A Carnot engine operates with source at 127°C and sink at 27°C. If the source supplies 40 kJ of heat energy, the work done by the engine is

30 kJ

10 kJ

4 kJ

1 kJ

Solution

10 kJ

Efficiency, η = 1 $- \frac{T_{2}}{T_{1}}$

η = $1 - \frac{(273 + 27)}{(273 + 127)}$

η = 1 $- \frac{300}{400}$

= 1 $- \frac{3}{4}$

η = $\frac{1}{4}$

Also, The efficiency defination given by

η = $\frac{Work done by engine}{Heat supplied by source}$

= $\frac{W}{40 kJ}$

∴ W = 40 η

= 40 × $\frac{1}{4}$

W = 10 kJ

Some More Questions From Thermodynamics Chapter

For Daily Free Study Material Join wiredfaculty