Chapter 21, Problem 27PQ

To determine

The energy required to transform an ice cube to steam.

Answer to Problem 27PQ

The energy required to transform an ice cube to steam is $\underset{\_}{1.38\times {10}^5\text{J}}$.

Explanation of Solution

Given that the mas of the ice is $45.0\text{g}$, the initial temperature of the ice is $-5.00°\text{C}$ and the final temperature is $120.0°\text{C}$.

The energy required to transform ice to steam is to be found. It is the sum of the energy required to reach the melting point, energy required to melt the ice, energy required to reach the boiling point, energy required to vaporize the water, and the energy required to raise the temperature of the vapour to $120.0°\text{C}$.

Write the expression for energy required to reach the melting point of the ice.

  $Q_1=m{c}_i\left(T_f-T_i\right)$                                                                                                  (I)

Here, $Q_1$ is the heat required to reach the melting point, $m$ is the mass of ice, $c_i$ is the specific heat capacity of ice, $T_f$ is the melting point of ice, and $T_f$ is the initial temperature of the ice.

Write the expression for energy required for the melting of ice.

  $Q_2=m{L}_f$                                                                                                             (II)

Here, $Q_2$ is the heat required for fusion of ice, and $L_f$ is the latent heat of fusion of ice.

Write the expression for the energy required to reach the boiling point of water.

  $Q_3=m{c}_w\left(T_f-T_i\right)$                                                                                             (III)

Here, $Q_3$ is the heat required to reach the boiling point of water, $c_w$ is the specific heat of water, $T_f$ is the boiling point of water, and $T_i$ is the initial temperature of the water.

Write the expression for the energy required to vaporize the water at its boiling point.

  $Q_4=m{L}_V$                                                                                                       (IV)

Here, $Q_4$ is the energy required to vaporize the water at its boiling point, and $L_v$ is the latent heat of vaporization.

Write the expression for the heat energy required to raise the vapour to $120.0°\text{C}$.

  $Q_5=m{c}_v\left(T_f-T_i\right)$                                                                                     (V)

Here, $Q_5$ is the heat energy required to raise the temperature of the vapour, $c_v$ is the specific heat capacity of the vapour or steam, $T_f$ is the final temperature of the vapour, and $T_i$ is the initial temperature of the vapour.

Write the expression for the total energy required to transform the ice to steam.

  $Q_t=Q_1+Q_2+Q_3+Q_4+Q_5$                                                                              (VI)

Conclusion:

Substitute $45.0\text{g}$ for $m$, $2100\text{J}/\text{kg}\cdot \text{°C}$ for $c_i$, $0°\text{C}$ for $T_f$, and $-5.00°\text{C}$ for $T_i$ in equation (I) to find $Q_1$.

  $\begin{array}{c}{Q}_1=\left(45.0\text{g}\right)\left(2100\text{J}/\text{kg}\cdot \text{°C}\right)\left[0°\text{C}-\left(-5.00°\text{C}\right)\right]\\ =\left(45.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right)\left(2100\text{J}/\text{kg}\cdot \text{°C}\right)\left[0°\text{C}-\left(-5.00°\text{C}\right)\right]\\ =472.5\text{J}\end{array}$

Substitute $45.0\text{g}$ for $m$, and $3.33\times {10}^5\text{J}/\text{kg}$ for $L_f$ in equation (II) to find $Q_2$.

  $\begin{array}{c}{Q}_2=\left(45.0\text{g}\right)\left(3.33\times {10}^5\text{J}/\text{kg}\right)\\ =\left(45.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right)\left(3.33\times {10}^5\text{J}/\text{kg}\right)\\ =1.49\times {10}^4\text{J}\end{array}$

Substitute $45.0\text{g}$ for $m$, $4190\text{J}/\text{kg}\cdot \text{°C}$ for $c_w$, $100°\text{C}$ for $T_f$, and $0°\text{C}$ for $T_i$ in equation (III) to find $Q_3$.

  $\begin{array}{c}{Q}_3=\left(45.0\text{g}\right)\left(4190\text{J}/\text{kg°C}\right)\left(100°\text{C}-0°\text{C}\right)\\ =\left(45.0\text{g}\times \frac{1\text{g}}{1000\text{kg}}\right)\left(4190\text{J}/\text{kg°C}\right)\left(100°\text{C}-0°\text{C}\right)\\ =1.86\times {10}^4\text{J}\end{array}$

Substitute $45.0\text{kg}$ for $m$,and $2.26\times {10}^6\text{J}/\text{kg}$ for $L_v$ in equation (IV) to find $Q_4$.

  $\begin{array}{c}{Q}_4=\left(45.0\text{g}\right)\left(2.26\times {10}^6\text{J}/\text{kg}\right)\\ =45.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}\left(2.26\times {10}^6\text{J}/\text{kg}\right)\\ =1.02\times {10}^5\text{J}\end{array}$

Substitute $45.0\text{g}$ for $m$, $2010\text{J}/\text{kg°C}$ for $c_v$, $120.0°\text{C}$ for $T_f$, and $100.0°\text{C}$ for $T_i$ in equation (V) to find $Q_5$.

  $\begin{array}{c}{Q}_5=\left(45.0\text{g}\right)\left(2010\text{J}/\text{kg}\cdot \text{°C}\right)\left(120.0°\text{C}-100.0°\text{C}\right)\\ =\left(45.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right)\left(2010\text{J}/\text{kg}\cdot \text{°C}\right)\left(120.0°\text{C}-100.0°\text{C}\right)\\ =1809\text{J}\end{array}$

Substitute $472.5\text{J}$ for $Q_1$, $1.49\times {10}^4\text{J}$ for $Q_2$, $1.86\times {10}^4\text{J}$ for $Q_3$, $1.02\times {10}^5\text{J}$ for $Q_4$, $1809\text{J}$ for $Q_5$ in equation (VI) to find $Q_t$.

  $\begin{array}{c}{Q}_t=472.5\text{J}+\left(1.49\times {10}^4\text{J}\right)+\left(1.86\times {10}^4\text{J}\right)+\left(1.02\times {10}^5\text{J}\right)+1809\text{J}\\ \text{=1}\text{.38}\times {\text{10}}^5\text{J}\end{array}$

Therefore, the energy required to transform an ice cube to steam is $\underset{\_}{1.38\times {10}^5\text{J}}$.