How much energy is required to melt a piece of lead weighing 0.5 kg taken at a temperature of 27 degrees?

univerkov | May 7, 2021 | education

Given: m (mass of lead) = 0.5 kg; t (the temperature that the taken lead has) = 27 ºС.

Reference values: C (specific heat of lead) = 140 J / (kg * K); tmelt (temperature of the beginning of lead melting) ≈ 327 ºС (327.46 ºС); λ (specific heat of fusion of lead) = 25 * 10 ^ 3 J / kg.

The consumption of heat energy for melting is determined by the formula: Q = Q1 + Q2 = C * m * (tmelt – t) + λ * m.

Calculation: Q = 140 * 0.5 * (327 – 27) + 25 * 10 ^ 3 * 0.5 = 21000 J + 12500 J = 33.5 kJ.

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