What amount of heat will be released during crystallization and cooling to a temperature t = 17C

| April 30, 2021 | education

What amount of heat will be released during crystallization and cooling to a temperature t = 17C of lead with a mass of m = 100g. at the melting point? (Specific heat of lead c = 1.2 * 10 ^ 2 J \ kg * C, melting point of lead t = 327 C, specific heat of fusion of lead lambda = 2.47 * 10 ^ 4 J \ kg).

t = 17 “C. m = 100 g = 0.1 kg. C = 1.2 * 10 ^ 2 J / kg *” C. tmelt = 327 “C. λ = 2.47 * 10 ^ 4 J / kg.
Q -?
The amount of heat Q will be the sum of the amount of heat Q1, which is released during the crystallization of lead, and the amount of heat Q2, which is released when the lead is cooled: Q = Q1 + Q2. The amount of heat during crystallization is determined by the formula: Q1 = λ * m. The amount of heat during cooling is determined by the formula: Q2 = C * m * (tmelt – t). Q = λ * m + C * m * (tmelt – t). Q = 2.47 * 10 ^ 4 J / kg * 0.1 kg + 1.2 * 10 ^ 2 J / kg * “C * 0.1 kg * (327” C – 17 “C) = 39670 J.
Answer: during crystallization and cooling of lead, an amount of heat Q = 39670 J.



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