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A Carnot air conditioner maintains the temperature in a house at 297 \(\mathrm{K}\) on a day when the temperature outside is 311 \(\mathrm{K}\) . What is the coefficient of performance of the air conditioner?

Short Answer

Expert verified
The coefficient of performance (COP) is approximately 21.21.

Step by step solution

01

Understand the Concepts

The problem asks for the coefficient of performance (COP) of a Carnot air conditioner. The COP for a refrigerator or air conditioner is defined as the ratio of the heat removed from the cold reservoir (house) to the work done by the system.
02

Know the Formula for COP of Carnot Air Conditioner

The coefficient of performance (COP) of a Carnot air conditioning system is given by the formula: \[COP = \frac{T_c}{T_h - T_c}\] where \(T_c\) is the temperature of the cold reservoir (inside the house) in Kelvin, and \(T_h\) is the temperature of the hot reservoir (outside) in Kelvin.
03

Assign Known Values

From the problem, we know:\(T_c = 297 \ \mathrm{K}\) (inside the house) \(T_h = 311 \ \mathrm{K}\) (outside the house)
04

Substitute Values into the COP Formula

Substitute the given temperatures into the formula: \[COP = \frac{297}{311 - 297} = \frac{297}{14}\]
05

Calculate the COP

Calculate the value of the COP using the substituted values:\[COP = \frac{297}{14} \approx 21.21\]

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Key Concepts

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

Coefficient of Performance (COP)
The Coefficient of Performance (COP) is a crucial concept in evaluating the efficiency of refrigeration and air conditioning systems. It's defined as the ratio of heat removed from the cold space to the work input needed.
This allows us to determine how efficient a system is at cooling or heating per unit of energy input.
In simpler terms, COP gives us an idea of how much cooling you get versus how much energy you put in to get that cooling. For a Carnot cycle air conditioner, the COP is given by the formula:
  • \( COP = \frac{T_c}{T_h - T_c} \)
where:
  • \( T_c \) is the temperature of the cold reservoir (inside the area you want cool, like in a house).
  • \( T_h \) is the temperature of the hot reservoir (outside the environment, like the air outside).
The higher the COP, the more efficient the system. This means more cooling with less energy. This measure is particularly important for understanding how economical a refrigeration system would be in terms of energy use.
Thermodynamics
Thermodynamics is the science of heat and energy transfer. It explores how energy moves and changes from one form to another in a system. An essential pillar of thermodynamics is the study of Carnot cycles, which embody the ideal behavior of refrigerators or air conditioners. The Carnot cycle, an idealized model, operates between two heat reservoirs:
  • The hot reservoir, which is the outside temperature in the case of an air conditioner.
  • The cold reservoir, which refers to the target indoor temperature.
Understanding these reservoirs clarifies why we use Kelvin in thermodynamic calculations. Kelvin allows precise and absolute measurement of thermal energy since it begins at absolute zero. Moreover, the second law of thermodynamics indicates that energy will always flow from a hotter to a cooler area. In air conditioning systems, energy is used to reverse this natural flow, pushing heat from inside a cooler indoor area to the warmer outdoors.
Heat Transfer
Heat transfer is the essential process of moving thermal energy from one place to another. This movement is a critical function in the operation of air conditioners and refrigerators. The three primary methods of heat transfer are:
  • Conduction: Transfer of heat through a material, such as the walls of your home.
  • Convection: Transfer of heat through fluids like air; for example, when air circulates inside a room being cooled.
  • Radiation: Transfer of heat through electromagnetic waves, although it is less significant in air conditioning systems.
In an air conditioner, focus is usually on how efficiently it can remove heat from the indoor air and discharge it outside. This involves the practical application of thermodynamic principles. By understanding heat transfer, we gain insight into the design and operation of systems focused on regulating temperatures within a specific space. Efficient heat transfer processes lead to better energy use and system efficiency.

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Most popular questions from this chapter

The internal energy of a system changes because the system gains 165 \(\mathrm{J}\) of heat and performs 312 \(\mathrm{J}\) of work. In returning to its initial state, the system loses 114 \(\mathrm{J}\) of heat. During this return process, (a) what work is involved, and \((\mathrm{b})\) is the work done by the system or on the system?

When a .22 -caliber rifle is fired, the expanding gas from the burning gunpowder creates a pressure behind the bullet. This pressure causes the force that pushes the bullet through the barrel. The barrel has a length of 0.61 \(\mathrm{m}\) and an opening whose radius is \(2.8 \times 10^{-3} \mathrm{m} . \mathrm{A}\) bullet (mass \(=2.6 \times 10^{-3} \mathrm{kg}\) ) has a speed of 370 \(\mathrm{m} / \mathrm{s}\) after passing through this barrel. Ignore friction and determine the average pressure of the expanding gas.

Six grams of helium (molecular mass \(=4.0 \mathrm{u} )\) expand isothermally at 370 \(\mathrm{K}\) and does 9600 \(\mathrm{J}\) of work. Assuming that helium is an ideal gas, determine the ratio of the final volume of the gas to the initial volume.

Heat is added to two identical samples of a monatomic ideal gas. In the first sample the heat is added while the volume of the gas is kept constant, and the heat causes the temperature to rise by 75 \(\mathrm{K}\) . In the second sample, an identical amount of heat is added while the pressure (but not the volume) of the gas is kept constant. By how much does the temperature of this sample increase?

A system gains 1500 \(\mathrm{J}\) of heat, while the internal energy of the system increases by 4500 \(\mathrm{J}\) and the volume decreases by 0.010 \(\mathrm{m}^{3}\) . Assume that the pressure is constant and find its value.

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