Chapter 20, Problem 20.21P

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 10.0 g and the mass of ice is 50.0 g. (b) What If? Repeat when the mass of steam is 1.00 g and the mass of ice is 50.0 g.

To determine

The amount of ice melted and final temperature when the mass of steam is $10.0\text{g}$ and the mass of ice is $50.0\text{g}$.

Answer to Problem 20.21P

The amount of ice melted is $50\text{g}$ and final temperature is $40.4°\text{C}$.

Explanation of Solution

Given info: Steam at $100°\text{C}$ is added to ice at $0°\text{C}$.

The expression for the energy needed to melt $50\text{g}$ of ice is,

$Q_1=m_{\text{ice}}{L}_{\text{f}}$

Here,

$m_{\text{ice}}$ is the mass of ice.

$L_{\text{f}}$ is the latent heat of fusion of ice.

Substitute $50\text{g}$ for $m_{\text{ice}}$ and $3.33\times {10}^5\text{J}/\text{kg}$ for $L_{\text{f}}$ in above equation.

$\begin{array}{c}{Q}_1=50\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times 3.33\times {10}^5\text{J}/\text{kg}\\ =16650\text{J}\end{array}$

Thus, the energy needed to melt $50\text{g}$ of ice is $16650\text{J}$.

The expression for the energy needed to warm $50\text{g}$ of melted ice water from $0°\text{C}$ to $100°\text{C}$ is,

$Q_2=m_{\text{ice}}{c}_{\text{w}}\Delta T_{\text{water}}$

Here,

$c_{\text{w}}$ is the specific heat capacity of water.

$\Delta T_{\text{water}}$ is the change in temperature of water.

Substitute $50\text{g}$ for $m_{\text{ice}}$, $4186\text{J}/\text{kg-C}$ for $c_{\text{w}}$ and $100°\text{C}$ for $\Delta T_{\text{water}}$ in above equation.

$\begin{array}{c}{Q}_2=50\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times 4186\text{J}/\text{kg-C}\times 100°\text{C}\\ =20930\text{J}\end{array}$

Thus, the energy needed to warm $50\text{g}$ of melted ice water from $0°\text{C}$ to $100°\text{C}$ is $20930\text{J}$.

The expression for the energy released by $10\text{g}$ of steam when it condenses to water at $100°\text{C}$ is,

$Q_3=m_{\text{steam}}{L}_{\text{v}}$

Here,

$m_{\text{steam}}$ is the mass of steam.

$L_{\text{v}}$ is the latent heat of vaporization of steam.

Substitute $10\text{g}$ for $m_{\text{steam}}$ and $2.26\times {10}^6\text{J}/\text{kg}$ for $L_{\text{v}}$ in above equation.

$\begin{array}{c}{Q}_3=10\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times 2.26\times {10}^5\text{J}/\text{kg}\\ =22600\text{J}\end{array}$

Thus, the energy released by $10\text{g}$ of steam when it condenses to water at $100°\text{C}$ is $22600\text{J}$.

The expression for the total amount of heat required to convert $50\text{g}$ of ice to melted ice water at $100°\text{C}$ is,

$Q_{\text{absolute}}=Q_1+Q_2$

Substitute $16650\text{J}$ for $Q_1$ and $20930\text{J}$ for $Q_2$ in above equation.

$\begin{array}{c}{Q}_{\text{absolute}}=16650\text{J}+20930\text{J}\\ =37580\text{J}\end{array}$

Thus, the total amount of heat required to convert $50\text{g}$ of ice to melted ice water at $100°\text{C}$ is $37580\text{J}$.

From the above result it is clear that the amount of energy released by the steam to condense is more than the amount of energy needed to melt $50\text{g}$ of ice to water at $0°\text{C}$.

So, the all the ice will melt to water.

For the final common temperature of ice-steam system:

$\begin{array}{c}{Q}_{\text{released}}=Q_{\text{absorved}}\\ Q_3+m_{\text{steam}}{c}_{\text{w}}\left(T_{\text{is}}-T\right)=\left[Q_1+m_{\text{ice}}{c}_{\text{w}}\left(T-T_{\text{iw}}\right)\right]\end{array}$

Here,

$T$ is the common temperature ice-steam system.

$T_{\text{is}}$ is the initial temperature of steam.

$T_{\text{iw}}$ is the initial temperature of water.

Substitute $22600\text{J}$ for $Q_3$, $16650\text{J}$ for $Q_1$, $50\text{g}$ for $m_{\text{ice}}$, $4186\text{J}/\text{kg-C}$ for $c_{\text{w}}$, $100°\text{C}$ for $T_{\text{is}}$, $10\text{g}$ for $m_{\text{steam}}$ and $0°\text{C}$ for $T_{\text{iw}}$ in above equation.

$\begin{array}{l}22600\text{J}+10\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times 4186\text{J}/\text{kg-C}\times \left(100°\text{C}-T\right)=\\ \left[16650\text{J}+50\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times 4186\text{J}/\text{kg-C}\times \left(T-0°\text{C}\right)\right]\\ T=40.4°\text{C}\end{array}$

Conclusion:

Therefore, the amount of ice melted is $50\text{g}$ and final temperature is $40.4°\text{C}$.

To determine

The amount of ice melted and final temperature when the mass of steam is $1.0\text{g}$ and the mass of ice is $50.0\text{g}$.

Answer to Problem 20.21P

The amount of ice melted is $\text{8}\text{.04}\text{g}$ and final temperature is $0°\text{C}$.

Explanation of Solution

The expression of the energy released by $1.0\text{g}$ of steam in the condensation of water at $100°\text{C}$ is,

$Q_4=m_{\text{steam}}{L}_{\text{v}}$

Substitute $1.0\text{g}$ for $m_{\text{steam}}$ and $2.26\times {10}^5\text{J}/\text{kg}$ for $L_{\text{v}}$ in above equation.

$\begin{array}{c}{Q}_4=1.0\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times 2.26\times {10}^5\text{J}/\text{kg}\\ =2260\text{J}\end{array}$

As the energy released by the steam is much less than the amount of heat required melting the ice to water. So, the final temperature of the ice steam system will be $0°\text{C}$.

For the final common temperature of ice-steam system:

$\begin{array}{c}{Q}_{\text{released}}=-Q_{\text{absorved}}\\ Q_4+m_{\text{steam}}{c}_{\text{w}}\left(T-T_{\text{is}}\right)=-m_{\text{melt}}{L}_{\text{v}}\end{array}$

Here,

$T$ is the final temperature ice-steam system.

$T_{\text{is}}$ is the initial temperature of steam.

$m_{\text{melt}}$ is the mass of ice melt.

Substitute $2260\text{J}$ for $Q_4$, $4186\text{J}/\text{kg-C}$ for $c_{\text{w}}$, $100°\text{C}$ for $T_{\text{is}}$, $1.0\text{g}$ for $m_{\text{steam}}$ and $2.26\times {10}^5\text{J}/\text{kg}$ for $L_{\text{v}}$ in above equation.

$\begin{array}{c}2260\text{J}+1.0\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times 4186\text{J}/\text{kg-C}\times \left(0°\text{C}-100°\text{C}\right)=-m_{\text{melt}}\times 2.26\times {10}^5\text{J}/\text{kg}\\ =814.78\times {10}^{-5}\text{kg}\left(\frac{1000\text{g}}{1\text{kg}}\right)\\ =8.14\text{g}\\ \approx \text{8}\text{.04}\text{g}\end{array}$

Conclusion:

Therefore, the amount of ice melted is $\text{8}\text{.04}\text{g}$ and final temperature is $0°\text{C}$.