Page 1 25.1 CHAPTER – 25 CALORIMETRY 1. Mass of aluminium = 0.5kg, Mass of water = 0.2 kg Mass of Iron = 0.2 kg Temp. of aluminium and water = 20°C = 297°k Sp heat o f Iron = 100°C = 373°k. Sp heat of aluminium = 910J/kg-k Sp heat of Iron = 470J/kg-k Sp heat of water = 4200J/kg-k Heat again = 0.5 × 910(T – 293) + 0.2 × 4200 × (343 –T) = (T – 292) (0.5 × 910 + 0.2 × 4200) Heat lost = 0.2 × 470 × (373 – T) ? Heat gain = Heat lost ? (T – 292) (0.5 × 910 + 0.2 × 4200) = 0.2 × 470 × (373 – T) ? (T – 293) (455 + 8400) = 49(373 – T) ? (T – 293) ? ? ? ? ? ? 94 1295 = (373 – T) ? (T – 293) × 14 = 373 – T ? T = 15 4475 = 298 k ? T = 298 – 273 = 25°C. The final temp = 25°C. 2. mass of Iron = 100g water Eq of caloriemeter = 10g mass of water = 240g Let the Temp. of surface = 0 ?C S iron = 470J/kg°C Total heat gained = Total heat lost. So, 1000 100 × 470 ×( ? – 60) = 1000 250 × 4200 × (60 – 20) ? 47 ? – 47 × 60 = 25 × 42 × 40 ? ? = 4200 + 47 2820 = 47 44820 = 953.61°C ? 3. The temp. of A = 12°C The temp. of B = 19°C The temp. of C = 28°C The temp of ? A + B = 16° The temp. of ? B + C = 23° In accordance with the principle of caloriemetry when A & B are mixed M CA (16 – 12) = M CB (19 – 16) ? CA4 = CB3 ? CA = 4 3 CB …(1) And when B & C are mixed M CB (23 – 19)= M CC (28 – 23) ? 4CB = 5CC ? CC = 5 4 CB …(2) When A & c are mixed, if T is the common temperature of mixture M CA (T – 12) = M CC (28 – T) ? ? ? ? ? ? ? 4 3 CB(T – 12) = ? ? ? ? ? ? 5 4 CB(28 – T) ? 15T – 180 = 448 – 16T ? T = 31 628 = 20.258°C = 20.3°C ? ? ? ? ??Read More

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