Question

4-10. The latent heat of fusion of ice is 333 kJ/kg. In the freezer compartment of a refrigerator, a tray holds 12 cubes of water each of mass 8.0 g. If the C.O.P. 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) into ice cubes (at 0°C)?

Answer 1

In the refrigerator to transfer heat we need to do external work and for this we need electrical energy to the supplied. That's what is parked here. C.Oop of the refrigerator means co-efficient of performance of the refrigerator = 9 = 3:3 (giren) o where 8 = heat to be extracted for where g cooling w = workdone on the refrigerator for the desired result,

Total mass of water = (8812) g = 96 g out or, m=0.096 kg] (m) The heat extracted from the water at oe to form ice at oºe will be the latent heat only because here temperature is not Changing but phase is changing

Total heat Q = mL L=Latent heat of fersion -0-096 kg x 333 kJ/kg € 333 Kolka = 31.968 *5 Nowe rom o 9.3.3 , 31: 88 e KJ = 9.6873 kJ - 3.3 31.g. 3 3.2

& Workdone required for the case here is = 9.6873 KJ = 9.68734103T @ But, the answer here is asked for Kilowatt - hours of energy (kWh) The kilowatt - hour (kWh) is a unit of energy equivalent to lkw power expanded for 1 hour of time. i 1 kwha 1000 watt x 3600 see ar, IkWw = 36X106 So 6: Wat Joule)

- 13) So, from ② and ③ we can write 9.6873 x 103 5=11 x 9.6873) ku (3.68106 X 103 = 2.6909 410 kWh 0.00 269 kwa ~ 0.0027 kWh 2 8 0.0027 kwh of electrical energy would be required to run the refrigerator to change the water cubes (at poe) into ice cubes (at 0°c)