How much heat is required to melt 200g of lead at 37 ° C?
July 29, 2021 | education|
Data: m (mass of lead taken for smelting) = 200 g (0.2 kg); t0 (initial temperature) = 37 ºС.
Constants: C (specific heat) = 140 J / (kg * ºС); tmelt (temperature of the beginning of melting) = 327.4 ºС; λ (specific heat of fusion of lead) = 25 * 10 ^ 3 J / kg.
1) Heating lead to the melting point: Q1 = C * m * (tmelt – t0) = 140 * 0.2 * (327.4 – 37) = 8131.2 J.
2) Melting lead: Q2 = λ * m = 25 * 10 ^ 3 * 0.2 = 5000 J.
3) Required amount of heat: Q = Q1 + Q2 = 8131.2 + 5000 = 13131.2 J.
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