ThermodynamicsLesson 2 of 5 in the course The Second Law of Thermodynamics

Reversed-Cycle Refrigerators and Heat-Pump Performance

5 flashcards · 5 quiz questions · 5 match pairs · 3 fill-in sentences · English

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Summary

A refrigerator and a heat pump are heat engines operated in reverse. Instead of using heat flow to produce work, they require work to transfer heat from a colder reservoir to a hotter one. This lesson explains how their energy transfers work and how to measure their effectiveness.

In a heat engine, heat flows from a hot reservoir to a cold reservoir, and some energy is converted into work. The cycles discussed in the preceding material are reversible: their steps can be performed in the opposite direction. Run backward, the cycle uses work to move heat from the cold reservoir to the hot reservoir.

All flashcards

How does a refrigerator or heat pump operate relative to a heat engine?
It runs a heat-engine cycle in reverse, using supplied work to transfer heat from a cold reservoir to a hot reservoir.
What do QcQ_c, QhQ_h, and WW represent in a reversed cycle?
QcQ_c is heat absorbed from the cold reservoir, QhQ_h is heat released to the hot reservoir, and WW is work supplied to the device.
What is the refrigerator coefficient of performance?
The heat removed from the cold reservoir per unit of work supplied: KR=Qc/WK_R = Q_c/W.
What is the heat-pump coefficient of performance?
The heat delivered to the hot reservoir per unit of work supplied: KP=Qh/WK_P = Q_h/W.
How are the refrigerator and heat-pump COPs related?
For the same cycle, KP=KR+1K_P = K_R + 1.

Quiz questions

  1. What distinguishes the purpose of a refrigerator from that of a heat pump?Answer: A refrigerator removes heat from a cold region; a heat pump delivers heat to a warm region.
  2. In a reversed-cycle device, what does the work WW represent?Answer: Work input done on the working substance.
  3. Which equation expresses energy conservation over one complete refrigerator or heat-pump cycle?Answer: Qh=Qc+WQ_h = Q_c + W
  4. A device removes 300 J300\,\text{J} of heat from a cold region and receives 100 J100\,\text{J} of work. How much heat does it release to the hot reservoir?Answer: 400 J400\,\text{J}
  5. If a device has refrigerator COP KR=3K_R = 3, what is its heat-pump COP for the same cycle?Answer: KP=4K_P = 4

Based on University Physics Volume 2 by Samuel J. Ling, William Moebs, Jeff Sanny (OpenStax), licensed CC BY 4.0. Adapted into a summary, flashcards and quizzes.

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