Question

An ideal gas is taken through a cyclic thermodynamic process involving four steps. The amounts of heat involved in these steps are $Q_1=5960J,Q_2=-5585J,Q_3=-2980J$ and $Q_4=3645J$, respectively. The corresponding amounts of work done are $W_1=2200J,W_2=-825J$ and $W_3=-1100J$ and $W_4=respectively$.
What is the efficiency of the cycle?

Answer 1

Efficiency of the cycle is defined as
$\eta =\frac{\text{net work done by the gas}}{\text{total heat absorbed by the gas}}$
$=\frac{△W}{Q_1+Q_4}=\frac{275+765}{5980+3645}=\frac{1040}{9605}=0.1083=10.83\%$.