A refrigerator works between 4 and 30 . It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take 1 cal = 4.2 Joules)
- 236.5 W
- 2365 W
- 2.365 W
- 23.65 W
Answer
(1) 236.5 W
Solution
T2 = 273 + 4 = 277 K, T1 = 273 + 30 = 303 K
Coefficient of performance: = Q2W = T2T1 - T2 = 27726
Heat removed per second: Q2 = 600 4.2 = 2520 J s-1
⇒ W = Q2 T1 - T2T2 = 2520 26277
∴ P = 236.5 W
Answer: (1)