Q:
In a thermally insulated kitchen, a refrigerator with 2 kW motor for running the compressor provides 6000 kJ of cooling to the refrigerated space during 30 min operation. If the condenser coil placed behind the refrigerator rejects 8000 kJ of heat to the kitchen during the same period, calculate the change in internal energy of the kitchen.
Answer & Explanation
Answer: A) 3600 kJ
Explanation: QKitchen = 0 (Insulated!),
W(Electrical) = – P*?? = – 2 kW*30*60 sec = – 3600 kJ
(It is negative because work is done on to the system)
Change in internal energy of the kitchen (?UKitchen) = QKitchen – WElectrical
= 0 – (–3600) = 3600 kJ.
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