the latent heat of fusion of ice is 333kJ/kg. in the freezer compartment of a re
ID: 582658 • Letter: T
Question
the latent heat of fusion of ice is 333kJ/kg. in the freezer compartment of a refrigerator, a tray of ice holds 12 cubes of water each of mass 8.0 g. if the COP of the refrigerator is 3.3, how many kilowatt-hours of electrical energy would be required to run the refrigerator to change the water cubes (at 0°C) lnto ice cubes (at 0°C). the latent heat of fusion of ice is 333kJ/kg. in the freezer compartment of a refrigerator, a tray of ice holds 12 cubes of water each of mass 8.0 g. if the COP of the refrigerator is 3.3, how many kilowatt-hours of electrical energy would be required to run the refrigerator to change the water cubes (at 0°C) lnto ice cubes (at 0°C). the latent heat of fusion of ice is 333kJ/kg. in the freezer compartment of a refrigerator, a tray of ice holds 12 cubes of water each of mass 8.0 g. if the COP of the refrigerator is 3.3, how many kilowatt-hours of electrical energy would be required to run the refrigerator to change the water cubes (at 0°C) lnto ice cubes (at 0°C).Explanation / Answer
Total mass =m = 12*8 = 96g = 0.096 kg
latent heat of fusion = L = 333kJ/Kg
T2 = 0 degree celcius = 273 K
Q2 = mL = 0.096*333 = 31.968 KJ
COP = Q2 / W
W = Q2 / COP
= 31.968 / 3.3
= 9.687 KJ
= 9.687 /3600 (1KW-Hours = 3600KJ)
= 2.690 * 10^-3 kw-hr
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