Ch 21: Heat Engines and Refrigerators

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 21: Heat Engines and Refrigerators

Problem 21a

Chapter 21, Problem 21a

What are (a) the heat extracted from the cold reservoir and (b) the coefficient of performance for the refrigerator shown in FIGURE EX21.21?

Verified step by step guidance

Identify the key variables and concepts involved in the problem. For a refrigerator, the coefficient of performance (COP) is defined as \( \text{COP} = \frac{Q_c}{W} \), where \( Q_c \) is the heat extracted from the cold reservoir and \( W \) is the work done on the system. The first law of thermodynamics also applies: \( W = Q_h - Q_c \), where \( Q_h \) is the heat released to the hot reservoir.

From the problem or figure, determine the values of \( Q_h \) (heat released to the hot reservoir) and \( W \) (work done on the system). If these values are not directly given, use any provided data to calculate them.

Rearrange the first law of thermodynamics equation \( W = Q_h - Q_c \) to solve for \( Q_c \): \( Q_c = Q_h - W \). Substitute the known values of \( Q_h \) and \( W \) to calculate \( Q_c \), the heat extracted from the cold reservoir.

Use the definition of the coefficient of performance (COP): \( \text{COP} = \frac{Q_c}{W} \). Substitute the calculated value of \( Q_c \) and the given or calculated value of \( W \) to find the COP of the refrigerator.

Verify the units and ensure consistency throughout the calculations. The heat quantities \( Q_c \) and \( Q_h \) should be in joules (J), and the work \( W \) should also be in joules (J). The COP is a dimensionless quantity.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Heat Transfer

Heat transfer refers to the movement of thermal energy from one object or system to another due to a temperature difference. In the context of refrigeration, heat is extracted from the cold reservoir, which is the area being cooled, and transferred to the hot reservoir, typically the surrounding environment. Understanding the principles of heat transfer is essential for analyzing the efficiency and operation of refrigeration systems.

Refrigerator Coefficient of Performance (COP)

The Coefficient of Performance (COP) of a refrigerator is a measure of its efficiency, defined as the ratio of the heat removed from the cold reservoir to the work input required to remove that heat. A higher COP indicates a more efficient refrigerator, as it means more heat is extracted for each unit of work done. This concept is crucial for evaluating the performance of refrigeration systems and understanding their energy consumption.

Thermodynamic Cycles

Thermodynamic cycles describe the series of processes that a refrigerant undergoes in a refrigeration system, including compression, condensation, expansion, and evaporation. These cycles are fundamental to understanding how refrigerators operate, as they illustrate how energy is transferred and transformed within the system. Familiarity with these cycles helps in analyzing the performance and efficiency of refrigeration systems.

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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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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

At what cold-reservoir temperature (in ℃) would a Carnot engine with a hot-reservoir temperature of 427℃ have an efficiency of 60%?

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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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Textbook Question

A Carnot engine whose hot-reservoir temperature is 400℃ has a thermal efficiency of 40%. By how many degrees should the temperature of the cold reservoir be decreased to raise the engine's efficiency to 60%?

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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?