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214 PHYSICS temperature difference. What are the different ⊳Example 10.5 Calculate the heat required ways by which this energy transfer takes to convert 3 kg of ice at –12 °C kept in a place? There are three distinct modes of heat calorimeter to steam at 100 °C at transfer: conduction, convection and radiation atmospheric pressure. Given specific heat –1 –1 (Fig. 10.13). capacity of ice = 2100 J kg K , specific heat capacity of water = 4186 J kg – K , latent heat of fusion of ice = 3.35 × 10 J kg –1 6 –1 and latent heat of steam = 2.256 ×10 J kg . Answer We have Mass of the ice, m = 3 kg specific heat capacity of ice, ice = 2100 J kg –1K –1 specific heat capacity of water, swater = 4186 J kg –1K –1 latent heat of fusion of ice, f ice = 3.35×10 J kg –1 latent heat of steam, L Fig. 10.13 Heating by conduction, convection and 6team –1 = 2.256× 10 J kg radiation. 10.9.1 Conduction Now, Q = heat required to convert 3 kg of ice at –12°C to steam at 100 °C, Conduction is the mechanism of transfer of heat Q = heat required to convert ice at 1 between two adjacent parts of a body because –12 °C to ice at 0 °C. of their temperature difference. Suppose, one end = m s ∆T = (3 kg) (2100 J kg –1. –1 ice 1 of a metallic rod is put in a flame, the other end K ) [0–(–12)]°C= 75600 J of the rod will soon be so hot that you cannot Q 2 = heat required to melt ice at hold it by your bare hands. Here, heat transfer 0 °C to water at 0 °C takes place by conduction from the hot end of = m L f ice(3 kg) (3.35×10 J kg )1 the rod through its different parts to the other = 1005000 J end. Gases are poor thermal conductors, while Q 3 = heat required to convert water liquids have conductivities intermediate between at 0 °C to water at 100°C. solids and gases. –1 –1 = ms ∆w = 23kg) (4186J kg K ) Heat conduction may be described (100°C) quantitatively as the time rate of heat flow in a = 1255800 J material for a given temperature difference. Q 4 = heat required to convert water Consider a metallic bar of length L and uniform at 100 °C to steam at 100°C. cross-section A with its two ends maintained at 6 different temperatures. This can be done, for = m L steam= (3 kg) (2.256 ×10 J kg ) example, by putting the ends in thermal contact with large reservoirs at temperatures, say, C and = 6768000 J T , respectively (Fig. 10.14). Let us assume the So, Q = Q +1Q +2Q + 3 4 D = 75600J + 1005000 J ideal condition that the sides of the bar are fully insulated so that no heat is exchanged between + 1255800 J + 6768000 J the sides and the surroundings. = 9.1×10 J6 ⊳ After sometime, a steady state is reached; the temperature of the bar decreases uniformly with 10.9 HEAT TRANSFER distance from T to T ; (T >T ). The reservoir at C D C D We have seen that heat is energy transfer C supplies heat at a constant rate, which transfers through the bar and is given out at from one system to another or from one part of a system to another part, arising due to the same rate to the reservoir at D. It is found Reprint 2026-27