Give the basic features of a heat engine based on a cyclic process and obtain the formula for its efficiency.

(N/A) The basic features of a heat engine are that the engine takes heat $Q_{1}$ from a hot reservoir at temperature $T_{1}$,releases heat $Q_{2}$ to a cold reservoir at temperature $T_{2}$,and delivers work $W$ to the surroundings. The basic features of this engine are represented in the figure.
Since the process is cyclic,the change in internal energy of the working substance is zero. Therefore,the net amount of heat absorbed by the working substance is $Q = W$,where $W = Q_{1} - Q_{2}$.
The ratio of the net work done per cycle by the engine to the total amount of heat absorbed per cycle by the working substance from the source is known as the efficiency $\eta$ of the engine.
$\therefore \text{Efficiency} = \frac{\text{Net work done / cycle}}{\text{Heat absorbed / cycle}}$
$\therefore \eta = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}}$
$\therefore \eta = 1 - \frac{Q_{2}}{Q_{1}}$
Above equation suggests that the efficiency of an engine is $\eta = 1$ if $Q_{2} = 0$,which means the efficiency of the engine becomes $100 \%$.
The first law of thermodynamics does not rule out such an engine,but an ideal engine with $\eta = 1$ ($100 \%$ efficiency) is never possible even if we can eliminate various kinds of losses associated with an actual heat engine.
The second law of thermodynamics provides a fundamental limitation to the efficiency of a heat engine.