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Physics Principles with Applications
Found in: Page 412
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

Question:The COPs are defined differently for heat pumps and air conditioners. Explain why.

The COP of the heat pump is as follows:

\(COP = \frac{{{Q_{\rm{H}}}}}{W}\)

The COP of the air conditioners is as follows:

\(COP = \frac{{{Q_{\rm{L}}}}}{W}\)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Coefficient of performance (COP)

The term coefficient of performance describes the effectiveness of a certain device like heat pump or refrigerator at transferring rate and the quantity of power consumed by the device.

Step 2: Explaining the difference in COP of heat pumps and air conditioners

The relation of COP of the heat pumps is given by:

\(COP = \frac{{{Q_{\rm{H}}}}}{W}\)

The coefficient of performance of the heat pump relies on the amount of heat delivered. The main objective of this device is to heat instead of cool. In order to make the working of the heat pump better, more heat should be delivered per amount of input work.

The relation of COP of theair conditioners is given by:

\(COP = \frac{{{Q_{\rm{L}}}}}{W}\)

The coefficient of performance of the air conditioners relies on the amount of heat removed from the reservoir. The main objective of this device is cooling instead of heating.

In order to make the working of the air conditioners better, increase the amount of heat extracted per amount of input work.

Most popular questions for Physics Textbooks

Question: (II) A heat pump is used to keep a house warm at 22°C. How much work is required of the pump to deliver 3100 J of heat into the house if the outdoor temperature is (a) 0°C, (b) \({\bf{ - 15^\circ C}}\)? Assume a COP of 3.0. (c) Redo for both temperatures, assuming an ideal (Carnot) coefficient of performance \({\bf{COP = }}{{\bf{T}}_{\bf{L}}}{\bf{/}}\left( {{{\bf{T}}_{\bf{H}}}{\bf{ - }}{{\bf{T}}_{\bf{L}}}} \right)\).

Question: (II) A 1.0-L volume of air initially at 3.5 atm of (gauge) pressure is allowed to expand isothermally until the pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a PV diagram, including numbers and labels for the axes.

(I) An ideal gas expands isothermally, performing\({\bf{4}}{\bf{.30 \times 1}}{{\bf{0}}^{\bf{3}}}\;{\bf{J}}\) of work in the process. Calculate (a) the change in internal energy of the gas, and (b) the heat absorbed during this expansion.

A “Carnot” refrigerator (the reverse of a Carnot engine) absorbs heat from the freezer compartment at a temperature of -17°C and exhausts it into the room at 25°C.

(a) How much work would the refrigerator do to change 0.65 kg of water at 25°C into ice at -17°C.

(b) If the compressor output is 105 W and runs 25% of the time, how long will this take?

Question: Which is possible: converting (i) 100 J of work entirely into 100 J of heat, (ii) 100 J of heat entirely into 100 J of work?

(a) Only (i) is possible.

(b) Only (ii) is possible.

(c) Both (i) and (ii) are possible.

(d) Neither (i) nor (ii) is possible.
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