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What is the Clausius expression of the second law of thermodynamics?

Short Answer

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Answer: The Clausius inequality represents the expression of the second law of thermodynamics, stating that for any real process (reversible or irreversible) occurring in a cyclic manner, the cyclic integral of heat transfer divided by the temperature is either equal to or less than zero. This inequality prevents the creation of a perpetual motion machine and limits the efficiency of any process involving heat transfer, as it implies that any real process must be less than or equal to the efficiency of an ideal Carnot cycle (a reversible process).

Step by step solution

01

Introduce the Clausius Inequality

The Clausius expression of the second law of thermodynamics is known as the Clausius inequality, which states that for any real process (reversible or irreversible) occurring in a cyclic manner in a system, the following inequality holds true: \[ \oint \frac{dQ}{T} \le 0 \]
02

Describe the symbols and terms used in the inequality

In the Clausius inequality, each term has a specific meaning: - \(\oint\): This symbol represents integration around a cycle, meaning that the process must return to its initial state after the cycle is completed. - \(dQ\): This term represents the heat transfer in a given step of the cyclic process. - \(T\): This term represents the absolute temperature (in Kelvin) at which the heat transfer occurs.
03

Explain the meaning of the inequality

The Clausius inequality implies that the cyclic integral of the heat transfer divided by the temperature (at which the heat transfer occurs) is either equal to or less than zero. This indicates that for any real process, the process must be less than or equal to the efficiency of an ideal Carnot cycle (a reversible process). In simpler terms, this inequality prevents the creation of a perpetual motion machine, as it limits the efficiency of any process involving heat transfer.

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Key Concepts

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

Entropy
Entropy is a fundamental concept in thermodynamics, often referred to as the measure of a system's thermal energy per unit temperature that is unavailable for doing work. As a function of state, entropy provides crucial insight into the direction of spontaneous processes. According to the second law of thermodynamics, the total entropy of an isolated system can never decrease over time. In a more practical sense, entropy can be viewed as a measure of disorder: the higher the entropy, the greater the disorder and the more dispersed the energy is in a system.

When a process occurs, if the system is reversible, the entropy change of the system and the surrounding environment is zero. However, in all real, irreversible processes, entropy always increases, which is a restatement of the second law of thermodynamics. The Clausius inequality, through the integral \( \oint \frac{dQ}{T} \le 0 \), ties directly to this concept because it mathematically expresses how entropy changes around a thermodynamic cycle.
Thermodynamic Cycles
Thermodynamic cycles are a series of processes that involve heat and work transfer, ultimately returning a system to its initial state. These cycles are essential to understanding how thermal engines work, as they are a model for the operation of real engines and refrigerators. Examples of thermodynamic cycles include the Carnot cycle and the Rankine cycle, which model idealized versions of heat engines and steam power plants, respectively.

In examining these cycles, the Clausius inequality provides a vital check on the feasibility of the cycle itself. The inequality serves as a boundary for the efficiency that can be achieved by any engine using that cycle, insisting that it cannot exceed the efficiency of a reversible cycle. The improbability of achieving a perfect engine, where \( \oint \frac{dQ}{T} = 0 \), reflects the unavoidable reality of energy loss and inefficiency due to factors like friction and heat loss.
Carnot Cycle
The Carnot cycle is the most efficient thermodynamic cycle possible for an engine operating between two heat reservoirs. It is an idealized concept that serves as a benchmark for the performance of actual heat engines. The cycle consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. As such, over one complete cycle, the net entropy change for a Carnot engine is zero, since all processes are reversible.

Applying the Clausius inequality to the Carnot cycle, it becomes clear why it is deemed an ideal cycle. For a Carnot engine, the integral \( \oint \frac{dQ}{T} = 0 \) holds true, signifying that there is no entropy production and hence no wasted energy. This is in stark contrast to real engines where \( \oint \frac{dQ}{T} < 0 \), indicating entropy generation and less than perfect efficiency.
Heat Transfer
Heat transfer is the movement of thermal energy from one body or system to another due to a difference in temperature. It is a fundamental process in thermodynamics and occurs in three primary ways: conduction, convection, and radiation. Understanding heat transfer is necessary for analyzing thermodynamic cycles since it plays a crucial role in how energy is moved and transformed within these systems.

In the context of the Clausius inequality, the term \(dQ\) represents the incremental amount of heat transfer occurring at temperature \(T\) during a process in the cycle. The Clausius inequality asserts that for a cycle to be consistent with the second law of thermodynamics, the sum of these heat transfers (when divided by their respective temperatures) over the entire cycle, must be less than or equal to zero. This relationship elegantly encapsulates the inescapable inefficiency present in every real-world heat transfer process, highlighting the limitations set by the laws of thermodynamics.

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Most popular questions from this chapter

A Carnot heat engine receives heat from a reservoir at \(1700^{\circ} \mathrm{F}\) at a rate of \(700 \mathrm{Btu} / \mathrm{min}\) and rejects the waste heat to the ambient air at \(80^{\circ} \mathrm{F}\). The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at \(20^{\circ} \mathrm{F}\) and transfers it to the same ambicnt air at \(80^{\circ} \mathrm{F}\). Determine \((a)\) the maximum rate of heat removal from the refrigerated space and ( \(b\) ) the total rate of heat rejection to the ambient air.

The kitchen, bath, and other ventilation fans in a house should be used sparingly since these fans can discharge a houseful of warmed or cooled air in just one hour. Consider a \(200-\mathrm{m}^{2}\) house whose ceiling height is \(2.8 \mathrm{m} .\) The house is heated by a 96 percent efficient gas heater and is maintained at \(22^{\circ} \mathrm{C}\) and \(92 \mathrm{kPa}\). If the unit cost of natural gas is \(\$ 1.20 /\) therm \((1 \text { therm }=105,500 \mathrm{kJ})\) determine the cost of energy "vented out" by the fans in 1 h. Assume the average outdoor temperature during the heating season to be \(5^{\circ} \mathrm{C}\)

A window air conditioner that consumes \(1 \mathrm{kW}\) of electricity when running and has a coefficient of performance of 3 is placed in the middle of a room, and is plugged in. The rate of cooling or heating this air conditioner will provide to the air in the room when running is \((a) 3 \mathrm{kJ} / \mathrm{s},\) cooling \((b) 1 \mathrm{kJ} / \mathrm{s},\) cooling \((c) 0.33 \mathrm{kJ} / \mathrm{s},\) heating \((d) 1 \mathrm{kJ} / \mathrm{s},\) heating \((e) 3 \mathrm{kJ} / \mathrm{s},\) heating

\(6-76 \quad\) A Carnot heat engine receives \(650 \mathrm{kJ}\) of heat from a source of unknown temperature and rejects \(250 \mathrm{kJ}\) of it to a \(\operatorname{sink}\) at \(24^{\circ} \mathrm{C} .\) Determine \((a)\) the temperature of the source and (b) the thermal efficiency of the heat engine.

Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second one. If the source temperature of the first engine is \(1300 \mathrm{K}\) and the sink temperature of the \(\sec\) ond engine is \(300 \mathrm{K}\) and the thermal efficiencies of both engines are the same, the temperature of the intermediate reservoir is \((a) 625 \mathrm{K}\) (b) \(800 \mathrm{K}\) \((c) 860 \mathrm{K}\) \((d) 453 \mathrm{K}\) \((e) 758 \mathrm{K}\)

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