Chapter 4, Problem 11PEB

To determine

The amount of heat removed by the cooling tower for each kilogram of spent steam.

Answer to Problem 11PEB

The amount of heat removed by the cooling tower for each kilogram of spent steam is $560\text{kcal}$.

Explanation of Solution

To change the steam at $120.0°\text{C}$ into water at $90.0\text{°C}$, three different quantities of heat are required.

First is the heat required to lower the temperature of steam to $100.0\text{°C}$. Second is the heat required to cause a phase change from steam to water at $100.0°\text{C}$. Third is the heat required to lower the temperature of water to $90.0\text{°C}$.

Write the expression for the total heat removed.

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

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

Write the expression for the heat removed to lower the temperature of steam.

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

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.

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

Write the expression for the heat removed for phase change of steam.

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

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

Write the expression for the heat removed to lower the temperature of water.

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

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.

Write the expression for the change in temperature.

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

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

Conclusion:

Substitute $100.0°\text{C}$ for $T_1$ and $120.0°\text{C}$ for $T_2$ in equation (V) to find $\Delta T_{\text{steam}}$.

    $\begin{array}{c}\Delta T_{\text{steam}}=120.0°\text{C}-100.0°\text{C}\\ =20.0°\text{C}\end{array}$

Substitute $20.0°\text{C}$ for $\Delta T_{\text{steam}}$ and $1.0\text{kg}$ for $m$ and $0.480\text{cal}/\text{g°C}$ for $c_{\text{steam}}$ in equation (II) to find $Q_1$.

    $\begin{array}{c}{Q}_1=\left(1.0\text{kg}\times \frac{{10}^3\text{g}}{1\text{kg}}\right)\left(0.480\text{cal}/\text{g°C}\right)\left(20.0°\text{C}\right)\\ =9600\text{cal}\end{array}$

Substitute $1.0\text{kg}$ for $m$ and $540.0\text{cal}/\text{g}$ for $L_{\text{v}}$ in equation (III) to find $Q_2$.

    $\begin{array}{c}{Q}_2=\left(1.0\text{kg}\times \frac{{10}^3\text{g}}{1\text{kg}}\right)\left(540.0\text{cal}/\text{g}\right)\\ =540000\text{cal}\end{array}$

Substitute $90.0°\text{C}$ for $T_2$ and $100.0°\text{C}$ for $T_1$ in equation (V) to find $\Delta T_{\text{water}}$.

    $\begin{array}{c}\Delta T_{\text{water}}=100.0°\text{C}-90.0°\text{C}\\ =10.0°\text{C}\end{array}$

Substitute $10.0°\text{C}$ for $\Delta T_{\text{water}}$, $1.0\text{kg}$ for $m$ and $1.00\text{cal}/\text{g°C}$ for $c_{\text{water}}$ in equation (IV) to find $Q_3$.

    $\begin{array}{c}{Q}_3=\left(1.0\text{kg}\times \frac{{10}^3\text{g}}{1\text{kg}}\right)\left(1.00\text{cal}/\text{g°C}\right)\left(10.0°\text{C}\right)\\ =10000\text{cal}\end{array}$

Substitute $9600\text{cal}$ for $Q_1$, $540000\text{cal}$ for $Q_{\text{2}}$ and $10000\text{cal}$ for $Q_3$ in equation (I) to find $Q$.

    $\begin{array}{c}Q=9600\text{cal}+540000\text{cal}+10000\text{cal}\\ =\left(\text{559600}\text{cal}\times \frac{{10}^{-3}\text{kcal}}{1\text{cal}}\right)\\ \approx 560\text{kcal}\end{array}$

Therefore, the amount of heat removed by the cooling tower for each kilogram of spent steam is $560\text{kcal}$.