The 18 kW engine consumed 12 kg of kerosene in 3 hours. What is the efficiency of this engine.

η = (A / Q) * 100%, where A is the useful work of the engine (A = N * t, where N is the engine power (N = 18 kW = 18 * 10 ^ 3 W), t is the movement time (t = 3 h = 3 * 3600 = 10800 s)), Q is the energy expended due to the combustion of kerosene (Q = q * m, where q is the specific heat of combustion of kerosene (q = 4.6 * 107 J / kg), m – mass of spent kerosene (m = 12 kg)).

η = (A / Q) * 100% = (N * t / (q * m)) * 100% = (18 * 10 ^ 3 * 10800 / (4.6 * 10 ^ 7 * 12)) * 100% = 35.2%.