Question

A heat engine having thermal efficiency of 40% receives heat from a source at \(600 \, K\) and rejects heat to a sink at \(300 \, K\). The second-law efficiency (in %) of this engine is .................. (answer in integer).

Hint:

Second-law efficiency = Actual efficiency ÷ Carnot efficiency. It measures how close the device is to the ideal reversible cycle.

Answer 1

Correct Answer: 80

Step 1: Actual efficiency given.
\[ \eta_{actual} = 0.40 \]

Step 2: Carnot efficiency (reversible maximum).
\[ \eta_{Carnot} = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{600} = 0.5 \]

Step 3: Second law efficiency.
\[ \eta_{II} = \frac{\eta_{actual}}{\eta_{Carnot}} \times 100 \] \[ \eta_{II} = \frac{0.40}{0.50} \times 100 = 80 % \] Final Answer:
\[ \boxed{80 %} \]