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Chapter 21: Q. 33 (page 596)

The coefficient of performance of a refrigerator is5.0. The compressor uses10Jof energy per cycle.

a. How much heat energy is exhausted per cycle

b. If the hot-reservoir temperature is27掳C, what is the lowest possible temperature inC鈭�of the cold reservoir?

Short Answer

Expert verified

a.Heat energy is exhausted per cycle is60J.

b.Lowest possible temperatureTcis-23掳C.

Step by step solution

01

Calculation for heat energy is exhausted per cycle  (part a)

a.

As a result of the first law that was applied to refrigerators, we now have

Qc=Qh+W

K=QcW鈬�Qc=KW

Substituting this we get,

Qc=Qh-W=W(K+1)

Qh=10路(5+1)=60J

02

Calculation for lowest possible temperature (part b)

b.

When the coefficient of performance is that of a Carnot refrigerator, the lowest value of the temperature of the cold reservoir is reached.

That is,

K=TcTh-Tc=ThTh-Tc-1

The value of the lower temperature can be separated as follows:

Th=(K+1)Th-Tc鈬�-ThK=-(K+1)Tc

Tc=ThKK+1

The minimum temperature ,

Tc=55+1路(273+27)=250K

In degrees centigrade,

Tc=-23掳C

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Most popular questions from this chapter

A heat engine using 1.0 mol of a monatomic gas follows the cycle shown in FIGURE P21.55. 3750 J of heat energy is transferred to the gas during process 1 -> 2.
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