What energy is required to melt a lead bar weighing 50 kg taken at a temperature of 27?

m = 50 kg.

t1 = 27 ° C.

t2 = 327 ° C.

C = 140 J / kg * ° C.

r = 25000 J / kg.

Q -?

Since the melting point of lead is t2 = 327 ° C, it must first be heated to this temperature, and then transferred from a solid to a liquid state of aggregation. The required amount of heat Q will be the sum of the amount of heat Qn required for heating the lead and for the melting process Qp: Q = Qn + Qp.

Qн = C * m * (t2 – t1).

Qп = r * m.

We take the specific heat C and the specific heat of fusion r of lead from the corresponding tables: C = 140 J / kg * ° C, r = 25000 J / kg.

Qн = 140 J / kg * ° С * 50 kg * (327 ° С – 27 ° С) = 2,100,000 J.

Qп = r * m.

Qp = 25000 J / kg * 50 kg = 1250000 J.

Q = 2,100,000 J + 1,250,000 J = 3,350,000 J.

Answer: to melt lead you need Q = 3,350,000 J of thermal energy.