The efficiency of a particular car engine is 23% when the engine does 7.2 kJ of

Question

The efficiency of a particular car engine is 23% when the engine does 7.2 kJ of work per cycle. Assume the process is reversible. What is the energy the engine gains per cycle as heat Qgain from the fuel combustion? kJ What is the energy the engine loses per cycle as heat Qlost? kJ If a tune - up increases the efficiency to 30%, what is Qgain at the same work value? KJ If a tune - up increases the efficiency to 30%, what is Qlost, at the same work value? KJ

Explanation / Answer

a) efficiency, e = Workdone/Qgain => 23/100 = 7.2/Qgain =>Qgain = 31.3KJ b) change in internal energy = 0 (cyclic process) =>Q = W + deltaU = W => Qgain - Qlost = W => 7.2 = 31.3- Qlost =>Qlost = 24.1KJ c) efficiency, e = Workdone/Qgain => 30/100 = 7.2/Qgain =>Qgain = 24KJ d) Qgain - Qlost = W => 7.2 = 24- Qlost =>Qlost = 16.8KJ

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