Chapter 4, Problem 11PEA

To determine

The amount of heat absorbed when $100.0\text{g}$ of water at $20.0°\text{C}$ is heated to steam at $125.0°\text{C}$.

Answer to Problem 11PEA

The amount of heat absorbed when $100.0\text{g}$ of water at $20.0°\text{C}$ is heated to steam at $125.0°\text{C}$ is $63.20\text{kcal}$.

Explanation of Solution

To change the water at $20.0°\text{C}$ into steam at $125.0\text{°C}$, three different quantities of heat are required.

First is the heat required to raise the temperature of water to $100.0\text{°C}$. Second is the heat required to cause a phase change from water to steam at $100.0°\text{C}$. Third is the heat required to raise the temperature of steam to $125.0\text{°C}$.

Write the expression for the total heat absorbed.

    $Q=Q_1+Q_2+Q_3$                                                               (I)

Here, $Q$ is the total heat energy absorbed, $Q_1$ is the heat required to raise the temperature of water, $Q_2$ is the heat required for phase change of water and $Q_3$ is the heat required to raise the temperature of steam.

Write the expression for the heat required to raise the temperature of water.

    $Q_1=m{c}_{\text{water}}\Delta T_{\text{water}}$                                                           (II)

Here, $m$ is the mass of the water, $c_{\text{water}}$ is the specific heat of water and $\Delta T_{\text{water}}$ is the change in temperature of water.

Water completely changes to steam. Hence, the mass of water is same as the mass of steam.

Write the expression for the heat required for phase change of water.

    $Q_2=m{L}_{\text{v}}$                                                                      (III)

Here, $Q_2$ is the heat required for phase change of water, $m$ is the mass of water and $L_{\text{v}}$ is the latent heat of vaporization for water.

Write the expression for the heat required to raise the temperature of steam.

    $Q_3=m{c}_{\text{steam}}\Delta T_{\text{steam}}$                                                        (IV)

Here, $m$ is the mass of the steam, $c_{\text{steam}}$ is the specific heat of steam and $\Delta T_{\text{steam}}$ is the change in temperature of steam.

Write the expression for the change in temperature.

    $\Delta T=T_2-T_1$                                                                  (V)

Here, $T_2$ is the final temperature and $T_1$ is the initial temperature.

Conclusion:

Substitute $20.0°\text{C}$ for $T_1$ and $100.0°\text{C}$ for $T_2$ in equation (V) to find $\Delta T_{\text{water}}$.

    $\begin{array}{c}\Delta T_{\text{water}}=100.0°\text{C}-20.0°\text{C}\\ =80.0°\text{C}\end{array}$

Substitute $80.0°\text{C}$ for $\Delta T_{\text{water}}$, $100.0\text{g}$ for $m$ and $1.00\text{cal}/\text{g°C}$ for $c_{\text{water}}$ in equation (II) to find $Q_1$.

    $\begin{array}{c}{Q}_1=\left(100.0\text{g}\right)\left(1.00\text{cal}/\text{g°C}\right)\left(80.0°\text{C}\right)\\ =8000\text{cal}\end{array}$

Substitute $100.0\text{g}$ for $m$ and $540.0\text{cal}/\text{g}$ for $L_{\text{v}}$ in equation (III) to find $Q_2$.

    $\begin{array}{c}{Q}_2=\left(100.0\text{g}\right)\left(540.0\text{cal}/\text{g}\right)\\ =54000\text{cal}\end{array}$

Substitute $100.0°\text{C}$ for $T_1$ and $125.0°\text{C}$ for $T_2$ in equation (V) to find $\Delta T_{\text{steam}}$.

    $\begin{array}{c}\Delta T_{\text{steam}}=125.0°\text{C}-100.0°\text{C}\\ =25.0°\text{C}\end{array}$

Substitute $25.0°\text{C}$ for $\Delta T_{\text{steam}}$, $100.0\text{g}$ for $m$ and $0.480\text{cal}/\text{g°C}$ for $c_{\text{steam}}$ in equation (IV) to find $Q_3$.

    $\begin{array}{c}{Q}_3=\left(100.0\text{g}\right)\left(0.480\text{cal}/\text{g°C}\right)\left(25.0°\text{C}\right)\\ =1200\text{cal}\end{array}$

Substitute $8000\text{cal}$ for $Q_1$, $54000\text{cal}$ for $Q_{\text{2}}$ and $1200\text{cal}$ for $Q_3$ in equation (I) to find $Q$.

    $\begin{array}{c}Q=8000\text{cal}+54000\text{cal}+1200\text{cal}\\ =\left(\text{63200}\text{cal}\times \frac{{10}^{-3}\text{kcal}}{1\text{cal}}\right)\\ =63.20\text{kcal}\end{array}$

Therefore, the amount of heat absorbed when $100.0\text{g}$ of water at $20.0°\text{C}$ is heated to steam at $125.0°\text{C}$ is $63.20\text{kcal}$.