-22°\text{C}$.

The amount of energy released to raise the temperature of steam from $133°\text{C}$ to $100°\text{C}$ 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 steam.

• ${\text{T}}_{\text{2}}$ is the final temperature $\left(100°\text{C}\right)$.

• ${\text{T}}_{\text{1}}$ is the initial temperature $\left(133°\text{C}\right)$.

The specific heat of steam is $2.00\text{J}/\text{g}\cdot °\text{C}$.

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

$\begin{array}{rll}{\text{q}}_{\text{1}}& =& \left(50\text{g}\right)\left(2.00\text{J}/\text{g}\cdot °\text{C}\right)\left(100°\text{C}-133°\text{C}\right)\\ & =& \left(-3.3\times {10}^3\text{J}\right)\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ & =& -3.3\text{kJ}\end{array}$

The amount of energy released for phase transformation from gas to liquid is calculated by the formula shown below.

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

Where,

• is the heat of vaporization.

The heat of vaporization of water is $2.26\text{kJ}/\text{g}$.

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

$\begin{array}{rll}{\text{q}}_{\text{2}}& =& -\left(50\text{g}\right)\left(2.26\text{kJ}/\text{g}\right)\\ & =& -113\text{kJ}\end{array}$

The amount of energy released to raise the temperature of water from $100°\text{C}$ to $0°\text{C}$ 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 water.

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

• ${\text{T}}_{\text{1}}$ is the initial temperature $\left(100°\text{C}\right)$.

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).

$\begin{array}{rll}{\text{q}}_{\text{3}}& =& \left(50\text{g}\right)\left(4.18\text{J}/\text{g}\cdot °\text{C}\right)\left(0°\text{C}-100°\text{C}\right)\\ & =& \left(-20.9\times {10}^3\text{J}\right)\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ & =& -20.9\text{kJ}\end{array}$

The amount of energy released for phase transformation from solid to liquid is calculated by the formula shown below.

${\text{q}}_{\text{4}}=-\text{m}\Delta {\text{H}}_{\text{fus}}$…(4)

Where,

• is the heat of fusion.

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

Substitute the mass and heat of fusion in equation (4).

$\begin{array}{rll}{\text{q}}_{\text{4}}& =& -\left(50\text{g}\right)\left(333\text{J}/\text{g}\right)\\ & =& -\left(16.650\times {10}^3\text{J}\right)\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ & \approx & -16.7\text{kJ}\end{array}$

The amount of energy released to raise the temperature of ice from $0°\text{C}$ to $-22°\text{C}$ is calculated by the formula shown below.

${\text{q}}_{\text{3}}=\text{mc}\left({\text{T}}_2-{\text{T}}_1\right)$…(5)

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 $\left(-22°\text{C}\right)$.

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

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 (5).

$\begin{array}{rll}{\text{q}}_{\text{5}}& =& \left(50\text{g}\right)\left(2.06\text{J}/\text{g}\cdot °\text{C}\right)\left(-22°\text{C}-0°\text{C}\right)\\ & =& -\left(2.266\times {10}^3\text{J}\right)\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ & \approx & -2.3\text{kJ}\end{array}$

The total amount of energy sample released when temperature is changed from $133°\text{C}$ to $-22°\text{C}$ is shown below.

$\text{q}={\text{q}}_1+{\text{q}}_2+{\text{q}}_3+{\text{q}}_4+{\text{q}}_5$

Substitute the value of energies in the above equation.

$\begin{array}{rll}\text{q}& =& -3.3\text{kJ}-113\text{kJ}-20.9\text{kJ}-16.7\text{kJ}-2.3\text{kJ}\\ & =& -156.2\text{kJ}\\ & \approx & -156\text{kJ}\end{array}$

Therefore, the amount of energy released to lower the temperature of $50.0\text{-g}$ of sample of steam from $133°\text{C}$ to ice at $-22°\text{C}$ is $-156\text{kJ}$.

Conclusion

The amount of energy released to lower the temperature of $50.0\text{-g}$ of sample of steam from $133°\text{C}$ to ice at $-22°\text{C}$ is $-156\text{kJ}$.