Question

A Carnot?s 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

• A. 30 kJ
• B. 10 kJ
• C. 4 kJ
• D. 1 kJ

Answer 1

Answer: (B)

10 kJ

Answer 2

Efficiency, $\eta = \frac{T_{1}-T_{2}}{T_{1}} = 1 -\frac{T_{2}}{T_{1}}$
$\therefore \:\:\: \eta = 1 -\frac{\left(273 +27\right)}{\left(273 +127 \right)} = 1 - \frac{300}{400} =1 -\frac{3}{4} =\frac{1}{4}$
$\therefore \:\:\:\:\: \eta = \frac{work \,done \, by \, engine}{heat \, supplied \, by \, source} = \frac{W}{40(kJ)}$
or $W = 40 \eta = 40 \times \frac{1}{4} =$ 10 kJ
or Work done = 10 kJ