If the lower temperature fixed by the refrigeration application is high, the C.O.P. of the Carnot refrigerator will be ________

Explanation:

Refrigerator:

A refrigerator is a device that works on a reverse Carnot cycle and extracts heat from a lower temperature body to keep the temperature of the body lower than the surrounding temperature and by taking work input it transfers heat to the higher temperature body or surrounding.

Refrigerating Effect R.E. = Q1

Work input = Q2 - Q1

\({\bf{Coefficient}}\;{\bf{of}}\;{\bf{Performance}}\;\left( {{\bf{COP}}} \right) = \frac{{{\bf{Refrigerating}}\;{\bf{Effect}}}}{{{\bf{Work}}\;{\bf{Input}}}} = \frac{{{{\bf{Q}}_{\bf{1}}}}}{{\bf{W}}}\)

COP of a Carnot refrigerator is given by;

\({\left( {COP} \right)_{ref}} = \frac{{{T_1}}}{{{T_2} - {T_1}}} = \frac{{{Q_{ref}}}}{W}\)

T1 = Lower Temperature, T2 = Higher temperature

From this relation, it is apparent that if T1 (the temperature inside the refrigerator) is raised, while T2  is kept constant as per conditions, the COP will increase. Hence, if the lower temperature fixed by the application is high, the COP of the Carnot refrigerator would be high.