CBSE Class 9 Answered

Ice melting provides a classic example in which entropy increases in a small 'universe', a thermodynamic system consisting of the 'surroundings' (the warm room) and the 'system' of glass, ice, cold water which has been allowed to reach thermodynamic equilibrium at the melting temperature of ice. In this universe, some heat energy δQ from the warmer room surroundings at 298 K (77°F, 25°C) will spread out to the cooler system of ice and water at its constant temperature T of 273 K (32°F, 0°C), the melting temperature of ice. Thus, the entropy of the system, which is δQ/T, increases by δQ/273 K. (The heat δQ for this process is the energy required to change water from the solid state to the liquid state, and is called the enthalpy of fusion, i.e. the ΔH for ice fusion.)

It is important to realize that the entropy of the surrounding room decreases less than the entropy of the ice and water increases: the room temperature of 298 K is larger than 273 K and therefore the ratio, (entropy change), of δQ/298 K for the surroundings is smaller than the ratio (entropy change), of δQ/273 K for the ice+water system. This is always true in spontaneous events in a thermodynamic system and it shows the predictive importance of entropy: the final net entropy after such an event is always greater than was the initial entropy.

As the temperature of the cool water rises to that of the room and the room further cools imperceptibly, the sum of the δQ/T over the continuous range, “at many increments”, in the initially cool to finally warm water can be found by calculus. The entire miniature ‘universe’, i.e. this thermodynamic system, has increased in entropy. Energy has spontaneously become more dispersed and spread out in that ‘universe’ than when the glass of ice + water was introduced and became a 'system' within it.