How much work does an ideal Carnot refrigerator require to remove 1.00 J of energy from liquid h

Name: Problem 22, Chapter 22: Physics for Scientists and Engineers with Modern Physics
Uploaded: 2020-06-13T00:57:28-08:00
Duration: 59 s
Channel: Henrique Saito
Description: How much work does an ideal Carnot refrigerator require to remove 1.00 J of energy from liquid helium at 4.00 Kand expel this energy to a room-temperature (293-K) environment?

step 1 So to approach this question, we can use our two COP formulas, one where it's like the practical

Question

How much work does an ideal Carnot refrigerator require to remove 1.00 $\mathrm{J}$ of energy from liquid helium at 4.00 $\mathrm{Kand}$ expel this energy to a room-temperature $(293-\mathrm{K})$ environment?

   How much work does an ideal Carnot refrigerator require to remove 1.00 $\mathrm{J}$ of energy from liquid helium at 4.00 $\mathrm{Kand}$ expel this energy to a room-temperature $(293-\mathrm{K})$ environment?

Physics for Scientists and Engineers with Modern Physics

Raymond A. Serway,… 8th Edition

Chapter 22, Problem 22

Instant Answer

Solved by Expert Henrique Saito

06/13/2020

Step 1

Step 1: We know that the work done by a Carnot refrigerator is given by the formula: \[W = \frac{Q_C}{\eta}\] where $Q_C$ is the heat absorbed from the cold reservoir and $\eta$ is the efficiency of the refrigerator. Show more…

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How much work does an ideal Carnot refrigerator require to remove 1.00 $\mathrm{J}$ of energy from liquid helium at 4.00 $\mathrm{Kand}$ expel this energy to a room-temperature $(293-\mathrm{K})$ environment?

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Key Concepts

Carnot Refrigerator

A Carnot refrigerator is an idealized thermodynamic cycle that represents the maximum possible performance of a refrigeration system operating between two thermal reservoirs. It assumes reversible processes that allow it to achieve the highest theoretical coefficient of performance, serving as a benchmark for real-world refrigerators.

Coefficient of Performance (COP)

The coefficient of performance (COP) measures the efficiency of a refrigeration system by comparing the amount of heat extracted from the cold reservoir to the amount of work input. For a Carnot refrigerator, the COP is defined as the ratio of the low temperature (cold reservoir) to the difference between the high temperature (hot reservoir) and the low temperature, and it is used to calculate the minimum work required for a given amount of heat removal.

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