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210 PHYSICS The amount of heat lost by the aluminium sphere = m 1 AlT= 0.047kg × sAl× 77 °C Mass of water (m )2= 0.25 kg Mass of calorimeter (m ) = 0.14 kg 3 Initial temperature of water and calorimeter=20 °C Final temperature of the mixture = 23 °C Change in temperature (∆T ) = 23°C – 20 C = 3 °C 2 Specific heat capacity of water (sw) = 4.18 × 10 J kg K1 –1 Specific heat capacity of copper calorimeter 3 –1 –1 = 0.386× 10 J kg K The amount of heat gained by water and calorimeter = m 2 ∆w + 2 s ∆3 cu 2 = (m s + m s ) (∆T ) Fig. 10.9 A plot of temperature versus time showing 2 w 3 cu 2 the changes in the state of ice on heating = (0.25 kg × 4.18×10 J kg K + 0.14 kg× (not to scale). 3 –1 –1 0.386× 10 J kg K ) (23 °C – 20 °C) In the steady state heat lost by the aluminium Thechangeofstatefromsolidtoliquidiscalled meltingorfusionandfromliquidtosolidiscalled sphere = heat gained by water + heat gained by calorimeter. freezing. It is observed that the temperature remains constant until the entire amount of the So, 0.047 kg×s ×Al°C solidsubstancemelts.Thatis,both the solid and = (0.25 kg × 4.18× 10 J kg K + 0.14 kg × the liquid states of the substance coexist in 0.386 × 10 J kg K )(3 °C) thermal equilibrium during the change of –1 –1 sAl 0.911 kJ kg K ⊳ states from solid to liquid. The temperature at which the solid and the liquid states of the substance is in thermal equilibrium with each 10.8 CHANGE OF STATE other is called its melting point. It is Matter normally exists in three states: solid, characteristic of the substance. It also depends liquid and gas. A transition from one of these on pressure. The melting point of a substance states to another is called a change of state. Two at standard atomspheric pressure is called its common changes of states are solid to liquid normal melting point. Let us do the following and liquid to gas (and, vice versa). These changes activity to understand the process of melting can occur when the exchange of heat takes place of ice. between the substance and its surroundings. Take a slab of ice. Take a metallic wire and To study the change of state on heating or fix two blocks, say 5 kg each, at its ends. Put cooling, let us perform the following activity. the wire over the slab as shown in Fig. 10.10. Take some cubes of ice in a beaker. Note the You will observe that the wire passes through temperature of ice. Start heating it slowly on a the ice slab. This happens due to the fact that constant heat source. Note the temperature after just below the wire, ice melts at lower every minute. Continuously stir the mixture of temperature due to increase in pressure. When the wire has passed, water above the wire freezes water and ice. Draw a graph between temperatureandtime(Fig.10.9).Youwillobserve again. Thus, the wire passes through the slab no change in the temperature as long as there and the slab does not split. This phenomenon is ice in the beaker. In the above process, the of refreezing is called regelation. Skating is temperature of the system does not change even possible on snow due to the formation of water though heat is being continuously supplied. The under the skates. Water is formed due to the heat supplied is being utilised in changing the increase of pressure and it acts as a state from solid (ice) to liquid (water). lubricant. Reprint 2026-27