Question

A heat engine with 20.0% efficiency does 0.100 kJ of work during each cycle. How much heat is absorbed from the hot reservoir during each cycle?

• A. 500 J
• B. 400 J
• C. 600 J
• D. 300 J

Answer 1

The correct option is A 500 J

$efficiency\left(\eta \right)=\frac{W}{Q}$

W=work done pre cycle, Q=heat absorbed from the reservoir during each cycle

$W=100J\eta =0.2Q=\frac{100}{0.2}=500J$