Question

Consider a Carnot’s cycle operating between source temperature $750\,K$ and sink temperature $350\,K$ producing $1.25\,kJ$ of mechanical work per cycle, the heat transferred by the engine to the reservoir is

A. $1.34\,kJ$

B. $3.24\,kJ$

C. $2.34\,kJ$

D. $4.34\,kJ$

Answer 1

First find the efficiency of the engine and then find the heat transferred by the heat engine to the reservoir. The efficiency of a Carnot’s heat engine depends only on the temperature of source ($T_1$), temperature of sink ($T_2$) and heat supplied ($Q_1$).

Complete step by step solution:

Here temperature of the source ${T_1} = 750\,K$

Temperature of the sink ${T_2} = 350\,K$

Efficiency of the Carnot heat engine $\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$

$\Rightarrow \eta = 1 - \dfrac{{350}}{{750}} = \dfrac{{400}}{{750}} = \dfrac{8}{{15}}$

Work done per cycle $W\, = \,1.25\,kJ\, = 1250\,J$

Heat transferred by the engine to the reservoir, Q$_1$ =?

We know that $\eta = \dfrac{W}{{{Q_1}}} \Rightarrow {Q_1} = \dfrac{W}{\eta } \Rightarrow {Q_1} = \dfrac{{1250}}{8} \times 15J = 2.34\,kJ$

Hence, the correct option is (C).

Note: Efficiency of an engine is defined as the ratio of the net mechanical work done per cycle to the amount of heat energy absorbed per cycle from the source. Carnot showed that no heat energy working between two given temperatures of source and sink can be more efficient than a perfectly reversible heat engine. This statement came to be known as the Carnot theorem. 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%. The equation $\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$ for efficiency reveals that efficiency of a Carnot heat engine depends upon the temperatures of source and sink. It does not depend upon the nature of the working substance.