200 g of lead, having a temperature of 27 C, was heated to the melting point

200 g of lead, having a temperature of 27 C, was heated to the melting point a) how much heat is required for this? b) What amount of heat must be spent for heating and subsequent complete melting of lead. c) what is the efficiency of the heating installation if 100 g of acetylene was burned to heat and melt the lead?

t1 = 27 ° C.

t2 = 327 ° C.

mw = 200 g = 0.2 kg.

C = 140 J / kg * ° C.

λ = 25 * 10 ^ 3 J / kg.

q = 5 * 10 ^ 7 J / kg.

ma = 100 g = 0.1 kg.

Efficiency -?

Q1 -?

Q2 -?

The required amount of heat Q1 for heating the lead is expressed by the formula: Q1 = C * mw * (t2 – t1).

Q1 = 140 J / kg * ° C * 0.2 kg * (327 ° C – 27 ° C) = 8400 J.

Q2 = Q1 + λ * mw.

Q2 = 8400 J + 0.2 kg * 25 * 10 ^ 3 J / kg = 13400 J.

Efficiency = Q2 * 100% / q * ma.

Efficiency = 13400 J * 100% / 5 * 10 ^ 7 J / kg * 0.1 kg = 0.268%.

Answer: Q1 = 8400 J, Q2 = 13400 J, the unit has an efficiency of 0.268.