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Coefficient of Performance is the ratio of useful heat produced by the pump to the energy intake of the pump. It is the reciprocal of the efficiency of the pump.

Temperature of cold environment\({T_c} = - 10\;^\circ {\rm{C}} = {\rm{263}}\;{\rm{K}}\)

Coefficient of performance of ideal refrigerator \({\beta _{ref}} = 7\)

The efficiency of the Carnot engine \[\eta = 1 - \frac{{{T_c}}}{{{T_h}}}\]

Heat pump’s performance coefficient \({\beta _{hp}} = \frac{1}{\eta }\)

Refrigerator’s performance coefficient\({\beta _{ref}} = {\beta _{hp}} - 1\)

Here,

\(\eta \)- efficiency of the engine.

\({T_c}\)- the temperature of the cold environment.

\({T_h}\)- temperature of the hot environment

\({\beta _{hp}}\)- heat pump’s performance coefficient.

\({\beta _{ref}}\) - refrigerator’s performance coefficient.

Heat pump’s performance coefficient is

\[\begin{array}{c}{\beta _{ref}} = {\beta _{hp}} - 1\\ = {\beta _{ref}} + 1\\ = 7 + 1\\ = 8\end{array}\]

efficiency can be calculated as

\[\begin{array}{l}{\beta _{hp}} = 1/\eta \\ \eta = 1/ {\beta _{hp}}\\\eta = 1/ 8\end{array}\]

Temperature of hot reservoir can be calculated as

\[\begin{array}{l}\eta = 1 - \frac{{{T_c}}}{{{T_h}}}\\\frac{1}{8} = 1 - \frac{{263\;{\rm{K}}}}{{{T_h}}}\\\frac{{263\;{\rm{K}}}}{{{T_h}}} = 1 - \frac{1}{8}\end{array}\]

\[\begin{array}{c}{T_h} = \frac{{263\;{\rm{K}} \times 8}}{7}\\ = 300.57 {\rm{K}}\\ = 27.57{\;^{\rm{o}}}{\rm{C}}\end{array}\]

Therefore, the temperature of hot reservoir is \(27.57\;^\circ {\rm{C}}\).