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
as a Carnot engine, the specific heat of ice is 2.09 kJ/kg ·C and the specific heat of wateris 4.186 kJ/kg ·C.
Answer in units of kJ
i have no idea how to do this please help
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.