Final answer:
The question concerns thermodynamics and the workings of a refrigerator, focusing on how to calculate the coefficient of performance and the minimum work required for heat extraction. A related real-world application is determining the work needed for a cold vaporizer based on its power consumption and efficiency.
Step-by-step explanation:
The question pertains to the concept of refrigeration and how a refrigerator works in the context of thermodynamics, particularly relating to the calculation of the coefficient of performance (COP) and the work required for heat transfer. Refrigerators and heat pumps operate on the principle of extracting heat from a colder area and releasing it into a warmer one, with the COP being a measure of the efficiency of this process.
Understanding the Coefficient of Performance
The coefficient of performance of a refrigerator is the ratio of the heat extracted from the cold reservoir (Qc) to the work input (W) necessary to transfer that heat. The higher the COP, the more efficient the refrigerator is. For example, if a refrigerator has a COP of 3.0 and requires 200 J of work per cycle, it means it can remove 600 J of heat from the cold reservoir because COP = Qc/W.
Calculating the Minimum Work Required
To calculate the minimum work required for a refrigerator to extract a certain amount of heat per cycle from the inside of a freezer, one must understand the temperatures involved. Using the Kelvin scale for temperature conversions and understanding the second law of thermodynamics allows for such calculations. For instance, extracting 50 J per cycle from a freezer at -10 °C (263 K) to the surrounding air at 25 °C (298 K) would require a specific amount of work that can be assessed by applying the concepts of thermodynamics and the COP formula.
Real-World Application
In a real-world application, such as determining how much work is needed for a home device like a cold vaporizer rated at 3.50 A and 120 V AC, the efficiency of the device must be accounted for. The device would have a certain vaporization rate in grams per minute and knowing the energy consumption, one can determine how much water is required for operation over a given time, such as 8 hours overnight.