Explanation Given data: COP is 5.0 , Number of students are 250, and Each student using metabolic power is 125 W. Formula used: Write the expression for coefficient of performance for cooling ( COP ) . COP = Q C W in (1) Here, Q C is the heat removed from the cold side, and W in is the input work. Find the total amount of energy ( Q C ) . Q C = ( Number of students ) ( Metabolic power of each student ) Substitute 250 for number of students and 125 W for metabolic power of each student, Q C = ( 250 student ) ( 125

Mastering Physics with Pearson eText -- Standalone Access Card -- for College Physics: A Strategic Approach (3rd Edition)

3rd Edition

ISBN: 9780321908803

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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First law of Thermodynamics

The word thermodynamics consists of two words: thermo- and dynamics. Dynamics means the study of motion. The field of thermodynamics is all about the study of the movement of heat. In a more meaningful way, it is stated as the study of the transfer of en…

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Chapter 11, Problem 63GP

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To find: The minimum electric power used to operate the air conditioner.

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Chapter 11, Problem 62GP

Chapter 11, Problem 64GP

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A 4-ton air conditioner removes 5.06×107 J (48,000 British thermal units) from a cold environment in 1.00 h. (a) What energy input in joules is necessary to do this if the air conditioner has an energy efficiency rating (EER) of 12.0? (b) What is the cost of doing this if the work costs 10.0 cents per 3.06×106 J (one kilowatthour)? (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 Carnot air conditioner takes energy from the thermal energy of a room at 70 F and transfers it as heat to the outdoors, which is at 96 F.For each joule of electric energy required to operate the air conditioner, how many joules are removed from the room?

A 4-ton air conditioner removes 5.06×107 J (48,000 British thermal units) from a cold environment in 1.00 h. (a) What energy input in joules is necessary to do this if the airconditioner has an energy efficiency rating ( EER ) of 12.0?(b) What is the cost of doing this if the work costs 10.0 centsper 3.60×106 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.

Chapter 11 Solutions

Mastering Physics with Pearson eText -- Standalone Access Card -- for College Physics: A Strategic Approach (3rd Edition)

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The minimum electric power used to operate the air conditioner.

Solution Summary: The author explains how the minimum electric power used to operate the air conditioner is 6,250 J. The expression for coefficient of performance for cooling is (COP).

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To find: The minimum electric power used to operate the air conditioner.

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Chapter 11 Solutions