Data: m (mass of a piece of lead for smelting) = 500 g (0.5 kg); t0 (temperature at which the piece of lead was located) = 20 ºС.
Reference data: С (specific heat) = 140 J / (kg * ºС); tmelt (temperature of the beginning of melting) = 327.4 ºС; λ (specific heat of fusion) = 25 * 10 ^ 3 J / kg.
1) Heating a piece of lead: Q1 = C * m * (tm – t) = 140 * 0.5 * (327.4 – 20) = 21504 J.
2) Melting a piece of lead: Q2 = λ * m = 25 * 10 ^ 3 * 0.5 = 12500 J
3) Heat consumption: Q = Q1 + Q2 = 21504 + 12500 = 34004 J (≈ 34 kJ).
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