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?

C. O. P = 1.49

W = 335.57 joules

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