91Ó°ÊÓ

Two Camot air conditioners, \(\mathrm{A}\) and \(\mathrm{B}\), are removing heat from different rooms. The outside temperature is the same for both, but the room temperatures are different. The room serviced by unit \(\mathrm{A}\) is kept colder than the room serviced by unit B. The heat removed from both rooms is the same, (a) Which unit requires the greater amount of work? (b) Which unit deposits the greater amount of heat outside? Explain. The outside temperature is \(309.0 \mathrm{~K}\). The room serviced by unit \(\mathrm{A}\) 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 \(4330 \mathrm{~J}\). For both units, find the magnitude of the work required and the magnitude of the heat deposited outside. Verify that your answers are consistent with your answers to the Concept Questions.

Step by step solution

01

Identify the Carnot Efficiency for Each Unit

For a Carnot air conditioner, the coefficient of performance (COP) is given by \[ \text{COP} = \frac{T_\text{cold}}{T_\text{hot} - T_\text{cold}} \]where \( T_\text{cold} \) is the temperature of the room and \( T_\text{hot} \) is the outside temperature.For Unit A: \[ \text{COP}_A = \frac{294.0}{309.0 - 294.0} = \frac{294.0}{15.0} = 19.6 \]For Unit B:\[ \text{COP}_B = \frac{301.0}{309.0 - 301.0} = \frac{301.0}{8.0} = 37.625 \]

02

Calculate Work Done by Each Unit

The work done \( W \) by a Carnot refrigerator is related to the heat removed \( Q_\text{cold} \) and COP by:\[ W = \frac{Q_\text{cold}}{\text{COP}} \]Given \( Q_\text{cold} = 4330 \mathrm{~J} \) for both units:For Unit A:\[ W_A = \frac{4330}{19.6} = 220.92 \mathrm{~J} \]For Unit B:\[ W_B = \frac{4330}{37.625} = 115.05 \mathrm{~J} \]

03

Determine the Heat Deposited Outside for Each Unit

The heat deposited outside, \( Q_\text{hot} \), is the sum of the heat removed and the work done:\[ Q_\text{hot} = Q_\text{cold} + W \]For Unit A:\[ Q_{\text{hot},A} = 4330 + 220.92 = 4550.92 \mathrm{~J} \]For Unit B:\[ Q_{\text{hot},B} = 4330 + 115.05 = 4445.05 \mathrm{~J} \]

04

Compare Results with Conceptual Understanding

From the results:(a) Unit A requires more work, \( W_A = 220.92 \mathrm{~J} \) compared to \( W_B = 115.05 \mathrm{~J} \), due to the lower room temperature creating a larger temperature difference.(b) Unit A deposits more heat outside, \( Q_{\text{hot},A} = 4550.92 \mathrm{~J} \) compared to \( Q_{\text{hot},B} = 4445.05 \mathrm{~J} \), as it converts more of the input work into heat to the environment.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Coefficient of Performance

The coefficient of performance (COP) is a key metric in thermodynamics for measuring the efficiency of devices like refrigerators and air conditioners. It tells us how effectively a device uses work to transfer heat from one place to another. The higher the COP, the more efficient the system.
For a Carnot refrigerator, the COP is given by the formula: \[\text{COP} = \frac{T_\text{cold}}{T_\text{hot} - T_\text{cold}} \]where \(T_\text{cold}\) is the temperature inside the refrigerated space, and \(T_\text{hot}\) is the external, or ambient, temperature.
This formula shows that the COP depends on the temperatures between which the system is transferring heat. It helps in understanding how much work is needed based on the temperature difference between the cold and hot reservoirs.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It governs how energy can be transferred and transformed, particularly through systems like refrigerators and air conditioners.
The laws of thermodynamics provide a theoretical foundation for understanding why certain systems require energy inputs for operating between different temperatures. The second law of thermodynamics, for example, implies that energy is needed to move heat from a cooler area to a warmer area, as happens in refrigeration cycles. This is because natural processes tend to move towards increasing entropy, or disorder, and transferring heat against a temperature gradient goes against this natural tendency.

Carnot Cycle

The Carnot cycle is an idealized process that helps us understand the maximum efficiency possible for heat engines, refrigerators, and heat pumps. Named after Sadi Carnot, this cycle assumes no energy losses due to factors like friction or unrecoverable heat exchange.
A Carnot cycle consists of four reversible processes:

• Isothermal expansion: The system absorbs heat from a hot reservoir and does work on the surroundings.
• Adiabatic expansion: The system continues to do work, causing its temperature to drop without exchanging heat.
• Isothermal compression: The system releases heat to a cold reservoir and work is done on the system.
• Adiabatic compression: The system's temperature increases back to the original state without exchanging heat.

For a Carnot refrigerator, the cycle is operated in reverse, effectively allowing heat to be extracted from the cold reservoir and discarded into the hot reservoir. This theoretical cycle sets a benchmark, defining the limits of performance for any practical refrigeration system operating between two temperatures.

Heat Transfer

Heat transfer is the flow of thermal energy from a hotter object to a cooler one. It is central to the operation of refrigerators and describes how heat is moved against its natural direction, from cooler spaces to warmer ones.
Heat can be transferred in three ways: conduction, convection, and radiation.

• Conduction: Transfer through direct contact between molecules, like heat moving through a metal rod exposed to a flame.
• Convection: Transfer through movement of fluids, such as air or liquid, where warmer areas of a fluid rise and cooler areas sink.
• Radiation: Transfer through electromagnetic waves, like how the sun warms the earth.

In a refrigerating system, these principles allow for the movement of heat from the interior, out of the refrigerator space, to the external environment. Work is performed by compressors to accomplish this transfer.