Question

An ideal refrigerator extracts 500 joules of heat from a reservoir at 295 K and rejects heat to a reservoir at 493 K. What is the ideal coefficient of performance and how much work is done in each cycle?

Answer 1

C.O.P = 1.49

W = 335.57 joules

Step-by-step explanation:

C.O.P = coefficient of performance = (benefit/cost) = Qc/W ...equ 1 where C.O.P is coefficient of performance, Qc is heat from cold reservoir, w is work done on refrigerator.

Qh = Qc + W...equ 2

W = Qh - Qc ...equ 3 where What is heat entering hot reservoir.

Substituting for W in equ 1

Qh/(Qh - Qc) = 1/((Qh /Qc) -1) ..equ 4

Since the second law states that entropy dumped into hot reservoir must be already as much as entropy absorbed from cold reservoir which gives us

(Qh/Th)>= (Qc/Tc)..equ 5

Cross multiple equ 5 to get

(Qh/Qc) = (Th/Tc)...equ 6

Sub equ 6 into equation 4

C.O.P = 1/((Th/Tc) -1)...equ7

Where Th is temp of hot reservoir = 493k and Tc is temp of cold reservoir = 295k

C.O.P = 1/((493/295) - 1)

C.O.P = 1.49

To solve for W= work done on every cycle

We substitute C.O.P into equ 1

Where Qc = 500 joules

1.49 = 500/W

W = 500/1.49

W = 335.57 joules