Chapter 11, Problem 32P

To determine

The energy required to change a 40 g ice cube from ice at $-10°\text{C}$ to steam at $110°\text{C}$.

Answer to Problem 32P

The energy required is $0.12\text{MJ}$.

Explanation of Solution

Given Info: Mass of ice cube is 40 g, initial temperature of ice cube is $-10°\text{C}$, temperature of steam is $110°\text{C}$,

Explanation:

Energy supplied must lower the temperature of ice from $-10°\text{C}$  to $0°\text{C}$, then melt the ice to cold water at $0°\text{C}$, then it has to raise the temperature of water to $100°\text{C}$, then transform the water to steam at $100°\text{C}$ and then increase the temperature of steam to $110°\text{C}$.

Formula to calculate the heat required to raise the temperature of ice $0°\text{C}$ is,

$Q_{\text{ice}}=m_{\text{ice}}{c}_{\text{ice}}\left(T_f{,}_{\text{ice}}-T_i{,}_{\text{ice}}\right)$ (I)

• $Q_{\text{ice}}$ is the heat required to raise the temperature of ice to $0°\text{C}$,
• $m_{\text{ice}}$ is the mass of the ice,
• $T_i{,}_{\text{ice}}$ is the initial temperature of ice,
• $T_f{,}_{\text{ice}}$ is the final temperature of ice,

Formula to calculate the heat required to melt the ice to cold water is,

$Q_{\text{melt}}=m_{\text{ice}}{L}_f$ (II)

• $Q_{\text{melt}}$ is the energy required to melt the ice to cold water,
• $m_{\text{ice}}$ is the mass of the ice,
• $L_f$ is the latent heat of fusion of ice,

Formula to calculate the heat required to raise the temperature of cold water to hot water at $100°\text{C}$ is,

$Q_{\text{boil}}=m_{\text{water}}{c}_{\text{water}}\left(T_{f,\text{water}}-T_i{,}_{\text{water}}\right)$ (III)

• $Q_{\text{boil}}$ is the heat required to raise the temperature of cold water to hot water.
• $c_{\text{water}}$ is the specific heat of water,
• $T_i{,}_{\text{water}}$ is the initial temperature of cold water from ice,
• $T_f$ is the boiling temperature of water,

Formula to calculate the heat required to transform water to steam at $100°\text{C}$ is,

$Q_{\text{steam}}=m_{\text{steam}}{L}_v$ (IV)

• $Q_{\text{steam}}$ is the energy required to transform water to steam,
• $m_{\text{steam}}$ is the mass of the steam,
• $L_v$ is the latent heat of vaporization,

Formula to calculate the heat required to raise the temperature of steam to $110°\text{C}$ is,

${Q^{\prime }}_{\text{steam}}=m_{\text{steam}}{c}_{\text{steam}}\left(T_{f,\text{steam}}-T_i{,}_{\text{steam}}\right)$ (V)

• ${Q^{\prime }}_{\text{steam}}$ is the heat required to raise the temperature of steam,
• $m_{\text{steam}}$ is the mass of steam,
• $c_{\text{steam}}$ is the specific heat of steam,
• $T_i{,}_{\text{steam}}$ is the initial temperature of steam,
• $T_{f,\text{steam}}$ is the final temperature of steam,

Formula to calculate the energy required to change 40 g ice cube from ice at $-10°\text{C}$ to steam at $110°\text{C}$ is,

$Q=Q_{\text{ice}}+Q_{\text{melt}}+Q_{\text{boil}}+Q_{\text{steam}}+{Q^{\prime }}_{\text{steam}}$ (VI)

Substitute equation (I), (II), (III) (IV) (V) in equation (VI) to calculate $Q_{\text{tran}}$.

$Q=\left\{\begin{array}{l}{m}_{\text{ice}}{c}_{\text{ice}}\left(T_f{,}_{\text{ice}}-T_i{,}_{\text{ice}}\right)+m_{\text{ice}}{L}_f+m_{\text{steam}}{L}_v\\ +m_{\text{water}}{c}_{\text{water}}\left(T_{f,\text{water}}-T_{i,\text{water}}\right)+m_{\text{steam}}{c}_{\text{steam}}\left(T_{f,\text{steam}}-T_i{,}_{\text{steam}}\right)\end{array}\right\}$

Substitute 40 g for $m_{\text{ice}}$, $2090\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{ice}}$, $0°\text{C}$ for $T_f{,}_{\text{ice}}$, $-10°\text{C}$ for $T_{i,ice}$, $3.33\times {10}^5\text{J}/\text{kg}$ for $L_f$ , 40 g for $m_{\text{water}}$, $4186\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{water}}$, $0°\text{C}$ for $T_{i,\text{water}}$, $100°\text{C}$ for $T_{f,\text{water}}$,  40 g from $m_{\text{steam}}$ , $2010\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{steam}}$, $100°\text{C}$ for $T_i{,}_{\text{steam}}$, $110°\text{C}$ for $T_{f,\text{steam}}$,  and $2.26\times {10}^6\text{J}/\text{kg}$ for $L_v$ to find $Q$.

$\begin{array}{c}Q=\left\{\begin{array}{l}\left(40\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)2090\text{J}/\text{kg}\cdot \text{°C}\left(10°-0°\right)\\ +\left(40\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\left(3.33\times {10}^5\text{J}/\text{kg}\right)\\ +\left(40\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\left(4186\text{J}/\text{kg}\cdot \text{°C}\right)\left(100°\text{C}-0°\text{C}\right)\\ +\left(40\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\left(2.26\times {10}^6\text{J}/\text{kg}\right)\\ +\left(40\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\left(2010\text{J}/\text{kg}\cdot \text{°C}\right)\left(110°-100°\right)\end{array}\right\}\\ =122104\text{J}\left(\frac{1\text{MJ}}{{10}^6\text{J}}\right)\\ \approx 0.12\text{MJ}\end{array}$

Conclusion:

Therefore, the energy required is $0.12\text{MJ}$.