Question

(a) What is the best coefficient of performance for a refrigerator that cools an environment at -29.5°C and has heat transfer to another environment at 45.5°C? (b) How much work in joules must be done for a heat transfer of 4186 kJ from the cold environment? (c) What is the cost (in cents) of doing this if the work costs 15.0 cents per 3.60 x 106 J (a kilowatt-hour)? (d) How many kJ of heat transfer occurs into the warm environment? kJ

Answer 1

• a) for refrigeraton coefficient of performance (w) = I2' T2 = -29.5+273 Ta 45.5 +273 = 243.5k 318.5k :w = 243.5 318.5-243.5 = 3.2467 Q2 = wow , A. & amount of heat transferred from cold environment 82 = 3.2467*W winork done Given Q = 4186 kJ = 4186 x 103 J. wa 4186 x 103 3 = 1289322.382 J 3.2467 . work done = 1289 322.382 ] > Cost of work a 15 cents per 3.600105 3. : cost of work done = 1289 322. 382 3.6x106 x 15 cents 25-372176591 canh 5.37 28 cents d) Amount of heat transferred to warm environment dia Qq tw 2 4186 kJ + 1289322.382 J 4186 + 1289. 3224) kJ 2 5475.322382 KJ ~ 5475. 322 kJ.