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College Physics (Urone)
Found in: Page 551

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Short Answer

Suppose you have an ideal refrigerator that cools an environment at −20.0ºC and has heat transfer to another environment at 50.0ºC . What is its coefficient of performance?

The refrigerator’s performance coefficient is 3.61.

See the step by step solution

Step by Step Solution


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.

Step 2:  Given parameters & formula for efficiency of heat engine and performance coefficient

Temperature of cold environment \({T_c} = - 20\;^\circ {\rm{C}} = {\rm{253}}\;{\rm{K}}\)

The temperature of hot environment \({T_h} = 50\;^\circ {\rm{C}} = {\rm{323}}\;{\rm{K}}\)

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 \({\beta _{ref}} = {\beta _{hp}} - 1\)


\(\eta \) - 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.

Step 3:  Calculate efficiency

Maximum efficiency can be calculated by calculating Carnot efficiency as

\(\begin{aligned}{}\eta & = 1 - \frac{{{T_c}}}{{{T_h}}}\\ & = 1 - \frac{{253\;{\rm{K}}}}{{323\;{\rm{K}}}}\\ &= 0.217\end{aligned}\)

Step 4:  Calculate coefficient of performance of ideal heat pump and of refrigerator

Heat pump’s performance coefficient is

\(\begin{aligned}{}{\beta _{hp}} &= 1/\eta \\ &= 1/0.217\\ &= 4.61\end{aligned}\)

Refrigerator’s performance coefficient is

\(\begin{aligned}{}{\beta _{ref}} &= {\beta _{hp}} - 1\\ &= 4.61 - 1\\ &= 3.61\end{aligned}\)

Therefore, the refrigerator’s performance coefficient is \(3.61\).

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