The temperatures of inside and outside of a room are 22 degr

The temperatures of inside and outside of a room are 22 degree C and 32 degree C. The conditioner performs 1.00 times 10^4 J work to cool the room. Calculate a) how much heat is taken away from the room (assume that the conditioner has the efficiency of Carno\'s reversed cycle); b) how much energy is transferred to surrounding; and c) if setting the room temperature to 26 degree C and transferring the same amount of heat from inside to outside, how much energy we will save. (7 percentage)

Solution

(a) efficiency of the air conditioner n = 1 - T_cold / T_hot = 0.0328

Work done W = nQ_out = 10^4 J

Q_extracted_from_room = (1-n) Q_out = (1-n)/n * W = T_cold / (T_hot - T_cold) * W = 29.5 X 10^4 J

(b) Q_out = W/n = 30.5 x 10^4 J

(c) new efficiency, n\' = 0.0197

Q_extracted_from_room = T\'_cold / (T_hot - T\'_cold) * W\' = 29.5 X 10^4 J => W\' = 0.59 x 10^4 J

Energy saved = (W - W\') / W x 100 = 41 %