Chapter 11, Problem 39P

Steam at 100.°C is added to ice at 0°C. (a) Find the amount of ice melted and the final temperature when the mass of steam is 10. g and the mass of ice is 50. g. (b) Repeat with steam of mass 1.0 g and ice of mass 50. g.

To determine

The amount of ice melted and the final temperature when the mass of steam is 10 g.

Answer to Problem 39P

Complete ice melts and the final temperature is $40°\text{C}$

Explanation of Solution

Section1:

To determine: Amount of ice melts.

Answer: All the ice melts.

Given Info: Mass of ice is50 g, mass of steam is 10 g, initialtemperature of iceis $0°\text{C}$, initial temperature of steam is $100°\text{C}$,

Formula to calculate the energy required to melt 50 g of ice is,

  $Q_A=m_{\text{ice}}{L}_f$

• $Q_A$ is the heat required to melt 50 g of ice,
• $m_{\text{ice}}$ is the mass of ice,
• $L_f$ is the latent heat of fusion of ice,

Substitute 50 g for $m_{\text{ice}}$ and $333\text{kJ}/\text{kg}$ for $L_f$ to find $Q_A$.

  $\begin{array}{c}{Q}_A=\left(50\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(333\text{kJ}/\text{kg}\right)\\ =16650\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ \approx \text{17}\text{kJ}\end{array}$

Formula to calculate the energy required to warm 50 g of ice-water from $0°\text{C}$ to $100°\text{C}$ is,

  $Q_B=m_{\text{ice}}{c}_{\text{water}}\left(T_f-T_{i,\text{ice-water}}\right)$

• $Q_B$ is the heat required to warm water,
• $c_{\text{water}}$ is the specific heat of water,
• $T_{i,\text{ice-water}}$ is the initial temperature of ice-water.
• $T_f$ is the boiling temperature of water,

  $\begin{array}{c}{Q}_B=\left(50\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(4186\text{J}/\text{kg}\cdot °\text{C}\right)\left(100°\text{C}-0°\text{C}\right)\\ =209300\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ \approx 21\text{kJ}\end{array}$

Formula to calculate energy given out by stem to condense to water is,

  $Q_C=m_{\text{steam}}{L}_v$

• $Q_C$ is the energy given out by stem to condense to water,
• $m_{\text{steam}}$ is the mass of the steam,
• $L_v$ is the latent heat of vaporization of water,

Substitute 10 g for $m_{\text{steam}}$ and $2260\text{kJ}/\text{kg}$ for $L_v$ to find $0°\text{C}$ for $Q_C$.

  $\begin{array}{c}{Q}_{\text{C}}=\left(10\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(2260\text{k}\text{J}/\text{kg}\right)\\ =22600\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ \approx 23\text{kJ}\end{array}$

The heat released by the steam as it condenses to water at $100°\text{C}$ is greater than the energy required to melt the ice that is $Q_C>Q_A$. Ice melts completely.

Section2:

To determine: Final temperature.

Answer: Final temperature is $40.0°\text{C}$.

Given Info:

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

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

• $Q_{\text{melt}}$ 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 is,

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

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

Formula to calculate the heat lost by the steam when it condensed to hot water at $100°\text{C}$ is,

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

• $Q_{\text{steam}}$ is the energy given out when steam condenses to water,
• $m_{\text{steam}}$ is the mass of the steam,
• $L_v$ is the latent heat of vaporization,

Formula to calculate the heat lost by the steam to reduces its temperature from $100°\text{C}$ to equilibrium temperature is,

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

• ${Q^{\prime }}_{\text{steam}}$ is the heat lost by the steam to reduces its temperature from $100°\text{C}$ to equilibrium temperature,
• $m_{\text{steam}}$ is the mass of steam,
• $c_{\text{steam}}$ is the specific heat of steam,
• $T_i{,}_{\text{steam}}$ is the initial temperature of steam,

Energy supplied by the steam when it condenses to water at $100°\text{C}$ and then reduces it temperature to $T_f$ . This energy is gained melts the ice to cold water at $0°\text{C}$, then it has to raise the temperature of water to to $T_f$.

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

Substitute equation (I), (II), (III) (IV) in equation (V) to calculate $T_f$.

$\begin{array}{c}{m}_{\text{ice}}{L}_f+m_{\text{ice}}{c}_{\text{water}}\left(T_f-T_i{,}_{\text{ice}}\right)=-m_{\text{steam}}{L}_v-m_{\text{steam}}{c}_{\text{water}}\left(T_f-T_i{,}_{\text{steam}}\right)\\ m_{\text{ice}}{L}_f+m_{\text{ice}}{c}_{\text{water}}{T}_f-m_{\text{ice}}{c}_{\text{water}}{T}_i{,}_{\text{ice}}=-m_{\text{steam}}{L}_v-m_{\text{steam}}{c}_{\text{water}}{T}_f+m_{\text{steam}}{c}_{\text{water}}{T}_i{,}_{\text{steam}}\\ m_{\text{ice}}{c}_{\text{water}}{T}_f+m_{\text{steam}}{c}_{\text{water}}{T}_f=m_{\text{steam}}{c}_{\text{steam}}{T}_i{,}_{\text{steam}}-m_{\text{steam}}{L}_v+m_{\text{ice}}{c}_{\text{water}}{T}_i{,}_{\text{ice}}-m_{\text{ice}}{L}_f\\ T_f=\frac{m_{\text{steam}}\left(c_{\text{water}}{T}_i{,}_{\text{steam}}-L_v\right)+m_{\text{ice}}\left(c_{\text{water}}{T}_i{,}_{\text{ice}}-L_f\right)}{c_{\text{water}}\left(m_{\text{ice}}+c_{\text{water}}\right)}\end{array}$

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

$\begin{array}{c}{T}_f=\frac{\left\{\begin{array}{l}\left\{\left(10\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(\left[\left(4186\text{J}/\text{kg}\cdot \text{°C}\right)\left(100°\text{C}\right)\right]+\left(2.26\times {10}^6\text{J}/\text{kg}\right)\right)\right\}\\ +\left\{\left(50\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(\left[\left(4186\text{J}/\text{kg}\cdot \text{°C}\right)\left(0°\text{C}\right)\right]-\left(3.33\times {10}^5\text{J}/\text{kg}\right)\right)\right\}\end{array}\right\}}{\left[\left(50\text{g}\right)+\left(10\text{g}\right)\right]\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(4186\text{J}/\text{kg}\cdot \text{°C}\right)}\\ =40.356°\text{C}\\ \approx \text{40}°\text{C}\end{array}$

Therefore, the final temperature of the mixture is $\text{40}°\text{C}$.

Conclusion:

Therefore, the ice melts completely and the final temperature of the mixture is $\text{40}°\text{C}$.

To determine

The amount of ice melted and the final temperature when the mass of steam is 1.0 g.

Answer to Problem 39P

Final temperature of mixture is $0°\text{C}$ and the mass of ice melts is $\text{8}\text{.0}\text{g}$.

Explanation of Solution

Section1:

To determine: Amount of ice melts.

Answer: Partial the ice melts.

Given Info: Mass of ice is 50 g, mass of steam is 1.0 g, initial temperature of ice is $0°\text{C}$, initial temperature of steam is $100°\text{C}$,

Formula to calculate energy given out by stem to condense to water is,

  $Q_C=m_{\text{steam}}{L}_v$

Substitute 1.0 g for $m_{\text{steam}}$ and $2260\text{kJ}/\text{kg}$ for $L_v$ to find $0°\text{C}$ for $Q_C$.

  $\begin{array}{c}{Q}_{\text{C}}=\left(1.0\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(2260\text{k}\text{J}/\text{kg}\right)\\ =2260\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ \approx 2.3\text{kJ}\end{array}$

Formula to calculate the heat lost by the steam to reduces its temperature from $100°\text{C}$ to $0°\text{C}$ is,

  ${Q^{\prime }}_C=m_{\text{steam}}{c}_{\text{water}}\left|T_f-T_i{,}_{\text{steam}}\right|$

Substitute 1.0 g for $m_{\text{steam}}$, $4186\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{water}}$, $100°\text{C}$ for $T_i{,}_{\text{steam}}$, and $0°\text{C}$ for $T_f$ to find ${Q^{\prime }}_C$.

  $\begin{array}{c}{Q^{\prime }}_C=\left(1.0\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1.0\text{g}}\right)\left(4186\text{J}/\text{kg}\cdot \text{°C}\right)\left|0°\text{C}-100°\text{C}\right|\\ =0.42\text{kJ}\end{array}$

The heat released by the steam as it condenses to water at $100°\text{C}$ and its subsequent cools to $0°\text{C}$ islesser than the energy required to melt the ice that is $Q_C+{Q^{\prime }}_C<Q_A$. Ice melts partially. Final temperature of the mixture is $0°\text{C}$.

Section2:

To determine: The amount of ice melted.

Answer: The amount of ice melted is 8.0 g.

Given Info:

The amount of heat released as the steam condenses to water $100°\text{C}$ and subsequent cooled to $0°\text{C}$ is supplied to ice to melt.

Formula to calculate the amount of ice melt is,

  $\begin{array}{c}{Q}_C+{Q^{\prime }}_C=m{L}_f\\ m=\frac{Q_C+{Q^{\prime }}_C}{L_f}\end{array}$

• m is the mass of ice melted

Substitute 2.3 kJ for $Q_C$ and 0.42 kJ for $Q^{\prime }$, and $3.33\times {10}^5\text{J}/\text{kg}$ for $L_f$ to find m.

  $\begin{array}{c}m=\frac{2.3\text{kJ}+0.42\text{kJ}}{3.33\times {10}^5\text{J}/\text{kg}}\\ =8.0168\text{g}\\ \approx \text{8}\text{.0}\text{g}\end{array}$

Therefore, the mass of ice melt is $\text{8}\text{.0}\text{g}$.

Conclusion:

Therefore, the final temperature of mixture is $0°\text{C}$ and the mass of ice melts is $\text{8}\text{.0}\text{g}$.