To determine the amount of heat that must be removed by the refrigerator to freeze the water in the 2L bottle, we need to consider the specific heat capacity and the latent heat of fusion of water.
- Heating the water from 20°C to 0°C: The specific heat capacity of water is approximately 4.18 J/g°C. Since we have 2,000 grams of water (2L), the heat required to cool it from 20°C to 0°C can be calculated using the formula:
Heat = mass × specific heat capacity × temperature change
Heat = 2,000 g × 4.18 J/g°C × (0°C - 20°C) = -83,600 J (Negative sign indicates heat removal)
- Freezing the water at 0°C: The latent heat of fusion of water is approximately 334,000 J/kg. Since we have 2,000 grams of water (2 kg), the heat required to freeze it at 0°C is given by:
Heat = mass × latent heat of fusion
Heat = 2,000 g × 334,000 J/kg = 668,000 J
Therefore, the total heat that must be removed by the refrigerator to freeze the water is the sum of the heat required to cool the water to 0°C and the heat required for freezing:
Total Heat = -83,600 J + 668,000 J = 584,400 J
Hence, approximately 584,400 Joules of heat must be removed by the refrigerator to freeze the 2L bottle of water starting at 20°C.