Question

A Carnot engine takes $3 \times {10^6}cal$ of heat from a reservoir at ${627^ \circ }C$, and gives it to a sink at ${27^ \circ }C$ . The work done by the engine is
A. $8.4 \times {10^6}J$
B. $16.8 \times {10^6}J$
C. $0$
D. $4.2 \times {10^6}J$

Answer 1

Hint:In the case, when a problem is based on Carnot Engine in a thermodynamic system, we know that all the parameters such as pressure, volume, temperature, etc., vary with the given conditions of the system and surroundings hence, use the scientific formula to calculate work done in Carnot engine ${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}} = \dfrac{W}{{{Q_1}}}$ to give the solution for the given problem.



Formula Used:
The efficiency of Carnot’s Heat Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$
where ${T_L} = $Lower Absolute Temperature = Temperature of the Sink
and, ${T_H} = $Higher Absolute Temperature = Temperature of the source





Complete answer:
We know that, the efficiency of Carnot Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}$ … (1)
where ${T_2} = $Lower Absolute Temperature = Temperature of the Sink
and, ${T_1} = $Higher Absolute Temperature = Temperature of the Source
Also, Efficiency of Carnot Engine in terms of work done can be given as: - ${\eta _{carnot}} = \dfrac{W}{{{Q_1}}}$ … (2)
where $W = $Work Done in the process
and, ${Q_1} = $Heat taken up from the Source
From eq. (1) and (2), we get
$\dfrac{W}{{{Q_1}}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}$ … (3)
Temperature of the hot reservoir (source) ${T_1} = {627^ \circ }C = 900K$ (given) $\left( {^ \circ C + 273 = K} \right)$
and Temperature of the Sink ${T_2} = {27^ \circ }C = 300K$ (given)
Heat taken up by the engine from the Reservoir ${Q_1} = 3 \times {10^6}cal$ (given)
Substituting these values in eq. (3), we get
$ \Rightarrow \dfrac{W}{{3 \times {{10}^6}}} = 1 - \dfrac{{300}}{{900}}$
$ \Rightarrow W = 3 \times {10^6}\left( {\dfrac{2}{3}} \right) = 2 \times {10^6}cal$
Converting this value into Joule, we get
As, $1cal = 4.184J$
Therefore, $W = 2 \times {10^6} \times 4.184 = 8.368 \times {10^6}J \approx 8.4 \times {10^6}J$
Thus, the work done by the engine is $8.4 \times {10^6}J$.
Hence, the correct option is (A) $8.4 \times {10^6}J$.



Thus, the correct option is A.



Note:Since this is a multiple-choice question (numerical-based), it is essential that given conditions are analyzed carefully to give an accurate solution. While writing an answer to this kind of numerical problem, always keep in mind to use the mathematical proven relations to find the solution.