A refrigerator does 18kJ of work while moving 115 kJ of thermal enegy from inside the refrigerator. What is its coefficient of performance?
Refrigerators do not do work. Work has to be done UPON them to transfer heat out. I will assume that they meant to say 18 kJ of work IS CONSUMED by the refrigerator.
Refrigerator C.O.P. = Qout/Win
= 115/18 = 6.38
To find the coefficient of performance (COP) of a refrigerator, we need to use the formula:
COP = (useful work output) / (energy input)
In this case, the useful work output is given as 18 kJ, and the energy input is the thermal energy being moved from inside the refrigerator, which is given as 115 kJ.
Now let's substitute the values into the formula:
COP = (18 kJ) / (115 kJ)
To simplify this, we can convert both values to the same unit, such as joules:
COP = (18,000 J) / (115,000 J)
Now, we can divide both numbers:
COP = 0.1565
Therefore, the coefficient of performance of the refrigerator is approximately 0.1565.