Answer
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Hint: The Carnot cycle is only suited for the cyclical devices like a heat engine, which tells the efficiency of the irreversible process of an irreversible heat engine is always less than that of the efficiency of the reversible process or reversible heat engine operating between the same two reservoirs. The Carnot COP is determined by using the Carnot COP formula.
Formula used:
The Coefficient of performance of the Carnot cycle is given by,
$COP = \dfrac{{{T_c}}}{{{T_c} - {T_e}}}$
Where, $COP$ is the coefficient of the performance, ${T_c}$ is the temperature of the condenser, and ${T_e}$ is the temperature of the evaporator.
Complete step by step answer:
Given that,
The temperature of the condenser is, ${T_c} = {27^ \circ }\,C$,
The temperature of the evaporator is, ${T_e} = {23^ \circ }\,C$.
Then the Carnot COP is given by,
$COP = \dfrac{{{T_c}}}{{{T_c} - {T_e}}}\,.................\left( 1 \right)$
By substituting the temperature of the condenser and the temperature of the evaporator in the equation (1), then the equation (1) is written as,
$\Rightarrow COP = \dfrac{{27}}{{27 - 23}}$
By subtracting the terms in the denominator in the RHS, then the above equation is written as,
$\Rightarrow COP = \dfrac{{27}}{4}$
On dividing the above equation, then the above equation is written as,
$\Rightarrow COP = 6.75 \simeq 6$
Thus, the above equation shows the COP of the Carnot cycle for the given condenser temperature and the evaporator temperature. Hence, option (C) is the correct answer.
Note:
A distinction is made between the theoretical COP and the actual COP of a refrigerating machine. The theoretical COP for all other refrigeration cycles does not exceed ${\varepsilon _c}$ for machines operating under the same temperature conditions. The COP of a real refrigerating machine is always less than the theoretical COP.
Formula used:
The Coefficient of performance of the Carnot cycle is given by,
$COP = \dfrac{{{T_c}}}{{{T_c} - {T_e}}}$
Where, $COP$ is the coefficient of the performance, ${T_c}$ is the temperature of the condenser, and ${T_e}$ is the temperature of the evaporator.
Complete step by step answer:
Given that,
The temperature of the condenser is, ${T_c} = {27^ \circ }\,C$,
The temperature of the evaporator is, ${T_e} = {23^ \circ }\,C$.
Then the Carnot COP is given by,
$COP = \dfrac{{{T_c}}}{{{T_c} - {T_e}}}\,.................\left( 1 \right)$
By substituting the temperature of the condenser and the temperature of the evaporator in the equation (1), then the equation (1) is written as,
$\Rightarrow COP = \dfrac{{27}}{{27 - 23}}$
By subtracting the terms in the denominator in the RHS, then the above equation is written as,
$\Rightarrow COP = \dfrac{{27}}{4}$
On dividing the above equation, then the above equation is written as,
$\Rightarrow COP = 6.75 \simeq 6$
Thus, the above equation shows the COP of the Carnot cycle for the given condenser temperature and the evaporator temperature. Hence, option (C) is the correct answer.
Note:
A distinction is made between the theoretical COP and the actual COP of a refrigerating machine. The theoretical COP for all other refrigeration cycles does not exceed ${\varepsilon _c}$ for machines operating under the same temperature conditions. The COP of a real refrigerating machine is always less than the theoretical COP.
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