Subtopic: Preservation of food by cooling and freezing
Question Statement:
A household refrigerator of 1 TR capacity operates half the time during 13-hour long days and 30% time during the nights. COP = 4.7, Rs. 3 per kWh, monthly (30 days) electricity bill in Rupees for the refrigerator is
Options:
(A) 110
(B) 220
(C) 440
(D) 660
Official Answer Key:
D
Step-by-Step Solution & Explanation
1 TR = 3.517 kW (cooling load). Compressor power drawn = Q_c/COP = 3.517/4.7 = 0.748 kW. Day = 13 h, running half-time = 6.5 h. Night = 24−13 = 11 h, running 30% = 3.3 h. Total run-time/day = 9.8 h. Energy/day = 0.748 × 9.8 = 7.33 kWh. Energy/month = 7.33 × 30 = 220.0 kWh. Bill = 220.0 × 3 = Rs. 660.
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