Chapter 4, Problem 10PEA

To determine

The amount of heat needed to change $250\text{g}$ of water at $80.0°\text{C}$ to steam at $100.0°\text{C}$.

Answer to Problem 10PEA

The amount of heat needed to change $250\text{g}$ of water at $80.0°\text{C}$ to steam at $100.0°\text{C}$ is $140.0\text{kcal}$.

Explanation of Solution

To change the water at $80°\text{C}$ into steam at $100.0\text{°C}$, two different quantities of heat are required.

One is the heat required to raise the temperature of water and the other one is the heat required to cause the phase change of water.

Write the expression for the total heat required.

    $Q=Q_1+Q_2$                                                                              (I)

Here, $Q$ is the total heat energy required, $Q_1$ is the heat required to raise the temperature and $Q_2$ is the heat required to change the phase.

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

    $Q_1=mc\Delta T$                                                                             (II)

Here, $Q_1$ is the heat required to raise the temperature, $m$ is the mass of water, $c$ is the specific heat and $\Delta T$ is the change in temperature.

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

Write the expression for the heat required to change the phase into steam.

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

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

Write the expression for the change in temperature.

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

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

Conclusion:

Substitute $80.0°\text{C}$ for $T_1$ and $100.0°\text{C}$ for $T_2$ in equation (IV) to find $\Delta T$.

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

Substitute $20.0°\text{C}$ for $\Delta T$, $250.0\text{g}$ for $m$ and $1.00\text{cal}/\text{g°C}$ for $c$ in equation (II) to find $Q_1$

    $\begin{array}{c}{Q}_1=\left(250.0\text{g}\right)\left(1.00\text{cal}/\text{g°C}\right)\left(20.0°\text{C}\right)\\ =5000\text{cal}\end{array}$

Substitute $250.0\text{g}$ for $m$ and $540.0\text{cal}/\text{g}$ for $L_{\text{v}}$ in equation (III) to find $Q_2$.

    $\begin{array}{c}{Q}_2=\left(250.0\text{g}\right)\left(540.0\text{cal}/\text{g}\right)\\ =135000\text{cal}\end{array}$

Substitute $5000\text{cal}$ for $Q_1$ and $135000\text{cal}$ for $Q_2$ in equation (I) to find $Q$.

    $\begin{array}{c}Q=5000\text{cal}+135000\text{cal}\\ \text{=}\left(\text{140000}\text{cal}\times \frac{{10}^{-3}\text{kcal}}{1\text{cal}}\right)\\ =140.0\text{kcal}\end{array}$

Therefore, the amount of heat needed to change $250\text{g}$ of water at $80.0°\text{C}$ to steam at $100.0°\text{C}$ is $140.0\text{kcal}$.