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Found in: Page 549

### College Physics (Urone)

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

# Why do refrigerators, air conditioners, and heat pumps operate most cost-effectively for cycles with a small difference between Th and Tc? (Note that the temperatures of the cycle employed are crucial to its COP.)

Refrigerators, air conditioners, and heat pumps work efficiently if the temperature difference is small because the work required for their performance is less.

See the step by step solution

## Step 1: Explain how the efficiencies of these devices are calculated.

The efficiency of heat pumps, refrigerators and air conditioners are determined by the coefficient of performance. It is calculated as follows:

$$\begin{array}{l}CO{P_{hp}} = \frac{{{Q_h}}}{W}\\CO{P_{ref}} = \frac{{{Q_c}}}{W}\\W = {Q_h} - {Q_c}\end{array}$$

Here, $$CO{P_{hp}}$$is the coefficient of performance of the heat pump, $${Q_h}$$is amount of heat transfer into the warmer space and $$W$$ is the work input. $$CO{P_{ref}}$$ is the coefficient of performance of refrigerator and $${Q_c}$$ is the heat in the environment. The expression for the coefficient of performance of an air conditioner is same as that of a refrigerator.

## Step 2: Explain how the efficiency depends on temperature differences.

The expressions of coefficient of performance indicates that it is inversely proportional to the work output. In a Carnot engine,

$$\begin{array}{l}{Q_c} = {T_c}\\{Q_h} = {T_h}\\W = {T_h} - {T_c}\end{array}$$

Here, $${T_c}$$is the temperature of the environment and $${T_h}$$ is the temperature at the system. So, if this difference is small, the work input required for the performance will be smaller as well. Therefore, the coefficient of performance increases.

Thus, air conditioners, refrigerators, and heat pumps operate most cost-efficiently while working with a small temperature difference between the system and its surroundings.