Answer:
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
