A refrigerator removes 400 J of heat from its interior per cycle and consumes 100 J of electrical work. (a) Find its COP. (b) How much heat is dumped into the kitchen per cycle? (c) If it were a Carnot refrigerator between 280 K (inside) and 300 K (kitchen), what would its maximum COP be? Solution: (a) COP = Q2/W = 400/100 = 4.0. (b) Q1 = Q2 + W = 500 J dumped into the kitchen - the fridge heats the room. (c) COP_Carnot = T2/(T1 - T2) = 280/(300 - 280) = 14. Real fridges achieve far less than the Carnot ideal because of friction and finite-temperature heat transfer.