A freezer with coefficient of performance 2.4 is used to convert 1.8 kg of water at 25 C to ice at -5 C. How much electrical energy is required?
For water Cw = 4190 J/Kg*K, Lf = 3.34 × 10^5 J/Kg & Cice = 2010 J/Kg*K
1 Answer | Add Yours
The freezer has to convert 1.8 kg of water at 25 C to ice at -5 C. This involves three processes; first the temperature of water has to be reduced from 25 C to 0 C. Additional heat has to be removed to allow the change of phase from liquid to solid. Finally the ice has to be cooled from 0 C to -5 C.
The heat of fusion of water is Lf = 3.34*10^5 J/kg, the specific heat of water is Cw = 4190 J/kg*K and the specific heat of ice Cice = 2010 J/kg*K.
Using the information provided, the heat that has to be removed from the 1.8 kg of water initially at 25 C is:
1.8*Cw*25 + 1.8*Lf + 1.8*Cice*5
=> 1.8*4190*25 + 1.8*3.34*10^5 + 1.8*2010*5
=> 188550+ 601200+ 18090
=> 807840 J
The freezer has a coefficient of performance of 2.4. This implies that the amount of heat removed from the cold junction is 2.4 times the input work. As the amount of heat to be removed is 807840 J, the electrical energy consumed is 336,600 J.
We’ve answered 319,865 questions. We can answer yours, too.Ask a question