A mass of 4.77 kg of chilled water at 0�C is placed in a freezer which is kept at −19.2�C. Approximately how much electric energy is needed to operate the compressor to cool this water to −19.2 �C if the room temperature is maintained at 24.2�C? The freezer works

To solve this problem, we need to determine the amount of energy required to cool the water from its initial temperature of 0°C to the final temperature of -19.2°C. We can calculate this energy by considering the heat transfer and the change in temperature.

The first step is to find the energy required to cool the water from 0°C to the freezing point of water (0°C). We can use the specific heat formula:

Q1 = mass * specific heat of water * change in temperature

Where:
- Q1 is the energy required to cool the water from 0°C to the freezing point
- mass is the mass of water (4.77 kg)
- specific heat of water is 4.186 kJ/kg·°C
- change in temperature is the difference between the initial temperature (0°C) and the freezing point (0°C)

Q1 = 4.77 kg * 4.186 kJ/kg·°C * (0°C - 0°C)

Since the change in temperature is zero, there is no energy required to cool the water from 0°C to the freezing point.

The second step is to find the energy required to freeze the water. When water freezes, it releases a certain amount of energy called the latent heat of fusion. We can calculate this energy using the formula:

Q2 = mass * latent heat of fusion of water

Where:
- Q2 is the energy required to freeze the water
- mass is the mass of water (4.77 kg)
- latent heat of fusion of water is the heat released when 1 kg of water freezes, which is 334 kJ/kg

Q2 = 4.77 kg * 334 kJ/kg

Next, we need to find the energy required to lower the temperature of the frozen water from its freezing point to the final temperature of -19.2°C. Since the water is already in solid form, we can use the specific heat of ice in this calculation.

Q3 = mass * specific heat of ice * change in temperature

Where:
- Q3 is the energy required to lower the temperature of the frozen water
- mass is the mass of water (4.77 kg)
- specific heat of ice is 2.09 kJ/kg·°C
- change in temperature is the difference between the freezing point (0°C) and the final temperature (-19.2°C)

Q3 = 4.77 kg * 2.09 kJ/kg·°C * (-19.2°C - 0°C)

Finally, the total energy required to cool the water to -19.2°C is obtained by adding the three calculated quantities:

Total Energy = Q1 + Q2 + Q3

Now you can substitute the values into the formulas and calculate the total energy required in kJ.