- #1

fluidistic

Gold Member

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## Homework Statement

Hello guys,

I've attempted a problem but I get a result that I find too low to make sense. Plus, I have a doubt. So here it comes:

I've put in red the part I'm unsure of. As far as I know the efficiency of a refrigerator could be any real number from 0 to infinity (1 is not the limit!). So what exactly is 15% of the "ideal" coefficient performance?! The ideal would be an efficiency of infinity?!A household refrigerator is maintained at a temperature of 2°C. Everytime the door is opened, warm material is placed inside, introducting an average of 50 kcal, but making only a small change in the temperature of the refrigerator.

The door is opened 15 times a day, and the refrigerator operates at 15% of the ideal coefficient performance. The cost of work is 2.5 cent per kW h. What is the monthly bill for operating this refrigerator?

## Homework Equations

Not many.

## The Attempt at a Solution

Per day there are 750 kcal =3139.5 kJ entering the refrigerator.

1kW h =##3.6 \times 10^ 6 J## which corresponds to a price of 0.025$.

So if by efficiency of 15% of the ideal one they mean that for each calory that enters the refrigerator I must pay as if I remove 1/0.15 =6.667 calories out of it, then I reach that I must pay as if I must remove 20930 kJ per day. This corresponds to approximately 0.145$ per day. So in a month, 4.36$ only.

The result seems too low to me and I don't understand why they give the data of "2°C" nor exactly what they mean by 15% of the ideal coefficient performance.

Thanks for any help.