Chapter 12, Problem 10STP

To determine

To Compare: The heat required to melt ice and that required to vaporize water.

Explanation of Solution

Given:

Mass of ice and water, $m=454\text{ g}$

Initial temperature of water, $T_1=0.0°\text{C}$

Final temperature of water, $100°\text{C}$

Formula Used:

  $Q=mc\Delta T$

Where, m is the mass, c is the specific heat capacity and $\Delta T$ is the difference in temperature.

Specific heat capacity of water is, $c_w=4.184\text{J/g}\cdot \text{°C}$

Heat required to melt the ice:

  $Q\text{'}=m{L}_f$

Latent heat of fusion of ice, $L_f=334\text{J/g}$

Heat required to vaporize water:

  $Q\text{'}\text{'}=m{L}_v$

Latent heat of vaporization of water, $L_v=2260\text{J/g}$

Calculations:

Heat required to melt the ice:

  $\begin{array}{c}Q\text{'}=\left(454\text{g}\right)\left(334\text{J/g}\right)\\ \approx 152\text{kJ}\end{array}$

Heat required to vaporise the water:

  $\begin{array}{c}Q"=m{L}_v\\ =454\text{g}\times 2260\text{J/g}\\ \approx \text{1026 kJ}\end{array}$

The difference in heat required to melt the ice and vaporize the water is:

  $\begin{array}{c}{Q}_d=Q"-Q\text{'}\\ =1026\text{kJ}-152\text{kJ}\\ =874\text{kJ}\end{array}$

Heat required to raise the temperature of water from $0.0°\text{C to}\text{100}°\text{C}$ :

  $\begin{array}{c}Q=mc\Delta T\\ =\left(454\text{g}\right)\left(4.18\text{J/g}\cdot \text{°C}\right)\left(100°\text{C-0}°\text{C}\right)\\ \approx 190\text{kJ}\end{array}$

Conclusion:

Thus, the heat required to vaporize the heat is much greater than that required to melt the ice or raise the temperature of water from $0.0°\text{C to}\text{100}°\text{C}$ .