A gas burner heats up an aluminum kettle weighing 1.2 kg, containing 2 liters of water at a temperature

A gas burner heats up an aluminum kettle weighing 1.2 kg, containing 2 liters of water at a temperature of 15 ° C. The water in the kettle has warmed up to 100 ° C. What is the efficiency of the burner if 0.1 m3 of natural gas is burnt?

mh = 1.2 kg.

Vw = 2 l = 0.002 m3.

ρw = 1000 kg / m3.

t1 = 15 ° C.

t2 = 100 ° C.

Cw = 4200 J / kg * ° С.

Ca = 920 J / kg * ° С.

Vg = 0.1 m3.

ρg = 0.8 kg / m3.

r = 4.4 * 107 J / kg.

Efficiency -?

Efficiency = Qp * 100% / Qz – the formula for determining the efficiency.

The useful amount of heat Qp is expressed by the formula: Qp = Cw * mw * (t2 – t1) + Ca * mh * (t2 – t1) = (Cw * mw + Ca * mh) * (t2 – t1) = (Cw * Vw * ρv + Ca * mh) * (t2 – t1).

Qp = (4200 J / kg * ° C * 0.002 m3 * 1000 kg / m3 + 920 J / kg * ° C * 1.2 kg) * (100 ° C – 15 ° C) = 807840 J.

The consumed amount of heat Qz is expressed by the formula: Qz = r * mg = r * Vg * ρg.

Qz = 4.4 * 107 J / kg * 0.1 m3 * 0.8 kg / m3 = 3,520000 J.

Efficiency = 807840 J * 100% / 3520000 J = 23%

Answer: a gas burner has an efficiency of 23%.