How much energy is required to heat and melt lead weighing 0.4 kg, having an initial temperature of 17 C
| August 16, 2021 | education
С = 140 J / kg * ° C.
m = 0.4 kg.
t1 = 17 ° C.
t2 = 327 ° C.
λ = 25 * 10 ^ 3 J / kg.
Q -?
Lead must be heated to its melting point, and then, at the melting point, it must be transferred from a solid to a liquid state.
Q = Q1 + Q2.
Q1 = C * m * (t2 – t1), where C is the specific heat of lead, m is the mass of lead, t2, t1 are the final and initial temperature of the lead.
Q2 = λ * m, where λ is the specific heat of fusion of lead.
The melting point t2, specific heat C, specific heat of fusion λ of lead are taken from the corresponding tables.
Q = C * m * (t2 – t1) + λ * m = (C * (t2 – t1) + λ) * m.
Q = (140 J / kg * ° C * (327 ° C – 17 ° C) + 25 * 10 ^ 3 J / kg) * 0.4 kg = 27360 J.
Answer: to melt lead, you need Q = 27360 J of thermal energy.
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