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.

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

   $P = \dfrac{W}{t}$

   $P = \dfrac{0.44}{0.296}$

   $P = 1.486\ kW$