When run backward, a Carnot cycle provides an ideal refrigerator: the cycle extr

Question

When run backward, a Carnot cycle provides an ideal refrigerator: the cycle extracts energy from the cold reservoir and dumps it into the hot reservoir. Work from an external source is required to run the refrigerator, and the energy associated with that work is also dumped into the hot reservoir. A room air conditioner is fundamentally a refrigerator.
(a) The temperature inside the room is Tinside=25oC, and the temperature outside the house is Toutside=32oC. The temperature difference causes energy to flow into the room (by conduction through the walls and window glass) at the rate of 3000J/s. To return this enrgy to the outside by running an ideal refrigerator, how much electrical energy must be supplied to the refrigerator (to perform the external work)?

(b) If the outside temperature grows to 39oC, so that T=Toutside-Tinside doubles, by what factor must the supply of electrical energy increase? Take the inflow of energy to be proportional to T.

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Explanation / Answer

Coeffient of performance K =    TL /    ( TH -  TL )   Here lower temperature    TL   = 25 + 273   =   298 K heigher temerature TH     = 32 + 273 =   305 K K   = 42.57 K = QC /   W     =  H t /   P  t     =     H /P Here   H =   QC   / t    heat removed in time t    = 3000 j/s given electrical energy suppied in time t Pold    =   H / K    = 3000 j/s   * 42.57 = 70.47 j /s --------------------------------- If   ( TH -  TL )     = 14 K   = 21.28 given electrical energy suppied in time t PNew   = 3000 j/s *   21.28   = 140. 97 j/s PNew     =   2*Pold  

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