Domanda del libro di testo
A gas following the pV trajectory of FIGURE EX21.11 does 60 J of work per cycle. What is Vmax?
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Ch 21: Heat Engines and Refrigerators
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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TOC.tableOfContents
- 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
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Capitolo 21, Problema 14
What are (a) Wout and QH and (b) the thermal efficiency for the heat engine shown in FIGURE EX21.14?
Guida verificata passo dopo passo1
Step 1: Understand the problem. The heat engine operates between two reservoirs, absorbing heat (QH) from the hot reservoir and performing work (Wₒᵤₜ) while expelling heat (QC) to the cold reservoir. The thermal efficiency (η) is defined as the ratio of work output to heat input: η = Wₒᵤₜ / QH.
Step 2: Identify the given values from FIGURE EX21.14. Extract the values for QH (heat absorbed from the hot reservoir) and QC (heat expelled to the cold reservoir). These values are typically provided in the figure or accompanying text.
Step 3: Calculate the work output (Wₒᵤₜ) using the first law of thermodynamics for a heat engine: Wₒᵤₜ = QH - QC. Subtract the heat expelled (QC) from the heat absorbed (QH).
Step 4: Determine the thermal efficiency (η) using the formula η = Wₒᵤₜ / QH. Divide the work output (Wₒᵤₜ) by the heat input (QH) to find the efficiency of the engine.
Step 5: Express the thermal efficiency as a percentage by multiplying the result by 100: η (%) = (Wₒᵤₜ / QH) × 100. This gives the efficiency in a more interpretable form.
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Concetti chiave
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Work Output (Wₒᵤₜ)
Work output (Wₒᵤₜ) refers to the useful work produced by a heat engine during its operation. It is calculated as the difference between the energy input from the heat source and the energy rejected to the heat sink. This concept is crucial for evaluating the performance of the engine, as it directly relates to the engine's ability to convert thermal energy into mechanical energy.
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Percorso guidato
06:58
Power Output of a Gasoline Engine
Heat Input (QH)
Heat input (QH) is the total amount of thermal energy absorbed by the heat engine from the heat source during one complete cycle. This energy is essential for the engine's operation, as it provides the necessary energy to perform work. Understanding QH is vital for calculating the thermal efficiency of the engine, as it serves as the baseline for comparing the work output.
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07:35Introduction to Heat Engines
Thermal Efficiency
Thermal efficiency is a measure of how effectively a heat engine converts heat energy into work. It is defined as the ratio of the work output (Wₒᵤₜ) to the heat input (QH), often expressed as a percentage. A higher thermal efficiency indicates a more effective engine, and it is a key performance indicator in thermodynamics, helping to assess the effectiveness of different engine designs.
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06:01Thermal Efficiency & The Second Law of Thermodynamics
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