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

Answer

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

 $\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

 $\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

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

   $P = \dfrac{W}{t}$

   $P = \dfrac{0.44}{0.296}$

   $P = 1.486\ kW$