1.Estimate the change in COP of your electric food freezer when it is removed fr

Question

1.Estimate the change in COP of your electric food freezer when it is removed from your kitchen to its new location in your basement, which is 8 C° cooler than your kitchen. (Assume that the temperature in your kitchen is 20°C and that the temperature of the interior of your freezer is -5°C.)

2.You are installing a heat pump whose COP is half the COP of a reversible heat pump. You plan to use the pump on chilly winter nights to increase the air temperature in your bedroom. Your bedroom's dimensions are 4.60 m 3.4 m 2.8 m. The air temperature should increase from 63° F to 68° F. The outside temperature is 35° F, and the temperature at the air handler in the room is 112° F. If the pump's electric-power consumption is 675 W, how long will you have to wait for the room's air to warm if the specific heat of air is 1.005 kJ/(kg · C°)? Assume you have good window draperies and good wall insulation so that you can neglect the release of heat through windows, walls, ceilings, and floors. Also assume that the heat capacity of the floor, ceiling, walls, and furniture are negligible. (Assume 1.293 kg/m3 for the density of air.)

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Explanation / Answer

COP= Tcold/(Thot - Tcold)

COP of freezer in kitchen = (-5+273)/(20 - (-5)) = 10.72

COP of freezer in basement = (-5+273)/(12 - (-5)) = 15.76

Change = 15.76 - 10.72 = 5.04

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