Relative performance comparison Compared with a typical domestic air-conditioner, the coefficient of performance (C.O.P.) of a domestic refrigerator is generally:

Introduction / Context:C.O.P. depends primarily on the temperature lift between evaporator and condenser. Domestic refrigerators and domestic air-conditioners operate across very different temperature ranges, leading to different typical C.O.P. values.

Given Data / Assumptions:

• Typical refrigerator: evaporator near −5°C to −15°C, condenser near 35–45°C.
• Typical room air-conditioner: evaporator near 10–15°C, condenser near 35–45°C.
• Similar technology and comparable component efficiencies.

Concept / Approach:For ideal cycles, C.O.P._R ≈ T_L / (T_H − T_L) in absolute units (Kelvin). A larger temperature lift (T_H − T_L) reduces C.O.P. Since refrigerators must maintain sub-zero or near-freezing temperatures, their lift is larger than that of comfort cooling systems, thus giving a lower C.O.P. in practice.

Step-by-Step Solution:Estimate for refrigerator: T_L ≈ 268 K, T_H ≈ 308 K → C.O.P. ≈ 268 / 40 ≈ 6.7 (ideal).Estimate for air-conditioner: T_L ≈ 295 K, T_H ≈ 308 K → C.O.P. ≈ 295 / 13 ≈ 22.7 (ideal).Real systems operate at a fraction of these ideal values, but the relative ordering (refrigerator lower than A/C) holds.

Verification / Alternative check:Seasonal efficiency ratings (e.g., EER/SEER vs. refrigerator labels) also reflect easier duty for A/C compared with low-temperature refrigeration.

Why Other Options Are Wrong:“Higher” contradicts the temperature-lift argument; “about the same” ignores the different duty points; “zero” and “always double” are non-physical generalizations.

Common Pitfalls:Comparing nameplate power rather than thermodynamic lift; C.O.P. must be compared at similar reference conditions in Kelvin.

Final Answer:Lower