The engine of a motor cycle delivers power of 12 KW by consuming petrol at the rate of 2.4 kg/hour. Calculate the efficiency of engine. If calorific value of petrol is 36 MJ/Kg

Heat received per second Q1t =   rate of consumption× calorific valueinput power=Q1t=2.43600×36×106Js−1∵t=1hour=60min=60×60=3600s, 36MJ/kg⇒1kg=36×106JPinput=Q1t=2.4×104=24kW;given P0utput=12kWefficiency =η=P0utputPinput=12kW24kW=12percentage efficiency=η×100=12×100=50%