Chapter 14, Problem 17QAP

The molar heats of fusion and vaporization for water are $\text{6}.0\text{2 kJ}/\text{mol}$and $\text{4}0.\text{6 kJ}/\text{mol}$respectively, and The specific heat capacity of liquid water is $\text{4}.\text{18 J}/\text{g }°\text{C}$. What quantity of heat energy is required to melt $\text{25}.0\text{ g}$of ice at $0°\text{C}$? What quantity of heat is required for vaporize $\text{37}.\text{5 g}$of liquid water at $\text{1}00°\text{C}$? What quantity of heat is required to warm $\text{55}.\text{2 g}$of liquid water from $0°\text{C}$to $\text{1}00°\text{C}$?

Interpretation Introduction

Interpretation:

The heat required to melt $25\text{g}$ of ice at $0^0\text{C}$, heat required to vaporize $37.5\text{g}$ liquid water at ${100}^0\text{C}$ and heat required to warm $55.2\text{g}$ water from $0^0\text{C}$ to ${100}^0\text{C}$ must be calculated.

Concept Introduction:

For melting and vaporization latent heat is used but for increasing temperature specific heat is used.

Answer to Problem 17QAP

The heat required to melt $25\text{g}$ of ice at $0^0\text{C}$ is $8.36\text{k}\text{J}$.

The heat required to vaporize $37.5\text{g}$ liquid water at ${100}^0\text{C}$ is $84.583\text{k}\text{J}$.

The heat required to warm $55.2\text{g}$ water from $0^0\text{C}$ to ${100}^0\text{C}$ is $23.073\text{k}\text{J}$.

Explanation of Solution

Given information:

Molar heat of fusion and vaporization of water is $6.02\text{k}\text{J}/\text{m}\text{o}\text{l}$ and $40.6\text{k}\text{J}/\text{m}\text{o}\text{l}$ respectively.

The specific heat capacity of water is $4.18\text{J}/{\text{g}}^0\text{C}$.

Heat of fusion of ice is $6.02\text{k}\text{J}/\text{m}\text{o}\text{l}$

So fusion of $18\text{g}$ ice needs $6.02\text{k}\text{J}$ heat energy.

Fusion of $1\text{g}$ ice needs $\frac{6.02\text{k}\text{J}}{18}$ heat energy.

Fusion of $25\text{g}$ ice needs $\frac{(6.02\times 25)\text{k}\text{J}}{18}$ heat energy $=\frac{150.5\text{k}\text{J}}{18}$ heat energy = $8.36\text{k}\text{J}$

Heat of vaporization of water is $40.6\text{k}\text{J}/\text{m}\text{o}\text{l}$

So vaporization of $18\text{g}$ water needs $40.6\text{k}\text{J}$ heat energy.

Vaporization of $1\text{g}$ water needs $\frac{40.6\text{k}\text{J}}{18}$ heat energy.

Vaporization of $37.5\text{g}$ water needs $\frac{(40.6\text{k}\text{J}\times 37.5)}{18}$ heat energy $=\frac{1522.5\text{k}\text{J}}{18}$

$=84.583\text{k}\text{J}$

Heat required to warm $55.2\text{g}$ water from $0^0\text{C}$ to ${100}^0\text{C}$ is $\text{H}=\text{m}\times \text{s}\times \text{t}$ where $\text{m},\text{s},\text{t}$ are mass of water, specific heat capacity of water and temperature rise of water respectively.

Now putting all the values we get $\text{H}=55.2\text{g}\times 4.18\text{J}/{\text{g}}^0\text{C}\times {(100-0)}^0\text{C}=23.073\text{k}\text{J}$

Conclusion

The heat required to melt $25\text{g}$ of ice at $0^0\text{C}$ is $8.36\text{k}\text{J}$.

The heat required to vaporize $37.5\text{g}$ liquid water at ${100}^0\text{C}$ is $84.583\text{k}\text{J}$.

The heat required to warm $55.2\text{g}$ water from $0^0\text{C}$ to ${100}^0\text{C}$ is $23.073\text{k}\text{J}$.