Question

A heat engine operates in a Carnot cycle between 75◦C and 492◦C. It absorbs 19300 J of energy per cycle from the hot reservoir. The duration of each cycle is 1.16 s. What is the mechanical power output of this engine? Answer in units of kW.

Answer 1

9.069 KW

Step-by-step explanation:

The heat engine operates in a carnot cycle between 75°C to 492°C so the lower temperature
`T_L=75^(\circ)C=273+75=348K` and the higher temperature
`T_H=492^(\circ)C=273+492=765K`

Efficiency of the carnot cycle
`\eta =1-(T_L)/(T_H)=1-(348)/(765)=0.545`

We know that
`\eta =(work\ done )/(heat\ absorbed)`

`work\ done=\eta * heat\ abosorbed=0.545* 19300=10520.392\ J`

it is given that duration of each cycle is 1.16 sec so power output
`P=(W)/(T)=(10520.392)/(1.16)=9069.30\ W=9.069\ KW`