A freezer has a coefficient of performance of 6.30. The freezer is advertised as

Question

A freezer has a coefficient of performance of 6.30. The freezer is advertised as using 446 kW-h/y. Note: One kilowatt-hour (kW-h) is an amount of energy equal to operating a 1-kW appliance for one hour. (A) On average, how much energy does the freezer use in a single day? (B) On average, how much thermal energy is removed from the freezer each day? (C) What maximum amount of water at 24.0

Explanation / Answer

latent heat of fusion of water (333000 J/kg). COP = heat removed/work ==> heat removed = COP*work Calculate work for 1 day as 446000*3600/365 = 439890.4 J. heat removed = 6.3*439890.4 = 27713095.89 J.

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