Correct Answer - Option 3 : 70 kW

__Concept:__

The COP of refrigerator is given as,

Ideal \(COP = \frac{{{T_L}}}{{{T_H} - {T_L}}}\) and Actual \(COP = \frac{Q}{W_{input}}\)

__Calculation:__

**Given:**

Q̇ = 20 × 3.5 = 70 kW, TL = 200 K, TH = 300 K

Ideal \(COP = \frac{{{200}}}{{{300} - {200}}}\) = 2

Actual COP = \(\frac{1}{2}\) × Ideal COP = \(\frac{1}{2}\) × 2 = 1

Actual COP = 1

Also, Actual COP = \(\frac{{\dot Q}}{{\dot W}} = \frac{{70}}{{\dot W}}\)

∴ \(\dot {W}\) = **70**** **kW.