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Found in: Page 551

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# 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

## Definition:

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 $${\beta _{ref}} = {\beta _{hp}} - 1$$

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.

## 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 $$3.61$$.