Matter normally exists in three states: solid, liquid, and gas. A transition from one of these states to another is called a change of state. Two common changes of states are solid to liquid and liquid to gas (and, vice versa). These changes can occur when the exchange of heat takes place between the substance and its surroundings.  Common changes of the state include melting, freezing, sublimation, deposition, condensation, and vaporization. These changes are shown in the figure given below.

Five Changes of State are:

• Melting
• Freezing
• Evaporation
• Condensation
• Sublimation

The process by which a substance changes from the solid phase to the liquid phase is known as melting. The process by which a substance changes from the liquid phase to the solid phase is known as freezing. The process by which a substance changes from the liquid phase to the gaseous phase is known as evaporation. The process by which a substance changes from the gaseous phase to the liquid phase is known as condensation. The transition of the solid phase to the gaseous phase without passing the intermediate liquid phase is known as sublimation.

Freezing

Heat transfer occurs between the warmer tray and the colder air in the freezer. The warm water loses heat to the cold air in the freezer. This heat transfer occurs until no energy is available for the particles to slide past each other. This forces them to remain in fixed positions, locked in place by the force of attraction between them. This way liquid water is changed into solid ice. The process of liquid water changing to solid ice is termed as freezing. The temperature at which it occurs is known as the freezing point.

Take some cubes of ice in a beaker. Note the temperature of ice. Start heating it slowly on a constant heat source. Note the temperature after every minute. Continuously stir the mixture of water and ice. Draw a graph between temperature and time (Fig. 11.9). You will observe no change in the temperature as long as there is ice in the beaker. In the above process, the temperature of the system does not change even though heat is being continuously supplied. The heat supplied is being utilised in changing the state from solid (ice) to liquid (water).

The temperature at which the solid and the liquid states of the substance is in thermal equilibrium with each other is called its melting point. It is characteristic of the substance. It also depends on pressure. The melting point of a substance at standard atmospheric pressure is called its normal melting point. Let us do the following activity to understand the process of melting of ice. 

The temperature at which the liquid and the vapour states of the substance coexist is called its boiling point. The boiling point of a substance at standard atmospheric pressure is called its normal boiling point.

Latent Heat

We have learnt that certain amount of heat energy is transferred between a substance and its surroundings when it undergoes a change of state. The amount of heat per unit mass transferred during change of state of the substance is called latent heat of the substance for the process. It might be from a gas to a liquid or from a liquid to a solid and back again.

Latent heat formula

The heat required during a change of state depends upon the heat of transformation and the mass of the substance undergoing a change of state. Thus, if mass \(m\) of a substance undergoes a change from one state to the other, then the quantity of heat required is given by

\(
\begin{aligned}
& Q=m L \\
\text { or } \quad L & =Q / m \dots(11.13)
\end{aligned}
\)

where \(L\) is known as latent heat and is a characteristic of the substance. Its SI unit is \(\mathrm{J} \mathrm{kg}^{-1}\). The value of \(L\) also depends on the pressure. Its value is usually quoted at standard atmospheric pressure. The latent heat for a solid-liquid state change is called the latent heat of fusion \(\left(L_{\mathrm{f}}\right)\), and that for a liquid-gas state, change is called the latent heat of vaporisation \(\left(L_v\right)\). These are often referred to as the heat of fusion and the heat of vaporisation. A plot of temperature versus heat for a quantity of water is shown in Fig. 11.12.

Note that when heat is added (or removed) during a change of state, the temperature remains constant. Note in Fig. \(11.12\) that the slopes of the phase lines are not all the same, which indicate that specific heats of the various states are not equal. For water, the latent heat of fusion and vaporisation are \(L_{\mathrm{f}}=3.33 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\) and \(L_{\mathrm{v}}=22.6 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\), respectively. That is, \(3.33 \times 10^5 \mathrm{~J}\) of heat is needed to melt \(1 \mathrm{~kg}\) ice at \(0^{\circ} \mathrm{C}\), and \(22.6 \times 10^5 \mathrm{~J}\) of heat is needed to convert \(1 \mathrm{~kg}\) water into steam at \(100^{\circ} \mathrm{C}\). So, steam at \(100^{\circ} \mathrm{C}\) carries \(22.6 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\) more heat than water at \(100^{\circ} \mathrm{C}\). This is why burns from steam are usually more serious than those from boiling water.

The latent heat of some substances, their freezing and boiling points, are given in Table 11.5.

Example 1: When \(0.15 \mathrm{~kg}\) of ice at \(0^{\circ} \mathrm{C}\) is mixed with \(0.30 \mathrm{~kg}\) of water at \(50^{\circ} \mathrm{C}\) in a container, the resulting temperature is \(6.7^{\circ} \mathrm{C}\). Calculate the heat of fusion of ice. \(\left(s_{\text {water }}=4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)\)

Solution:

Heat lost by water \(=m s_w\left(\theta_{\mathrm{f}}-\theta_1\right)_w\)
\(=(0.30 \mathrm{~kg})\left(4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)\left(50.0^{\circ} \mathrm{C}-6.7^{\circ} \mathrm{C}\right)\)
\(=54376.14 \mathrm{~J}\)
Heat required to melt ice \(=m_2 L_{\mathrm{f}}=(0.15 \mathrm{~kg}) L_{\mathrm{f}}\)
Heat required to raise temperature of ice
water to final temperature \(=m_{\mathrm{I}} \mathrm{s}_{\mathrm{w}}\left(\theta_{\mathrm{f}}-\theta_1\right)_{\mathrm{I}}\)
\(=(0.15 \mathrm{~kg})\left(4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)\left(6.7^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}\right)\)
\(=4206.93 \mathrm{~J}\)
Heat lost \(=\) heat gained
\(54376.14 \mathrm{~J}=(0.15 \mathrm{~kg}) L_{\mathrm{f}}+4206.93 \mathrm{~J}\)
\(L_{\mathrm{f}}=3.34 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\).

Example 2: Calculate the heat required to convert \(3 \mathrm{~kg}\) of \(i c e\) at \(-12{ }^{\circ} \mathrm{C}\) kept in a calorimeter to steam at \(100^{\circ} \mathrm{C}\) at atmospheric pressure. Given specific heat capacity of \(i c e=2100 \mathrm{~J} \mathrm{kg}^{-1} \mathrm{~K}^{-1}\), specific heat capacity of water \(=4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\), latent heat of fusion of \(i c e=3.35 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\) and latent heat of steam \(=2.256 \times 10^6 \mathrm{~J} \mathrm{~kg}^{-1}\).

Solution:
We have
Mass of the ice, \(m=3 \mathrm{~kg}\)
specific heat capacity of ice, \(s_{\text {ice }}\)
\(
=2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}
\)
specific heat capacity of water, \(s_{\text {water }}\)
\(
=4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}
\)
latent heat of fusion of ice, \(L_{\mathrm{f} \text { ice }}\)
\(
=3.35 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}
\)
latent heat of steam, \(L_{\text {steam }}\)
\(
=2.256 \times 10^6 \mathrm{~J} \mathrm{~kg}^{-1}
\)
Now, \(\quad Q=\) heat required to convert \(3 \mathrm{~kg}\) of ice at \(-12{ }^{\circ} \mathrm{C}\) to steam at \(100^{\circ} \mathrm{C}\),
\(Q_1=\) heat required to convert ice at \(-12^{\circ} \mathrm{C}\) to ice at \(0^{\circ} \mathrm{C}\).
\(=m s_{\text {ice }} \Delta T_1=(3 \mathrm{~kg})\left(2100 \mathrm{~J} \mathrm{~kg}^{-1}\right.\). \(\left.\mathrm{K}^{-1}\right)[0-(-12)]^{\circ} \mathrm{C}=75600 \mathrm{~J}\)
\(Q_2=\) heat required to melt ice at \(0^{\circ} \mathrm{C}\) to water at \(0^{\circ} \mathrm{C}\)
\(=m L_{\text {f ice }}=(3 \mathrm{~kg})\left(3.35 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\right)\)
\(=1005000 \mathrm{~J}\)
\(Q_3=\) heat required to convert water at \(0^{\circ} \mathrm{C}\) to water at \(100^{\circ} \mathrm{C}\).
\(=m s_w \Delta T_2=(3 \mathrm{~kg})\left(4186 \mathrm{~J} \mathrm{~kg} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)\) \(\left(100^{\circ} \mathrm{C}\right)\)
\(=1255800 \mathrm{~J}\)
\(Q_4=\) heat required to convert water at \(100^{\circ} \mathrm{C}\) to steam at \(100^{\circ} \mathrm{C}\).
\(=m L_{\text {steam }}=(3 \mathrm{~kg})\left(2.256 \times 10^6\right.\) \(\mathrm{J} \mathrm{kg}^{-1}\) )
\(=6768000 \mathrm{~J}\)
So, \(\quad Q=Q_1+Q_2+Q_3+Q_4\)
\(=75600 \mathrm{~J}+1005000 \mathrm{~J}\)
\(+1255800 \mathrm{~J}+6768000 \mathrm{~J}\)
\(=9.1 \times 10^6 \mathrm{~J}\)