Textbook Question
The heat engine shown in FIGURE P21.62 uses 2.0 mol of a monatomic gas as the working substance.c. What is the engine's thermal efficiency?
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Ch 21: Heat Engines and Refrigerators
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 21, Problem 19b
An air conditioner removes 5.0 x 10⁵ J/min of heat from a house and exhausts 8.0 x 10⁵ J/min to the hot outdoors. What is the air conditioner's coefficient of performance?
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The coefficient of performance (COP) for a cooling device is defined as the ratio of the heat removed from the cold reservoir (house) to the work input required to operate the device. Mathematically, it is expressed as: , where is the heat removed from the cold reservoir and is the work input.
The work input can be determined using the first law of thermodynamics, which states that the total energy input to the system is equal to the sum of the heat removed from the cold reservoir and the heat exhausted to the hot reservoir. Mathematically: , where is the heat exhausted to the hot reservoir.
Substitute the given values into the equation for work: . This will give the value of in joules per minute.
Now, substitute the values of and into the COP formula: .
Simplify the expression to calculate the coefficient of performance (COP). This will give you the final value of the air conditioner's COP.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Coefficient of Performance (COP)
The Coefficient of Performance (COP) is a measure of the efficiency of a heat pump or air conditioning system. It is defined as the ratio of the heat removed from the cooled space (useful output) to the work input (energy consumed) required to remove that heat. A higher COP indicates a more efficient system, as it means more heat is removed per unit of energy consumed.
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Refrigerators
Heat Transfer
Heat transfer refers to the movement of thermal energy from one physical system to another. In the context of air conditioning, it involves the removal of heat from the indoor environment and its release to the outdoor environment. Understanding the principles of heat transfer, including conduction, convection, and radiation, is essential for analyzing the performance of HVAC systems.
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Energy Units and Conversion
In physics, energy is often measured in joules (J), and understanding how to convert between different units of energy and power is crucial for solving problems related to energy consumption. In this case, the air conditioner’s heat removal and exhaust rates are given in joules per minute, which may need to be converted to a consistent unit when calculating the COP. Familiarity with these conversions ensures accurate calculations and comparisons.
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Related Practice
Textbook Question
What are (a) the heat extracted from the cold reservoir and (b) the coefficient of performance for the refrigerator shown in FIGURE EX21.21?
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Textbook Question
Which, if any, of the heat engines in FIGURE EX21.22 violate (a) the first law of thermodynamics or (b) the second law of thermodynamics? Explain.
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Textbook Question
What are (a) the thermal efficiency and (b) the heat extracted from the hot reservoir for the heat engine shown in FIGURE EX21.16?
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Textbook Question
What are (a) Wout and QH and (b) the thermal efficiency for the heat engine shown in FIGURE EX21.14?
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Textbook Question
A 15 kW electric generator burns 1.2 gal of diesel fuel per hour. The energy density of diesel fuel is 140 MJ/gal. What is the generator's thermal efficiency?