A refrigerator extracts 600 J from the cold compartment and rejects 800 J to the room. Its coefficient of performance is:
A0.25
B0.75
C3.0
D4.0
Answer & Solution
Correct answer: C. 3.0
1. Work done on the refrigerant W = Q₂ − Q₁ = 800 − 600 = 200 J.
2. COP for a refrigerator = Q₁ / W (heat removed per joule of work).
3. COP = 600 / 200 = 3.0.
4. Higher COP means better refrigerator. Real fridges: COP ≈ 2–4.
_Source: NCERT Class 11 Physics Ch 11, §11.11 Refrigerators and Heat Pumps_
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