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Find the relation between the efficiency of a reversible ideal heat engine and the coefficient of performance of the reversible refrigerator obtained by running the engine backwards.

Short Answer

Expert verified
The relation is \( \eta = \frac{1}{1 + \frac{1}{\text{COP}} } \) where \(\eta\) is the efficiency and COP is the coefficient of performance.

Step by step solution

01

Understanding the Heat Engine Efficiency

The efficiency of a reversible ideal heat engine is defined as the ratio of the work output to the heat input. For a heat engine operating between two reservoirs (hot and cold) at temperatures \(T_h\) and \(T_c\) respectively, its efficiency \(\eta\) can be expressed as: \[ \eta = 1 - \frac{T_c}{T_h} \] where \(T_h\) is the temperature of the hot reservoir and \(T_c\) is the temperature of the cold reservoir.
02

Coefficient of Performance for a Refrigerator

The coefficient of performance (COP) of a reversible refrigerator is defined as the ratio of the heat extracted from the cold reservoir to the work input required to extract that heat. For a refrigerator operating between the same two reservoirs (at temperatures \(T_c\) and \(T_h\)), the COP can be expressed as: \[ \text{COP} = \frac{T_c}{T_h - T_c} \] The value of COP indicates how much heat can be extracted per unit of work input.
03

Relating Efficiency to COP

To find the relation between the efficiency \(\eta\) of the heat engine and the COP of the refrigerator, we start from their definitions. The efficiency can be rewritten in terms of work \(W\) as \(\eta = \frac{W}{Q_h} = 1 - \frac{T_c}{T_h}\), while the COP is \( \text{COP} = \frac{T_c}{Q_h - Q_c} = \frac{T_c}{W} \). Since \(W = Q_h \cdot \eta\), substitute in the COP: \[ \text{COP} = \frac{T_c}{Q_h \cdot \eta} \] \[ \Rightarrow \eta \cdot \text{COP} = \frac{T_c}{Q_h} \] \[ \Rightarrow \eta \cdot \text{COP} = 1 - (1-\eta) = \eta \] Thus, the relation can be simplified to: \[ \eta = \frac{1}{1 + \frac{1}{\text{COP}} } \]
04

Concluding the Relation

By analyzing the expressions for both efficiency and COP, we see that there is an inverse relationship between them. This signifies that as the efficiency \(\eta\) increases, the COP for the reversible refrigerator decreases, and vice versa. This relationship holds true as long as both systems are reversible and ideal.

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Key Concepts

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

Thermodynamic Efficiency
Thermodynamic efficiency is a measure that evaluates how effectively a heat engine converts heat energy into work. It gives insight into the engine's performance when operating between two thermal reservoirs at different temperatures. The efficiency (\( eta \)) can be formulated as \[ eta = 1 - \frac{T_c}{T_h} \]. Here, \(T_h\) signifies the temperature of the hot reservoir, and \(T_c\) represents the cold reservoir. This equation highlights a fundamental aspect: no heat engine operating between two temperatures is 100% efficient unless \(T_c\) is absolute zero, which is practically unattainable.
  • Efficiency depends on the temperature difference between reservoirs.
  • The larger the difference, the higher the potential efficiency.
  • It serves as an upper limit for real-world engines, such as car engines or power plants, that can't achieve this ideal due to irreversibilities and losses.
Understanding thermodynamic efficiency is crucial, as it reflects the potential performance and limitations of thermal machines in real and theoretical scenarios.
Coefficient of Performance
The Coefficient of Performance (COP) is a measure used to gauge the effectiveness of refrigerators and heat pumps. It expresses how much heat is moved per unit of work input. For a refrigerator focusing on extracting heat from a cold reservoir, the COP is given by \[ \text{COP} = \frac{T_c}{T_h - T_c} \]. This equation assumes the system is ideal and reversible, and the temperatures \(T_c\) and \(T_h\) are for the cold and hot reservoirs, respectively.
  • The higher the COP, the more efficiency the system extracts heat using less work.
  • It characterizes the effectiveness and energy consumption of cooling systems.
  • COP provides a comparison point for different refrigeration or heat pump systems.
Importantly, COP interlinks with the thermodynamic efficiency of heat engines, with both optimally functioning in reversible conditions.
Reversible Processes
Reversible processes are an idealized concept in thermodynamics in which a system undergoes changes that can be completely reversed with no net change in both the system and the surroundings. These processes are characterized by their ability to be retraced in opposite directions, returning the system to its initial state without any entropy generation. In the context of the Carnot cycle, reversible processes are essential because:
  • They maximize efficiency, achieving the theoretical limits set by the Carnot theorem.
  • All steps of the Carnot cycle - isothermal and adiabatic processes, are reversible.
While no real process is truly reversible due to friction and dissipation, they serve as a benchmark. Reversible processes set the ideal standards that real systems strive to emulate. Understanding these processes helps in evaluating efficiency improvements and designing systems closer to ideal performance.

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Most popular questions from this chapter

The temperature of \(1.00 \mathrm{~mol}\) of a monatomic ideal gas is raised reversibly from \(300 \mathrm{~K}\) to \(400 \mathrm{~K},\) with its volume kept constant. What is the entropy change of the gas?

A cylindrical copper rod of length \(1.50 \mathrm{~m}\) and radius \(2.00 \mathrm{~cm}\) is insulated to prevent heat loss through its curved surface. One end is attached to a thermal reservoir fixed at \(300^{\circ} \mathrm{C} ;\) the other is attached to a thermal reservoir fixed at \(30.0^{\circ} \mathrm{C}\). What is the rate at which entropy increases for the rod- reservoirs system?

In the first stage of a two-stage Carnot engine, energy is absorbed as heat \(Q_{1}\) at temperature \(T_{1},\) work \(W_{1}\) is done, and energy is expelled as heat \(Q_{2}\) at a lower temperature \(T_{2}\). The second stage absorbs that energy as heat \(Q_{2},\) does work \(W_{2}\), and expels energy as heat \(Q_{3}\) at a still lower temperature \(T_{3}\). Prove that the efficiency of the engine is \(\left(T_{1}-T_{3}\right) / T_{1}\)

A heat pump is used to heat a building. The external temperature is less than the internal temperature. The pump's coefficient of performance is \(3.8,\) and the heat pump delivers \(7.54 \mathrm{MJ}\) as heat to the building each hour. If the heat pump is a Carnot engine working in reverse, at what rate must work be done to run it?

A \(500 \mathrm{~W}\) Carnot engine operates between constant-temperature reservoirs at \(100^{\circ} \mathrm{C}\) and \(60.0^{\circ} \mathrm{C}\). What is the rate at which energy is (a) taken in by the engine as heat and (b) exhausted by the engine as heat?

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