Question

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

(a) What is the engine's efficiency?
(b) How much work is done by the engine in each cycle?
(c) What is the power output of the engine if each cycle lasts 0.296 s?

Answer 1

Given,

Energy absorbed,
`Q_H = 1.69\ kJ`

Energy expels,
`Q_C =  1.25\ kJ`

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

`\eta = (Q_H - Q_C)/(Q_H)* 100`

`\eta = (1.69 - 1.25)/(1.69)* 100`

`\eta =26.03 %`

b) Work done by the engine

`W = Q_H- Q_C`

`W =1.69 - 1.25`

`W = 0.44\ kJ`

c) Power output

t = 0.296 s

`P = (W)/(t)`

`P = (0.44)/(0.296)`

`P = 1.486\ kW`