Example 15-8 Kitchen Refrigerator
A kitchen refrigerator has a coefficient of performance of 3.30. When the refrigerator pump is running, it removes energy
from the interior of the refrigerator at a rate of 760 J>s. (a) How much work must be done per second to extract this energy?
(b) How much energy is rejected into the kitchen per second in this process? (c) If the refrigerator operates between 2oC (the
interior of the refrigerator) and 25oC (the kitchen), what is the theoretical maximum coefficient of performance?
Set Up
The energy removed from the cold interior of
the refrigerator is represented by  QC  , and the
work done to remove that energy is represented
by  W  . We can relate these to the coefficient
of performance by using Equation 15-31, and
to the energy  QH  rejected into the warm
kitchen by using Equation 15-30. The theoretical maximum value of CP is the Carnot value,
given by Equation 15-32.
Coefficient of performance of a refrigerator:
CP =
 QC 
W
(a) Use Equation 15-31 to determine the
amount of work that must be done per second.
 QC 
 QH  -  QC 
(15-31)
Energy relationships for a refrigerator:
refrigerator
 QC  +  W  =  QH  or
 W  =  QH  2  QC  (15-30)
Coefficient of performance of a Carnot
refrigerator:
CPCarnot =
Solve
=
|QH| = ?
25°C
TC
TH - TC
|W| = ?
CP = 3.30
CPCarnot = ?
|QC| = 760 J
2°C
(15-32)
In 1 s, an amount of energy  QC  = 760 J is removed from the
interior of the refrigerator. The amount of work  W  that must be
done to make this happen depends on  QC  and the coefficient of
performance:
CP =
W =
 QC 
W
 QC 
760 J
=
= 230 J
CP
3.30
So the rate at which work must be done is 230 J>s.
(b) Find the rate at which energy is rejected into
the kitchen.
From Equation 15-30, the amount of heat  QH  rejected into the
kitchen in 1 s is
 QH  =  QC  +  W  = 760 J + 230 J = 990 J
So the rate at which the refrigerator rejects heat is 990 J>s.
(c) Calculate the coefficient of performance of a
Carnot refrigerator operating between the same
hot and cold temperatures.
The temperatures between which the refrigerator operates are
2°C or TC = (273.15 + 2) K = 275 K and
25°C or TH = (273.15 + 25) K = 298 K
The coefficient of performance of a Carnot refrigerator operating
between these temperatures is
CPCarnot =
TC
275 K
275 K
=
=
TH - TC
298 K - 275 K
23 K
= 12
Reflect
The coefficient of performance CP is substantially less than what could be achieved by an ideal Carnot refrigerator.
Notice that the higher the actual value of CP, the less work must be done to achieve the same amount of cooling
inside the refrigerator. Less work means less waste, and as a result, less heating of the kitchen. This actual refrigerator rejects heat into the kitchen at a rate of 990 J>s or 990 W, about the same as the power output of an electric space
heater.