Final answer:
To freeze 0.27 kg of water at 20°C, 22.572 kJ of energy must be removed to cool it to 0°C, and an additional 90.18 kJ to freeze it, totalling 112.752 kJ of energy required.
Step-by-step explanation:
To calculate the energy that must be removed from 0.27 kg of water at 20°C to freeze it, we need to perform two main calculations: the energy required to cool the water from 20°C to 0°C, and the energy required to change the state of the water from liquid to solid at 0°C. The specific heat capacity of water is typically about 4.18 kJ/kg°C for the cooling part, and the latent heat of fusion for water is about 334 kJ/kg for the phase change. The first step involves multiplying the mass of the water by the specific heat capacity and the temperature change (20°C). Then, we multiply the mass by the latent heat of fusion.
- Energy to cool water to 0°C: E_cooling = mass × specific heat capacity × temperature change = 0.27 kg × 4.18 kJ/kg°C × (20°C) = 22.572 kJ
- Energy to freeze water at 0°C: E_freezing = mass × latent heat of fusion = 0.27 kg × 334 kJ/kg = 90.18 kJ
The total energy that must be removed is the sum of E_cooling and E_freezing, which is 22.572 kJ + 90.18 kJ = 112.752 kJ.