How much heat is needed to transform a piece of ice with a mass of 200 g, taken at 0 degrees Celsius, into water at a temperature of 20 g Celsius.
Data: m (mass of a piece of ice) = 200 g (0.2 kg); t0 (initial temperature of a piece of ice) = 0 ºС (temperature of the beginning of melting); t (required water temperature) = 20 ºС.
Constants: λ (specific heat of melting of ice) = 34 * 10 ^ 4 J / kg; C (specific heat of water) = 4200 J / (kg * K).
1) Melting a piece of ice: Q = λ * m = 34 * 10 ^ 4 * 0.2 = 68000 J.
2) Water heating: Q = C * m * (t – t0) = 4200 * 0.2 * (20 – 0) = 16800 J.
3) Total heat consumption: Q = Q1 + Q2 = 68000 + 16800 = 84800 J (84.8 kJ).
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