Chapter 14, Problem 39P

To determine

The heat energy required.

Answer to Problem 39P

The heat energy required is $\text{3100}\text{kJ}$.

Explanation of Solution

Write the expression for energy needed to raise the temperature of ice to $0.0\text{°C}$.

$Q_1=m{c}_{ice}\left(T_{ice}^f-T_{ice}^i\right)$

Here, $Q_1$ is the heat energy, m is the mass, $c_{ice}$ is the specific heat capacity of ice, $T_{ice}^f$ is the final temperature and $T_{ice}^i$ is the initial temperature.

Write the expression for energy needed for melting of ice.

$Q_2=m{L}_f$

Here, $Q_2$ is the heat energy and $L_f$ is the latent heat of fusion.

Write the expression for energy needed to raise the temperature of water to $100\text{°C}$.

$Q_3=m{c}_{water}\left(T_{water}^f-T_{water}^i\right)$

Here, $Q_3$ is the heat energy, m is the mass, $c_{water}$ is the specific heat capacity of water, $T_{water}^f$ is the final temperature and $T_{water}^i$ is the initial temperature.

Write the expression for energy needed for evaporation of water.

$Q_4=m{L}_v$

Here, $Q_4$ is the heat energy and $L_v$ is the latent heat of vaporization.

Write the expression for energy needed to raise the temperature of steam to $110\text{°C}$.

$Q_5=m{c}_{steam}\left(T_{steam}^f-T_{steam}^i\right)$

Here, $Q_3$ is the heat energy, m is the mass, $c_{steam}$ is the specific heat capacity of water vapor, $T_{steam}^f$ is the final temperature and $T_{steam}^i$ is the initial temperature.

The total heat required is given by,

$Q_{total}=Q_1+Q_2+Q_3+Q_4+Q_5$

Use the corresponding expressions in the above equation.

$Q_{total}=m\left[c_{ice}\left(T_{ice}^f-T_{ice}^i\right)+L_f+c_{water}\left(T_{water}^f-T_{water}^i\right)+L_v+c_{steam}\left(T_{steam}^f-T_{steam}^i\right)\right]$

Conclusion:

Substitute $1.0\text{kg}$ for $m$, $0.0\text{°C}$ for $T_{ice}^f$,$-20.0\text{°C}$ for $T_{ice}^i$,$2.1\text{kJ/kg}\cdot \text{K}$ for $c_{ice}$,$333.7\text{kJ/kg}$ for $L_f$,$4.186\text{kJ/kg}\cdot \text{K}$ for $c_{water}$,$100\text{°C}$ for $T_{water}^f$,$0.0\text{°C}$ for $T_{water}^i$,$2256\text{kJ/kg}$ for $L_v$,$2.01\text{kJ/kg}\cdot \text{K}$ for $c_{steam}$,$110\text{°C}$ for $T_{water}^f$ and $100\text{°C}$ for $T_{water}^i$ to get $Q_{total}$.

$\begin{array}{c}{Q}_{total}=\left\{\left(1.0\text{kg}\right)\left[\begin{array}{l}\left(2.1\text{kJ/kg}\cdot \text{K}\right)\left(0.0\text{°C}-\left(-20.0\text{°C}\right)\right)+\left(333.7\text{kJ/kg}\right)+\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(100\text{°C}-0.0\text{°C}\right)\\ +\left(2256\text{kJ/kg}\right)+\left(2.01\text{kJ/kg}\cdot \text{K}\right)\left(110\text{°C}-100\text{°C}\right)\end{array}\right]\right\}\\ \approx \text{3100}\text{kJ}\end{array}$

Therefore, heat energy required is $\text{3100}\text{kJ}$.