Calculating the Efficiency of a Reversible Refrigerator

[ajmCane22]

Homework Statement

A reversible refrigerator has a coefficient of performance equal to 9.55

What is its efficiency?

Homework Equations

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)

This answer was incorrect. Can somebody please help?

 

[rock.freak667]

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.

 

[ajmCane22]

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.

 

[rock.freak667]

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 = Q_c/W

Applying simple heat balance will give Q_c=Q_h+W

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

 

[Andrew Mason]

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