
An engine is supposed to operate between two reservoirs at temperature ${727^ \circ }C$ and ${227^ \circ }C$ . The maximum possible efficiency of such an engine is
A. $\dfrac{1}{2}$
B. $\dfrac{1}{4}$
C. $\dfrac{3}{4}$
D. $1$
Answer
163.5k+ views
Hint:In this case, a problem is based on the Calculation of the efficiency in a thermodynamic system, we know that all the parameters such as temperature, heat exchange, work done, etc., vary with the given conditions of the system and surroundings hence, use the formula of efficiency of Carnot cycle for calculating the maximum possible efficiency in the given problem.
Formula used:
the efficiency of Carnot Heat Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$
where, ${T_L} = $Lower Absolute Temperature = Temperature of Cold Reservoir
and, ${T_H} = $Higher Absolute Temperature = Temperature of Hot Reservoir
Complete answer:
To find the maximum possible efficiency of an engine, the Efficiency of the given engine must be equal to the efficiency of the Carnot Heat Engine.
We know that, the efficiency of Carnot Heat Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$
As the given engine is operated between two reservoirs at temperature ${T_H} = {727^ \circ }C = 1000K$ and ${T_L} = {227^ \circ }C = 500K$ (given) $\left( {^ \circ C + 273 = K} \right)$
The efficiency of the given Engine is maximum as: -
${\eta _{engine}} = {\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}} = 1 - \dfrac{{500}}{{1000}}$
$ \Rightarrow {\eta _{engine}} = \dfrac{1}{2}$ … (1)
Thus, the maximum possible efficiency of the given engine is $\dfrac{1}{2}$ .
Hence, the correct option is (A) $\dfrac{1}{2}$.
Therefore, the answer is option (A)
Note:Since this is a multiple-choice question (numerical-based) hence, it is essential that given conditions must be analyzed very 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. While solving questions related to this topic the value of temperature should be put in the kelvin unit.
Formula used:
the efficiency of Carnot Heat Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$
where, ${T_L} = $Lower Absolute Temperature = Temperature of Cold Reservoir
and, ${T_H} = $Higher Absolute Temperature = Temperature of Hot Reservoir
Complete answer:
To find the maximum possible efficiency of an engine, the Efficiency of the given engine must be equal to the efficiency of the Carnot Heat Engine.
We know that, the efficiency of Carnot Heat Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$
As the given engine is operated between two reservoirs at temperature ${T_H} = {727^ \circ }C = 1000K$ and ${T_L} = {227^ \circ }C = 500K$ (given) $\left( {^ \circ C + 273 = K} \right)$
The efficiency of the given Engine is maximum as: -
${\eta _{engine}} = {\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}} = 1 - \dfrac{{500}}{{1000}}$
$ \Rightarrow {\eta _{engine}} = \dfrac{1}{2}$ … (1)
Thus, the maximum possible efficiency of the given engine is $\dfrac{1}{2}$ .
Hence, the correct option is (A) $\dfrac{1}{2}$.
Therefore, the answer is option (A)
Note:Since this is a multiple-choice question (numerical-based) hence, it is essential that given conditions must be analyzed very 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. While solving questions related to this topic the value of temperature should be put in the kelvin unit.
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