Question

660°C]? 2- How much heat must be removed by a refrigerator from 2 kg of water at 70 °C to convert it to ice cubes at -11°C? [Specific heat capacity of water = 4200J kg°C ; Specific latent heat of fusion of ice = 334,000 Jkg?, specific heat capacity of ice = 2100 J/(kg K)] = -1 - 9​

Answer 1

1302200J

Step-by-step explanation:

To bring water at 70°C to water at 0°C, we use MC∆T.

To convert water at 0°C to ice at 0°C, ML is applicable, where L is the latent heat of fusion and M equals mass. NB: Temperature is constant during fusion.

Again, to raise the temperature of ice to -11°c, we use MC∆T.

Total heat = MCw∆T + ML + MCi∆T

2 ×4200× 70 + 2×334000 + 2× 2100× 11 = 1302200J