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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 \(\mathrm{J}\) J. For both units, find the magnitude of the work required and the magnitude of the heat deposited outside.

Short Answer

Expert verified
Work for A is 220.92 J, and for B is 115.05 J. Heat outside is 4550.92 J for A and 4445.05 J for B.

Step by step solution

01

Understand the Problem

We know that both units, A and B, are Carnot air conditioners and they are removing 4330 J of heat from two separate rooms. Room A is kept at 294 K and room B at 301 K, with the same outdoor temperature of 309 K for both.
02

Identify the Formula for COP

The Coefficient of Performance (COP) for an ideal Carnot air conditioner is given by the formula: \[COP = \frac{T_{cold}}{T_{hot} - T_{cold}}\] where \(T_{cold}\) is the temperature of the room and \(T_{hot}\) is the outside temperature.
03

Calculate COP for Unit A

Using the temperatures for unit A, calculate its COP: \(T_{cold,A} = 294 \ K\), \(T_{hot} = 309 \ K\).\[COP_A = \frac{294}{309 - 294} = \frac{294}{15} = 19.6\]
04

Calculate COP for Unit B

Using the temperatures for unit B, calculate its COP: \(T_{cold,B} = 301 \ K\), \(T_{hot} = 309 \ K\).\[COP_B = \frac{301}{309 - 301} = \frac{301}{8} = 37.625\]
05

Determine Work for Unit A

For an air conditioner, the work done, \(W_A\), is related to the heat removed, \(Q_{cold} = 4330\ J\), by: \[COP = \frac{Q_{cold}}{W}\] Rearrange to find work for unit A: \[W_A = \frac{Q_{cold}}{COP_A} = \frac{4330}{19.6} = 220.92\ J\]
06

Determine Work for Unit B

Work done by unit B, \(W_B\), follows the same formula:\[W_B = \frac{Q_{cold}}{COP_B} = \frac{4330}{37.625} = 115.05\ J\]
07

Calculate Heat Deposited Outside for Unit A

The heat deposited outside, \(Q_{hot}\), is the sum of the heat removed and the work done: \[Q_{hot,A} = Q_{cold} + W_A = 4330 + 220.92 = 4550.92\ J\]
08

Calculate Heat Deposited Outside for Unit B

Similarly, calculate \(Q_{hot}\) for Unit B:\[Q_{hot,B} = Q_{cold} + W_B = 4330 + 115.05 = 4445.05\ J\]

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Key Concepts

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

Carnot cycle
The Carnot cycle is a fundamental concept in thermodynamics that describes the most efficient cycle possible for a heat engine or refrigerator. It involves reversible processes and sets the standard for the highest efficiency achievable between two thermal reservoirs.
The key steps of a Carnot cycle include:
  • Isothermal expansion
  • Adiabatic expansion
  • Isothermal compression
  • Adiabatic compression
For a Carnot air conditioner, as in the given exercise, the process is reversible. This means no energy is lost to entropy and the operation occurs between two specific temperatures: the room (cold reservoir) and the outside environment (hot reservoir).
The Carnot cycle principle ensures that the process has maximum efficiency, which is crucial for understanding the work and heat transfer during the cycle's operations.
Coefficient of Performance
The Coefficient of Performance (COP) is a measure of an air conditioner's efficiency. It is defined as the ratio of heat removed from a cooled space to the work input required to remove that heat. In the context of the Carnot cycle, COP is calculated using the formula:\[COP = \frac{T_{cold}}{T_{hot} - T_{cold}}\]
where:
  • \(T_{cold}\) is the temperature of the cooled room.
  • \(T_{hot}\) is the temperature of the outside environment.
A higher COP value indicates a more efficient air conditioning unit since it signifies that less work is needed to transfer a certain amount of heat. In the problem, the units A and B have different COP values due to their differing room temperatures. This leads to varying work requirements to maintain room temperatures despite having the same heat removal rate.
Work and heat transfer
In the operation of a Carnot air conditioner, both work and heat transfer play crucial roles. The work done by the air conditioner is the energy required to remove heat from the room and expel it outside. This is calculated using the relationship between COP and heat extracted:\[W = \frac{Q_{cold}}{COP}\]
where \(W\) is the work input and \(Q_{cold}\) is the heat removed from the cooled space. After determining the work, another essential calculation is the heat deposited outside:\[Q_{hot} = Q_{cold} + W\]
This represents the total heat transferred to the hot reservoir (the outside environment in this context).
Together, these calculations illustrate how efficiently energy is used and redistributed in air conditioning systems based on the principles of the Carnot cycle.

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Most popular questions from this chapter

The wattage of a commercial ice maker is 225 \(\mathrm{W}\) and is the rate at which it does work. The ice maker operates just like a refrigerator or an air conditioner and has a coefficient of performance of \(3.60 .\) The water going into the unit has a temperature of \(15.0^{\circ} \mathrm{C},\) and the ice maker produces ice cubes at \(0.0^{\circ} \mathrm{C}\) . Ignoring the work needed to keep stored ice from melting, find the maximum amount (in \(\mathrm{kg}\) ) of ice that the unit can produce in one day of continuous operation.

Beginning with a pressure of \(2.20 \times 10^{5}\) Pa and a volume of \(6.34 \times 10^{-3} \mathrm{m}^{3},\) an ideal monatomic gas \(\left(\gamma=\frac{5}{3}\right)\) undergoes an adiabatic expansion such that its final pressure is \(8.15 \times 10^{4} \mathrm{Pa}\) . An alternatice process leading to the same final state begins with an isochoric cooling to the final pressure, followed by an isobaric expansion to the final volume. How much more work does the gas do in the adiabatic process than in the alternative process?

The temperatures indoors and outdoors are 299 and \(312 \mathrm{K},\) respectively. A Carnot air conditioner deposits \(6.12 \times 10^{5} \mathrm{J}\) of heat outdoors. How much heat is removed from the house?

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?

A Carnot refrigerator is used in a kitchen in which the temperature is kept at 301 \(\mathrm{K}\) . This refrigerator uses 241 \(\mathrm{J}\) of work to remove 2561 \(\mathrm{J}\) of heat from the food inside. What is the temperature inside the refrigerator?

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