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The entropy of the working fluid of the ideal Carnot cycle (increases, decreases, remains the same) during the isothermal heat rejection process.

Short Answer

Expert verified
Answer: The entropy of the working fluid of the ideal Carnot cycle *increases* during the isothermal heat rejection process.

Step by step solution

01

Understand the Carnot Cycle

A Carnot cycle is an ideal thermodynamic cycle consisting of four reversible processes. The four processes are: 1. Isothermal heat addition at a high temperature (constant temperature T1) 2. Adiabatic (isentropic) expansion 3. Isothermal heat rejection at a low temperature (constant temperature T2) 4. Adiabatic (isentropic) compression The question concerns the third process – isothermal heat rejection.
02

Define Entropy

Entropy (S) is a measure of the disorder or randomness in a system, and it is a state function, meaning its change depends only on the initial and final states of the system irrespective of the path taken. For a reversible process, change in entropy (ΔS) can be described as: ΔS = Q_reversible/T where Q_reversible is the heat transfer during the reversible process, and T is the absolute temperature in Kelvin.
03

Examine Isothermal Heat Rejection Process

During the isothermal heat rejection process, the working fluid experiences a constant temperature (T2) and releases heat (Q2) to the surroundings. Because this is an ideal Carnot cycle, we consider the process to be reversible. Using the entropy formula from Step 2, the change in entropy during heat rejection is: ΔS = -Q2/T2 Since heat is released from the system, Q2 is negative. Therefore, -Q2 is positive.
04

Determine if Entropy Increases, Decreases or Remains the Same

As shown in Step 3, the change in entropy (ΔS) has a positive value due to the positive -Q2. Since the change in entropy is positive, the entropy of the working fluid during the isothermal heat rejection process increases. Hence, the entropy of the working fluid of the ideal Carnot cycle *increases* during the isothermal heat rejection process.

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Key Concepts

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

Understanding Isothermal Heat Rejection
Isothermal heat rejection is a key stage in the Carnot cycle, which involves a working fluid releasing heat at a constant temperature. Within this stage, the temperature of the system does not change, hence the term 'isothermal,' which comes from 'iso' meaning equal and 'thermal' meaning concerning heat.

During isothermal heat rejection, the working fluid interacts with a cooler reservoir, transferring heat to it. This is a crucial step to restoring the system to a lower energy state. According to the ideal Carnot cycle, which is a theoretical construct used to set an upper efficiency limit for heat engines, this process is reversible, meaning it can be conducted in the opposite direction without the loss of thermal energy.

To make these concepts clearer to students, it is important to visualize the Carnot cycle on a temperature-entropy diagram. This helps to show that during isothermal processes, the entropy change reflects the heat transfer, directly tying these two concepts together. When heat is rejected isothermally, the entropy decreases in the system but increases in the surrounding environment, since the process is fundamentally about heat transfer.
Entropy in Thermodynamics
Entropy is a fundamental concept in thermodynamics dealing with randomness or disorder within a system. Understanding entropy allows students to grasp why certain processes occur naturally and why others need energy input to happen.

In the context of the Carnot cycle, entropy helps to quantify the unavoidable inefficiencies or losses that occur when converting heat into work. During the isothermal heat rejection step, despite the system losing heat, the total entropy change accounts for a balance between the system and its surroundings. It's essential to recognize that in a reversible isothermal process, such as the ideal Carnot cycle, the system's entropy change is directly proportional to the heat exchange divided by the temperature at which the process occurs, expressed mathematically as \( \Delta S = \frac{Q_{reversible}}{T} \). This relationship can be a robust tool for understanding energy distributions in various thermodynamic processes.

To deepen students' understanding, emphasize the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time. This law underpins the unidirectional flow of time itself and is the basis for why heat flows from hot to cold bodies spontaneously.
Reversible Processes
Reversible processes are theoretical constructs in thermodynamics that require understanding subtle but powerful concepts. They are hypothetical processes that happen infinitely slowly, so both the system and its surroundings can adjust without any friction or unrestrained heat transfer. These conditions ensure that the system’s changes are at equilibrium at all times.

Why is the idea of reversible processes important? They represent the ideal, most efficient way a system can change because no energy is wasted or unrecoverably released into the surroundings. In the Carnot cycle, each stage is considered reversible, which is why this cycle is a benchmark for the maximum efficiency that any heat engine can achieve. Real-world processes, however, are always irreversible to some extent due to natural inefficiencies such as friction and rapid, uncontrolled heat transfer.

To help students appreciate the importance of reversible processes, explain that they provide a valuable baseline for understanding the thermodynamic efficiency of real-world systems. Although no true reversible process exists outside of theoretical physics, approaching this level of efficiency is a goal worth striving for in the development of engines and other systems that do work through heat exchange.

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Most popular questions from this chapter

In order to cool 1 -ton of water at \(20^{\circ} \mathrm{C}\) in an insulated tank, a person pours \(80 \mathrm{kg}\) of ice at \(-5^{\circ} \mathrm{C}\) into the water. Determine ( \(a\) ) the final equilibrium temperature in the \(\operatorname{tank}\) and \((b)\) the entropy generation during this process. The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg}\).

Air is expanded in an adiabatic turbine of 85 percent isentropic efficiency from an inlet state of \(2200 \mathrm{kPa}\) and \(300^{\circ} \mathrm{C}\) to an outlet pressure of \(200 \mathrm{kPa}\). Calculate the outlet temperature of air and the work produced by this turbine per unit mass of air.

Consider two bodies of identical mass \(m\) and specific heat \(c\) used as thermal reservoirs (source and sink) for a heat engine. The first body is initially at an absolute temperature \(T_{1}\) while the second one is at a lower absolute temperature \(T_{2}\). Heat is transferred from the first body to the heat engine, which rejects the waste heat to the second body. The process continues until the final temperatures of the two bodies \(T_{f}\) become equal. Show that \(T_{f}=\sqrt{T_{1} T_{2}}\) when the heat engine produces the maximum possible work.

Long cylindrical steel rods (\(\rho=7833 \mathrm{kg} / \mathrm{m}^{3}\) and \(\left.c_{p}=0.465 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) of \(10-\mathrm{cm}\) diameter are heat treated by drawing them at a velocity of \(3 \mathrm{m} / \mathrm{min}\) through a 7 -m-long oven maintained at \(900^{\circ} \mathrm{C}\). If the rods enter the oven at \(30^{\circ} \mathrm{C}\) and leave at \(700^{\circ} \mathrm{C}\), determine ( \(a\) ) the rate of heat transfer to the rods in the oven and \((b)\) the rate of entropy generation associated with this heat transfer process.

Steam expands in an adiabatic turbine from \(4 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) to \(0.1 \mathrm{MPa}\) at a rate of \(2 \mathrm{kg} / \mathrm{s}\). If steam leaves the turbine as saturated vapor, the power output of the turbine is \((a) 2058 \mathrm{kW}\) \((b) 1910 \mathrm{kW}\) \((c) 1780 \mathrm{kW}\) \((d) 1674 \mathrm{kW}\) \((e) 1542 \mathrm{kW}\)

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