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An ideal refrigerator or ideal heat pump is equivalent to a Carnot engine running in reverse. That is, energy \(Q_{c}\) is taken in from a cold reservoir and energy \(Q_{h}\) is rejected to a hot reservoir. (a) Show that the work that must be supplied to run the refrigerator or heat pump is $$W=\frac{T_{h}-T_{c}}{T_{c}} Q_{c}$$ (b) Show that the coefficient of performance of the ideal refrigerator is $$\operatorname{COP}=\frac{T_{c}}{T_{h}-T_{c}}$$

Short Answer

Expert verified
The work required to run the ideal refrigerator or heat pump is given by \(W = \frac{T_{h} - T_{c}}{T_{c}} Q_{c}\). The coefficient of performance of the ideal refrigerator is \(\operatorname{COP} = \frac{T_{c}}{T_{h} - T_{c}}\).

Step by step solution

01

Understand the Carnot Engine process

The Carnot cycle consists of two isothermal processes and two adiabatic processes. In a refrigerator or heat pump, the cycle runs in reverse. The work done on the system is used to extract heat (\(Q_{c}\)) from the cold reservoir at absolute temperature (\(T_{c}\) in Kelvin) and reject heat (\(Q_{h}\)) to the hot reservoir at absolute temperature (\(T_{h}\) in Kelvin). The efficiency of the Carnot engine operating in reverse can be used as a starting point.
02

Write down the definition of Carnot efficiency

The efficiency (\(\text{e}\)) for a Carnot engine operating forward is given by \[\text{e} = 1 - \frac{T_{c}}{T_{h}}\] To adapt this for a refrigerator or heat pump running in reverse, we consider the work input required.
03

Relating work to heat energy

Since the work (\(W\)) is used to transfer heat from the cold reservoir to the hot reservoir, we can relate the work supplied to the heat extracted from the cold reservoir as follows: \[W = Q_{h} - Q_{c}\] According to the Carnot cycle's conservation of energy.
04

Substitute the efficiency relation into the work equation

Using the definition of the Carnot efficiency in the context of a refrigerator or heat pump (i.e., the engine's efficiency running in reverse), we have \[Q_{h} = Q_{c} + W\] Substituting this into the efficiency expression, we get \[1 - \frac{T_{c}}{T_{h}} = \frac{W}{Q_{c}}\] which upon rearrangement gives the required work expression \[W = \frac{T_{h} - T_{c}}{T_{c}} Q_{c}\]
05

Derive the coefficient of performance (COP) for the refrigerator

The COP for a refrigerator or heat pump is defined as the ratio of heat extracted from the cold reservoir to the work input: \[\operatorname{COP} = \frac{Q_{c}}{W}\] By substituting the expression for work we just derived (Step 4), we get \[\operatorname{COP} = \frac{Q_{c}}{\frac{T_{h} - T_{c}}{T_{c}}Q_{c}}\] Simplifying this expression yields the COP \[\operatorname{COP} = \frac{T_{c}}{T_{h} - T_{c}}\]

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Key Concepts

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

Carnot Engine Efficiency
Carnot engine efficiency is a crucial concept in the fundamentals of thermodynamics, and it serves as a standard for the most efficient heat engine possible. Simply stated, it is a measure of how well an engine converts thermal energy into work. It is represented by the formula

\[ e = 1 - \frac{T_c}{T_h} \]

where \(e\) stands for efficiency, \(T_c\) is the absolute temperature of the cold reservoir, and \(T_h\) is the absolute temperature of the hot reservoir (both in Kelvin). In essence, the closer the cold reservoir temperature is to the hot reservoir temperature, the lower the efficiency. No real engine can be more efficient than a Carnot engine operating between the same two temperatures. Understanding this efficiency is key to comprehending the energy conservation within these systems and sets the stage for analyzing reversible cycles and their practical applications like refrigerators.
Coefficient of Performance (COP)
The coefficient of performance (COP) is a term used to evaluate the efficiency of refrigerators and heat pumps. It can be thought of as a ratio that tells us how effective these devices are at heating or cooling. The COP is given by the formula:

\[ \operatorname{COP} = \frac{T_c}{T_h - T_c} \]

where \(T_c\) is the temperature of the cold space (the inside of the fridge, for example) and \(T_h\) is the temperature of the surrounding environment. A higher COP value indicates a more efficient refrigeration process, meaning that for each unit of work input, more heat is extracted from the cold space. When considering an ideal refrigerator, the COP provides insight into the theoretical limits of performance and helps in the comparison of real-world refrigeration systems.
Thermodynamic Processes
In the study of thermodynamics, processes describe how systems go from one state to another. There are several kinds of processes, each with distinct characteristics and rules. For instance, isothermal processes occur at a constant temperature, where the system’s temperature does not change.

Conversely, adiabatic processes have no heat exchange with the surroundings; instead, work done on the system changes its internal energy directly. These concepts are integral in understanding how heat engines, like the Carnot engine, and refrigerators work. Recognizing and analyzing these thermodynamic processes is essential for students as they examine how energy is transferred and transformed in various thermodynamic cycles.
Heat Transfer in Refrigeration
Heat transfer is at the essence of how refrigeration systems function. In simple terms, refrigerators move heat from a cooler interior space to the warmer environment outside. This is in contrast to heat engines, which typically move heat from a hot source to a cooler sink to create work.

To understand heat transfer in refrigeration, it's key to grasp concepts like conduction, where heat transfers through a medium due to a temperature difference, and evaporation/condensation, mechanisms central to the refrigeration cycle. By examining these heat transfer techniques, one can appreciate the refrigeration processes’ sophistication and why certain materials and coolant fluids are chosen for optimal performance.

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Most popular questions from this chapter

A Carnot engine has a power output of 150 \(\mathrm{kW}\) . The engine operates between two reservoirs at \(20.0^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (a) How much energy does it take in per hour? (b) How much energy is lost per hour in its exhaust?

A gasoline engine has a compression ratio of 6.00 and uses a gas for which \(\gamma=1.40 .\) (a) What is the efficiency of the engine if it operates in an idealized Otto cycle? (b) What If ? If the actual efficiency is 15.0%, what fraction of the fuel is wasted as a result of friction and energy losses by heat that could by avoided in a reversible engine? (Assume complete combustion of the air–fuel mixture.)

In a cylinder of an automobile engine, just after combustion, the gas is confined to a volume of 50.0 \(\mathrm{cm}^{3}\) and has an initial pressure of \(3.00 \times 10^{6} \mathrm{Pa}\) . The piston moves outward to a final volume of \(300 \mathrm{cm}^{3},\) and the gas expands without energy loss by heat. (a) If \(\gamma=1.40\) for the gas, what is the final pressure? (b) How much work is done by the gas in expanding?

In 1993 the federal government instituted a requirement that all room air conditioners sold in the United States must have an energy efficiency ratio (EER) of 10 or higher. The EER is defined as the ratio of the cooling capacity of the air conditioner, measured in Btu/h, to its electrical power requirement in watts. (a) Convert the EER of 10.0 to dimensionless form, using the conversion 1 Btu ! 1055 J. (b) What is the appropriate name for this dimensionless quantity? (c) In the 1970s it was common to find room air conditioners with EERs of 5 or lower. Compare the operating costs for 10 000-Btu/h air conditioners with EERs of 5.00 and 10.0. Assume that each air conditioner operates for 1 500 h during the summer in a city where electricity costs 10.0\(€\) per \(\mathrm{kWh}\).

One of the most efficient heat engines ever built is a steam turbine in the Ohio valley, operating between \(430^{\circ} \mathrm{C}\) and \(1870^{\circ} \mathrm{C}\) on energy from West Virginia coal to produce electricity for the Midwest. (a) What is its maximum theoretical efficiency? (b) The actual efficiency of the engine is 42.0\(\%\) . How much useful power does the engine deliver if it takes in \(1.40 \times 10^{5} \mathrm{J}\) of energy each second from its hot reservoir?

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