Question

In a mechanical refrigerator, the low temperature coils are at a temperature of $ - {23^ \circ }C$ and the compressed gas in the condenser has a temperature of ${27^ \circ }C$ . The theoretical coefficient of performance is
A. $5$
B. $8$
C. $6$
D. $6.5$

Answer 1

Hint:This problem is based on Refrigeration System in thermodynamics, 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 scientific formula to calculate the coefficient of performance (C.O.P.)$C.O.P. = \dfrac{{{T_L}}}{{{T_H} - {T_L}}}$ to give the solution for the given problem.




Formula used:
The coefficient of performance (C.O.P.)$C.O.P. = \dfrac{{{T_L}}}{{{T_H} - {T_L}}}$


Complete answer:
We know that, the Coefficient of Performance (C.O.P.) of Refrigerator is given as: -
$C.O.P. = \dfrac{{{T_L}}}{{{T_H} - {T_L}}}$ … (1)
where ${T_L} = $Lower Absolute Temperature = Temperature of the Sink
and, ${T_H} = $Higher Absolute Temperature = Temperature of the Source
Temperature of the Condenser is ${T_H} = {27^ \circ }C = 300K$ (given) $\left( {^ \circ C + 273 = K} \right)$
And the Temperature of the Coil is ${T_L} = - {23^ \circ }C = 250K$ (given)
 From eq. (1), we get
$ \Rightarrow C.O.P. = \dfrac{{{T_L}}}{{{T_H} - {T_L}}} = \dfrac{{250}}{{300 - 250}}$
$ \Rightarrow C.O.P. = \dfrac{{250}}{{50}} = 5$
Thus, the theoretical Coefficient of Performance for the given mechanical refrigerator is $5$ .
Hence, the correct option is (A) $5$ .




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.