Question

A "Carnot" refrigerator (reverse of a Carnot engine) absorbs heat from the freezer compartment at a temperature of -17

Answer 1

The refrigerator absorbs this amount heat Qc at Tc and some work W and exhausts heat Qh at Th. Because net energy transfer to the fridge is zero, heat absorbed plus work done equals heat exhausted:
Qc + W = Qh
Furthermore the net entropy change of the refrigerator is zero. A Carnot refrigerator operates reversibly, that means no entropy is produced in it. So change in entropy is determined by the energy transfer alone. Since reversible work does not affect entropy. the entropy flow in due to heat absorption equals entropy flow our due to heat exhaust:
Qc/Tc = Qh/Th
=>
Qh = (Th/Tc)?Qc

Hence,
Qc + W = (Th/Tc)?Qc
=>
W = ((Th/Tc) - 1)?Qc

with
Tc = (-17 + 273) k = 256 K
Th = (24 + 273) K = 297 K

Qc= Q1+Q2+Q3

Q1= (.2)(4186)(24)= 20092.8

Q2=(.2)(3.33*10^5)=66600

Q3=(.2)(2100)(17)= 7140

Thus, Qc= 93832.8J
=>
W = ((297/256) - 1)?93832.8 J = 15027.9 J

Hence the answer....

Time required = total heat lost / rate of heat loss(compressor energy output)

t = 15027.9 / 200 = 75.2 seconds = 1.152 mins

hence, the solution.