(Figure 1) shows three refrigerators. Part A For each refrigerator calculate ? E

Question

(Figure 1) shows three refrigerators.

Part A

For each refrigerator calculate ?E=QC+Win?QH, where QH is the amount of heat transferred to the hot reservoir, QC is the amount of heat transferred from the cold reservoir, Win is the energy used by the refrigerator. QH, QC, Win are positive quantities.

Part B

Which, if any, of the refrigerators violate(s) the first law of thermodynamics.

Part C

For each refrigerator calculate the theoretical maximal coefficient of performance COPmax.

Part D

For each refrigerator calculate the actual coefficient of performance COP.

Part E

Which, if any, of the refrigerators violate(s) the second law of thermodynamics.

(a) Hot reservoir = 400 K (b) Hot reservoir = 400 K 60J 50J 20 J Refrigerator 10J Refrigerator 40 J 40 J Cold reservoir Tc=300 K Cold reservoir Tc=300 K (c) Hot reservoirTH400 K 40J 20J Refrigerator 30 J Cold reservoir Tc=300 K

Explanation / Answer

(A). For 1 st refrigerator Qc= 40 J

                               W in = 20 J

                                  QH=60 J

So, E=QC+WinQH

          = 40 + 20 -60

          = 0 J

For 2nd refrigerator Qc= 40 J

                               W in = 10 J

                                  QH=50 J

So, E=QC+WinQH

          = 40 + 10 -50

          = 0 J

For 3 rd refrigerator Qc= 30 J

                               W in = 20 J

                                  QH=40 J

So, E=QC+WinQH

          = 30 + 20 -40

          = 10 J

(B).3 rd refrigerator violate the first law of thermodynamics

(C).For 1 st refrigerator

the theoretical maximal coefficient of performance COPmax= TC/(TH-TC)

                                                                                      = 300 /(400-300)

                                                                                      = 3

For 2nd refrigerator

the theoretical maximal coefficient of performance COPmax= TC/(TH-TC)

                                                                                      = 300 /(400-300)

                                                                                      = 3

For 3rd refrigerator

the theoretical maximal coefficient of performance COPmax= TC/(TH-TC)

                                                                                      = 300 /(400-300)

                                                                                      = 3

(D).For 1 st refrigerator

the actual coefficient of performance COP = QC/Win

                                                             = 40 /20

                                                             = 2

For 2nd refrigerator

the actual coefficient of performance COP = QC/Win

                                                             = 40 /10

                                                             = 4

For 2nd refrigerator

the actual coefficient of performance COP = QC/Win

                                                             = 40 /20

                                                             = 2

(e).2nd refrigerator violates second law of thermodynamics

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