A freezer has a coefficient of performance of 6.30. The freezer is advertised as
ID: 583494 • Letter: A
Question
A freezer has a coefficient of performance of 6.30. The freezer is advertised as using 470 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?__________ J
(b) On average, how much thermal energy is removed from the freezer each day?_____________ J
(c) What maximum amount of water at 17.0°C could the freezer freeze in a single day? (The latent heat of fusion of water is 3.33 105 J/kg, and the specific heat of water is 4186 J/kg · K.)________ kg
Explanation / Answer
COP = 6.30
COP = heat removed/work
heat removed = COP*work
latent heat of fusion of water (334000 J/kg).
A ) Calculate work for 1 day as 470000*3600/365 = 4635616.438 J.
B) heat removed = COP*work = 6.30 * 4635616.438 = 29204383.56 J
C) Q = mst
29204383.56 = m* 3.33 105 J/kg * 17
m =5.1588 Kg
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