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College Physics (Urone)
Found in: Page 549
College Physics (Urone)

College Physics (Urone)

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

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Short Answer

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.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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.

Most popular questions for Physics Textbooks

Would you be willing to financially back an inventor who is marketing a device that she claims has 25 kJ of heat transfer at 600 K, has heat transfer to the environment at 300 K, and does 12 kJ of work? Explain your answer.

(a) What is the work output of a cyclical heat engine having a 22% efficiency and 6×109 J of heat transfer into the engine (b)How much heat transfer occur to the environment

A \({\bf{4}}\)-ton air conditioner removes \({\bf{5}}.{\bf{06}} \times {\bf{1}}{{\bf{0}}^{\bf{7}}}\;{\bf{J}}\) (\({\bf{48}},{\bf{000}}\) British thermal units) from a cold environment in \({\bf{1}}.{\bf{00}}\;{\bf{h}}\). (a) What energy input in joules is necessary to do this if the air conditioner has an energy efficiency rating ( EER ) of \({\bf{12}}.{\bf{0}}\)? (b) What is the cost of doing this if the work costs \({\bf{10}}.{\bf{0}}\) cents per \({\bf{3}}.{\bf{60}} \times {\bf{1}}{{\bf{0}}^{\bf{6}}}\;{\bf{J}}\) (one kilowatt-hour)? (c) Discuss whether this cost seems realistic. Note that the energy efficiency rating ( EER ) of an air conditioner or refrigerator is defined to be the number of British thermal units of heat transfer from a cold environment per hour divided by the watts of power input.

A woman shuts her summer cottage up in September and returns in June. No one has entered the cottage in the meantime. Explain what she is likely to find, in terms of the second law of thermodynamics.

(a) What is the best coefficient of performance for a refrigerator that cools an environment at −30.0ºC and has heat transfer to another environment at 45.0ºC ? (b) How much work in joules must be done for a heat transfer of 4186 kJ from the cold environment? (c) What is the cost of doing this if the work costs 10.0 cents per 3.60×106 J (a kilowatt-hour)? (d) How many kJ of heat transfer occurs into the warm environment? (e) Discuss what type of refrigerator might operate between these temperatures.
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