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A homeowner buys a new refrigerator and a new air conditioner. Which one of these devices would you expect to have a higher COP? Why?

Short Answer

Expert verified
Answer: An air conditioner generally has a higher COP compared to a refrigerator due to its relatively smaller temperature difference between the cold and warm reservoirs within which it operates.

Step by step solution

01

Understanding the operating conditions of both devices

Refrigerators and air conditioners have similar functions, as they both transfer heat from a colder to a warmer environment. However, their operating conditions differ. A refrigerator operates at moderately low temperatures (usually between 0°C to 10°C) to preserve food and prevent bacteria growth, while an air conditioner operates at higher temperatures, usually around 20°C to 30°C, to provide comfort to the occupants of the room.
02

Comparing the temperature difference between both devices

The Coefficient of Performance (COP) for both devices depends on the difference in temperature between the colder and warmer environments. When the temperature difference is smaller, the COP is higher, indicating that the device is more efficient in transferring heat. In this case, the temperature difference between the cold and warm reservoirs would be smaller for an air conditioner compared to a refrigerator.
03

Analyzing the efficiency factors for both devices

Both refrigerators and air conditioners usually use the same principle of operation, i.e., a vapor-compression refrigeration cycle for heat transfer. Therefore, the difference in their efficiencies depends on their operating temperature ranges rather than the mechanism itself. As the air conditioner operates within a smaller temperature difference, it would be expected to have a relatively higher efficiency, and therefore, a higher COP.
04

Conclusion

Based on the comparison of operating conditions and the temperature difference, we can conclude that the air conditioner would have a higher COP compared to the refrigerator. It would be more efficient in transferring heat due to the relatively smaller temperature difference between the cold and warm reservoirs within which it operates.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vapor-Compression Refrigeration Cycle
The vapor-compression refrigeration cycle is a concept crucial to understanding the workings of refrigerators and air conditioners. This cycle consists of four primary stages: evaporation, compression, condensation, and expansion.

In the evaporation stage, a low-pressure refrigerant absorbs heat from the interior of the refrigerator or the room that is to be cooled, turning the refrigerant from a liquid to a gas. During compression, this gas is pressurized by a compressor, which raises its temperature. The high-pressure, high-temperature gas then releases its heat in the condenser, typically located at the back of the refrigerator or on the exterior side of an air conditioner, transitioning back into a liquid state. Lastly, the refrigerant passes through an expansion device, which reduces its pressure, cooling it significantly and allowing the cycle to repeat.

Understanding this cycle is essential because the efficiency of both devices is influenced by how effectively they can carry out this process, with lower energy input for a given amount of heat transfer leading to a higher COP.
Refrigerator Efficiency
The efficiency of a refrigerator is often gauged by its Coefficient of Performance (COP). The COP measures the ratio of useful heating or cooling provided to the energy consumed by the system. Refrigerators are designed to maintain very low temperatures to preserve food, which typically creates a larger temperature difference between the inside of the refrigerator and the ambient conditions outside it. As the COP is inversely related to this temperature difference, refrigerators tend to have a lower COP compared to other cooling devices, such as air conditioners, which operate at relatively milder temperatures.

Another aspect influencing refrigerator efficiency is the insulation quality and door seal, as these affect the amount of heat that leaks into the cold space from the surroundings, thereby increasing the workload of the refrigeration cycle.
Air Conditioner Efficiency
Air conditioner efficiency, similar to refrigerators, can be evaluated using the COP. Air conditioners typically operate over smaller temperature differences as they aim to cool the air to comfortable levels rather than preserve perishable goods. This smaller temperature differential results in a higher COP because less work is needed to transfer heat. Furthermore, air conditioners are often equipped with features like variable speed fans and compressors, which allow them to adjust their operation based on cooling demand, enhancing their efficiency.

It's also important to note that the seasonal energy efficiency ratio (SEER) is another metric used to assess air conditioner efficiency over a cooling season and provides a broader view of its performance under varying conditions.
Heat Transfer Thermodynamics
Heat transfer thermodynamics is a field of study that focuses on how heat moves from one system to another and the energy conversions that accompany this process. Understanding these principles is key to analyzing the efficiency of refrigeration cycles in devices like refrigerators and air conditioners.

The first law of thermodynamics, or the conservation of energy, plays a crucial role here, stating that energy cannot be created or destroyed, only transferred. This law underpins the principle that the energy removed from the cool space must be equal to the energy expelled to the surroundings plus the work input to the refrigeration cycle. Another vital principle is the second law of thermodynamics, which dictates that heat naturally flows from hot to cold bodies. Refrigerators and air conditioners essentially reverse this natural flow through work input, which adds context to the COP's significance as it measures the effectiveness of this reversal process in terms of energy consumed versus heat transferred.

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Most popular questions from this chapter

\(6-76 \quad\) A Carnot heat engine receives \(650 \mathrm{kJ}\) of heat from a source of unknown temperature and rejects \(250 \mathrm{kJ}\) of it to a \(\operatorname{sink}\) at \(24^{\circ} \mathrm{C} .\) Determine \((a)\) the temperature of the source and (b) the thermal efficiency of the heat engine.

A cold canned drink is left in a warmer room where its temperature rises as a result of heat transfer. Is this a reversible process? Explain.

An inventor claims to have devised a cyclical engine for use in space vehicles that operates with a nuclear-fuel-generated energy source whose temperature is \(920 \mathrm{R}\) and a sink at \(490 \mathrm{R}\) that radiates waste heat to deep space. He also claims that this engine produces 4.5 hp while rejecting heat at a rate of \(15,000 \mathrm{Btu} / \mathrm{h}\). Is this claim valid?

In an effort to conserve energy in a heat-engine cycle, somebody suggests incorporating a refrigerator that will absorb some of the waste energy \(Q_{L}\) and transfer it to the energy source of the heat engine. Is this a smart idea? Explain.

It is often stated that the refrigerator door should be opened as few times as possible for the shortest duration of time to save energy. Consider a household refrigerator whose interior volume is \(0.9 \mathrm{m}^{3}\) and average internal temperature is \(4^{\circ} \mathrm{C} .\) At any given time, one-third of the refrigerated space is occupied by food items, and the remaining \(0.6 \mathrm{m}^{3}\) is filled with air. The average temperature and pressure in the kitchen are \(20^{\circ} \mathrm{C}\) and \(95 \mathrm{kPa}\), respectively. Also, the moisture contents of the air in the kitchen and the refrigerator are 0.010 and \(0.004 \mathrm{kg}\) per \(\mathrm{kg}\) of air, respectively, and thus \(0.006 \mathrm{kg}\) of water vapor is condensed and removed for each kg of air that enters. The refrigerator door is opened an average of 20 times a day, and each time half of the air volume in the refrigerator is replaced by the warmer kitchen air. If the refrigerator has a coefficient of performance of 1.4 and the cost of electricity is 11.5 cents per \(\mathrm{kWh}\), determine the cost of the energy wasted per year as a result of opening the refrigerator door. What would your answer be if the kitchen air were very dry and thus a negligible amount of water vapor condensed in the refrigerator?

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