How much heat is absorbed by an electric refrigerator in charging 2 Kg of water at 15°C to ice at 0°C?

Answer 1

2000 g *[(1.0 cal/C g)*15C +80 cal/g]
= 2000*95 = 190,000 calories
= 1.65*10^5 Joules
= 46.0 watt-hours
= 0.046 kWh