Question

A refrigerator with an average COP of 2.8 is used to cool a well-insulated container whose contents are equivalent to 7 kg of water from 25oC to 10oC. When running, the refrigerator consumes 600 W of electric power. The length of time the refrigerator must run to accomplish this cooling is and C=4.18 kJ/kg.C A) 6.33 min B) 1.12 min C) 2.23 min D) 4.35 min E) 1.76 min

Answer 1

Givens data : x Average cop 2.8 Equivalent mass (m) = 7kg Temperature difference is written as AT 25°C -10°c OT = 15°C Electic power (W) GooW that We know COP = W Q Copx W Q Q 2.8 X Goo = 1680W Now, the refrigerator effecting q q = 7K9 X 9 18 kJ (150) ДТ mcp at kg.°C q = 438.9 KJ Running time of refrigeraton q 438.9 KJ t = Q. 1680w ta 426.9x10J 1680 J/S 2= 261.25 sec t = 261.25 min 60 t=4.354 min