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Go Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms, \(309.0 \mathrm{K}\). The room serviced by unit \(A\) is kept at a temperature of \(294.0 \mathrm{K},\) while the room serviced by unit \(\mathrm{B}\) is kept at \(301.0 \mathrm{K}\). The heat removed from either room is 4330 J. For both units, find the magnitude of the work required and the magnitude of the heat deposited outside.

Short Answer

Expert verified
Unit A requires less work and deposits less heat outside compared to Unit B.

Step by step solution

01

Understanding the Carnot Cycle

A Carnot air conditioner operates on a reversible cycle and it is the most efficient cycle possible between two temperatures. The efficiency or coefficient of performance (COP) of a Carnot refrigerator (air conditioner) is given by the formula: \( \text{COP} = \frac{T_c}{T_h - T_c} \), where \( T_c \) is the cold reservoir temperature and \( T_h \) is the hot reservoir temperature.
02

Calculating the COP for Unit A

For unit A, the cold temperature \( T_c = 294.0 \text{ K} \) and the hot temperature \( T_h = 309.0 \text{ K} \). Substitute these values into the COP formula to find:\[\text{COP}_A = \frac{294.0}{309.0 - 294.0} = \frac{294.0}{15.0}\].
03

Calculating Work for Unit A

The work done by an air conditioner is given by \( W = \frac{Q_c}{\text{COP}} \), where \( Q_c \) is the heat removed from the cold reservoir (room). For unit A, \( Q_c = 4330 \text{ J} \). Substitute the values: \[W_A = \frac{4330}{\frac{294.0}{15.0}} = \frac{4330 \times 15}{294.0} \].Calculate the result to find the work done \( W_A \) for unit A.
04

Calculating Heat Deposited Outside for Unit A

The heat rejected to the outside, \( Q_h \), is given by \( Q_h = Q_c + W \). For unit A, substitute the values: \[Q_{hA} = 4330 + W_A\].Use the previously calculated value of \( W_A \) to find \( Q_{hA} \).
05

Calculating the COP for Unit B

For unit B, the cold temperature \( T_c = 301.0 \text{ K} \) and the hot temperature \( T_h = 309.0 \text{ K} \). Use the COP formula:\[\text{COP}_B = \frac{301.0}{309.0 - 301.0} = \frac{301.0}{8.0}\].
06

Calculating Work for Unit B

Use the formula \( W = \frac{Q_c}{\text{COP}} \) for unit B, where \( Q_c = 4330 \text{ J} \): \[W_B = \frac{4330}{\frac{301.0}{8.0}} = \frac{4330 \times 8}{301.0}\].Calculate to find the work done \( W_B \) for unit B.
07

Calculating Heat Deposited Outside for Unit B

Use the formula \( Q_h = Q_c + W \) for unit B: \[Q_{hB} = 4330 + W_B\].Substitute the previously calculated \( W_B \) to find \( Q_{hB} \).

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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 measure when evaluating the efficiency of thermodynamic systems like air conditioners and refrigerators. It can be understood as an indicator of how well a device uses energy to move heat around. For heat pumps and air conditioners, COP represents the ratio of the heat removed from the cold reservoir (inside your house) compared to the work required to remove that heat.

Mathematically, the COP for a Carnot air conditioner is given by the formula:
  • \[ \text{COP} = \frac{T_c}{T_h - T_c} \] where:
    • \(T_c\) is the temperature of the cold reservoir.
    • \(T_h\) is the temperature of the hot reservoir.
Its value is typically higher for systems operating between temperatures that are close to each other, as less work is needed to transfer the heat. A higher COP indicates a more efficient cooling cycle which is generally desired for cost-effective and sustainable energy use.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat and temperature, and their relation to energy and work. It is fundamental in explaining how thermal systems like air conditioners operate. There are four main laws in thermodynamics, but air conditioners mainly involve the second law, which states that heat will naturally flow from hot to cold unless energy is expended to move it in the opposite direction.

In the context of air conditioning, devices exploit the principles of the second law to perform work so heat is transferred against its natural direction, from a cooler interior to a warmer exterior, utilizing cycles like the Carnot cycle for maximum efficiency. The Carnot cycle is an idealized thermodynamic cycle that defines the maximum possible efficiency any real engine or heat pump can achieve during the heat transfer process between two reservoirs, making it a crucial benchmark in heating and cooling technology.
Heat Transfer
Heat transfer is the movement of thermal energy from one thing to another due to a temperature difference. This process is essential in the operation of air conditioners, which execute heat transfer from indoor spaces to the external environment.

There are three primary modes of heat transfer:
  • Conduction: Direct transfer of heat through materials touching each other.
  • Convection: Transfer of heat by the movement of fluids or gases, such as the coolant in an air conditioner cycle.
  • Radiation: Transfer of heat through electromagnetic waves, which plays less of a direct role in air conditioning but is crucial in many other thermal processes.
Effective heat transfer, primarily through convection in the context of air conditioning, means more of the heat from indoors is pushed outside, relying on the flow of refrigerant to achieve this process efficiently. Understanding these principles helps enhance the design and efficiency of thermal systems, ensuring comfortable temperatures while minimizing energy expenditure.

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Most popular questions from this chapter

Consider three engines that each use \(1650 \mathrm{J}\) of heat from a hot reservoir (temperature \(=550 \mathrm{K}\) ). These three engines reject heat to a cold reservoir (temperature \(=330 \mathrm{K}\) ). Engine I rejects \(1120 \mathrm{J}\) of heat. Engine II rejects \(990 \mathrm{J}\) of heat. Engine III rejects \(660 \mathrm{J}\) of heat. One of the engines operates reversibly, and two operate irreversibly. However, of the two irreversible engines, one violates the second law of thermodynamics and could not exist. For each of the engines determine the total entropy change of the universe, which is the sum of the entropy changes of the hot and cold reservoirs. On the basis of your calculations, identify which engine operates reversibly, which operates irreversibly and could exist, and which operates irreversibly and could not exist.

A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignite the fuel occurs because the pistons compress the air in the cylinders. Suppose that air at an initial temperature of \(21^{\circ} \mathrm{C}\) is compressed adiabatically to a temperature of \(688^{\circ} \mathrm{C} .\) Assume the air to be an ideal gas for which \(\gamma=\frac{7}{5} .\) Find the compression ratio, which is the ratio of the initial volume to the final volume.

A Carnot engine has an efficiency of 0.40. The Kelvin temperature of its hot reservoir is quadrupled, and the Kelvin temperature of its cold reservoir is doubled. What is the efficiency that results from these changes?

The hot reservoir for a Carnot engine has a temperature of \(890 \mathrm{K}\), while the cold reservoir has a temperature of \(670 \mathrm{K}\). The heat input for this engine is 4800 J. The \(670-\mathrm{K}\) reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of \(420 \mathrm{K}\). Find the total work delivered by the two engines.

Argon is a monatomic gas whose atomic mass is 39.9 u. The temperature of eight grams of argon is raised by \(75 \mathrm{K}\) under conditions of constant pressure. Assuming that argon behaves as an ideal gas, how much heat is required?

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