# Calculating the Efficiency of a Reversible Refrigerator

• ajmCane22
In summary, the question asks for the efficiency of a reversible refrigerator with a coefficient of performance equal to 9.55. The efficiency can be calculated as COP = Qc/W or W/Qh, depending on whether the output is considered to be the heat removed or the heat delivered. Further information is needed to determine the correct calculation for efficiency.
ajmCane22

## Homework Statement

A reversible refrigerator has a coefficient of performance equal to 9.55

What is its efficiency?

COP = Qh/W

e = W/Qh

## The Attempt at a Solution

I thought e would simply be 1/9.55 = 0.1047 =0.105 (Sig. fig)

For one I am not sure how you are calculating efficiency since the refrigerator is producing no work output but using work input to move heat from a region of low temperature to a region of high temperature.

I don't know. That's the question in its entirety, and the answer is not 0, so there must be a way of figuring it out.

ajmCane22 said:
I don't know. That's the question in its entirety, and the answer is not 0, so there must be a way of figuring it out.

The problem will come in that your efficiency from a COP calculation can turn out to be more than 100%.

COP = Qc/W

Applying simple heat balance will give Qc=Qh+W

But W is your work input and Qh is your heat output.

ajmCane22 said:

## Homework Statement

A reversible refrigerator has a coefficient of performance equal to 9.55

What is its efficiency?

## Homework Equations

COP = Qh/W
For a refrigerator, output is considered to be the heat removed from the inside. So COP (output/input) is: Qc/W. For a heat pump, output is the heat delivered, Qh, so COP is Qh/W.

So if they say the COP of the refrigerator is 9.55, this means Qc/W = 9.55

Like rock.freak I am not clear on what efficiency means for a refrigerator. Does the question provide any further information? Try W/Qh using W+Qc = Qh

AM

## 1. How is the efficiency of a reversible refrigerator calculated?

The efficiency of a reversible refrigerator can be calculated using the Carnot cycle formula, which is the ratio of the difference between the temperatures of the hot and cold reservoirs to the temperature of the hot reservoir.

## 2. What is the ideal efficiency of a reversible refrigerator?

The ideal efficiency of a reversible refrigerator is 100%, which means that all of the energy input is converted into useful work. This is only achievable in theory, as no real system can be 100% efficient.

## 3. What factors can affect the efficiency of a reversible refrigerator?

The efficiency of a reversible refrigerator can be affected by the temperature difference between the hot and cold reservoirs, the quality and type of insulation used, and the efficiency of the compressor.

## 4. Can the efficiency of a reversible refrigerator be improved?

Yes, the efficiency of a reversible refrigerator can be improved by increasing the temperature difference between the hot and cold reservoirs, using better insulation materials, and using a more efficient compressor.

## 5. How does the efficiency of a reversible refrigerator compare to other types of refrigerators?

The efficiency of a reversible refrigerator is higher than other types of refrigerators, such as vapor-compression refrigerators, as it operates on a reversible cycle and does not have the same energy losses. However, it is still limited by the laws of thermodynamics and cannot achieve 100% efficiency.

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