Question

Find out the most efficient engine in the following
A. An engine converts 80 kJ of heat energy into 20 kJ of work
B. An engine converts 50 kJ of heat energy into 15 kJ of work
C. An engine converts 30 kJ of heat energy into 6 kJ of work
D. An engine converts 60 kJ of heat energy into 24 kJ of work

Answer 1

Efficiency of a heat engine is defined as the ratio of net work done per cycle by the engine to the total amount of heat absorbed from the source. If $\eta$ is the efficienciency of the engine, W is the work done and ${Q_1}$is the amount of heat absorbed by the working substance from the source, then $\eta = \dfrac{W}{{{Q_1}}}$.

Complete step by step answer:
The work done by a heat engine is equal to ${Q_2} - {Q_1}$, where ${Q_1}$is the amount of heat absorbed by the working substance from the source and ${Q_2}$ is the amount of heat rejected to the sink. Efficiency of a heat engine is defined as the ratio of net work done per cycle by the engine to the total amount of heat absorbed from the source. It is denoted by $\eta$ $\Rightarrow \eta = \dfrac{W}{{{Q_1}}}$
In option A, Work done = 20 kJ and Heat Input = 80 kJ
Hence efficiency =$\dfrac{{20}}{{80}} = \dfrac{1}{4} = 0.25$
In option B, Work done = 15 kJ and Heat Input = 50 kJ
Hence efficiency = $\dfrac{{15}}{{50}} = \dfrac{3}{{10}} = 0.3$
In option C, Work done = 6 kJ and Heat Input = 30 kJ
Hence efficiency = $\dfrac{6}{{30}} = \dfrac{1}{5} = 0.2$
In option D, Work done = 24 kJ and Heat Input = 60 kJ
Hence efficiency = $\dfrac{{24}}{{60}} = \dfrac{2}{5} = 0.4$
Hence, an engine that converts 60 kJ of heat energy into 24 kJ of work will be more efficient.
Hence, the correct option is (D).

So, the correct answer is “Option D”.

Note:
In a Carnot engine, source at infinite temperature or sink at zero temperature is not attainable. Hence it is not possible to convert heat energy into mechanical work unless source and sink of heat are at different temperatures. In other words, efficiency of a heat engine can never be 100%.