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Given: T2 = 49°C = 4 × 273 = 277 K, T1 = 30°C = 30 + 273 = 303 K.

1. β = Q2/W = T2/(T1 − T2)
2. β = 600 × 4.2/303 − 277
3. β = W = 236.5 joule
4. Power P = W/t = 236.5 joule/1 sec = 236.5 watt.

Correct answer: β = Q2/W = T2/(T1 − T2)

Solution

Option A is correct because the formula for the coefficient of performance (β) of a refrigerator is given by β = Q2/W = T2/(T1 − T2), where T1 and T2 are the temperatures of the hot and cold reservoirs, respectively.

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