Question

A cyclic process ABCDA has 4 steps. Heats involved in all processes are given by $Q_{AB}=4800Joules$, $Q_{BC}=1000Joules$, $Q_{CD}=500Cal$ and $Q_{DA}=-1000Cal$. Find efficiency of engine based on above cyclic process. (1 Cal = 4.2 Joules)

• A. $23\%$
• B. $47\%$
• C. $36\%$
• D. $11\%$

Answer 1

The correct option is B $47\%$

$Q_1=4800+1000+(500\times 4.2)=7900$
$Q_2=1000\times 4.2=4200$

Efficiency = $1-\frac{Q_2}{Q_1}=1-\frac{4200}{7900}=0.47$