Part A A freezer has a coefficient of performance of 2.58. The freezer is to con

Question

Part A A freezer has a coefficient of performance of 2.58. The freezer is to convert 1.98 kg of water at 27.0 C to 1.98 kg of ice at -5.00 ° C in 1 hour. What amount of heat must be removed from the water at 27.0°C to convert it to ice at-5.00 °C ? Submit My Answers Give Up Part B How much electrical energy is consumed by the freezer during this hour? Submit My Answers Give Up Part C How much wasted heat is rejected to the room in which the freezer sits? Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

given COP of freezer = 2.58 = Qc/W

Qc = heat removed from cold reservoir

W = work required by the system

m = 1.98 kg water

Ti = 27 C

Tf = -5 C ice

so heat removed form water = mCTi + mL + mc(0 - Tf)

where C is heat capacity of water = 4184 J/kg K

c is heat capacity of ice = 2108 J/Kg K

L = latent heat of fusion of water = 334,000 J/ kg

hence

a. Qc = m[CTi + L + c(0 - Tf)] = 1.98[4184*27 + 334000 + 2108*5] = 905865.84 J

b. 2.58 = Qc/W

W = 351110.790 J

c. heat rejected = Qh = Qc + W = 1256976.630 J

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