/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 The hot reservoir for a Carnot e... [FREE SOLUTION] | 91Ó°ÊÓ

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The hot reservoir for a Carnot engine has a temperature of \(890 \mathrm{~K},\) while the cold reservoir has a temperature of \(670 \mathrm{~K}\). The heat input for this engine is \(4800 \mathrm{~J}\). The \(670-\mathrm{K}\) reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of \(420 \mathrm{~K}\). Find the total work delivered by the two engines.

Short Answer

Expert verified
The total work delivered by the two engines is 2535.33 J.

Step by step solution

01

Determine Efficiency of First Carnot Engine

The efficiency \( \eta \) of a Carnot engine is calculated using the temperatures of its hot (\( T_h \)) and cold (\( T_c \)) reservoirs: \[ \eta = 1 - \frac{T_c}{T_h} \] For the first engine, \( T_h = 890 \mathrm{~K} \) and \( T_c = 670 \mathrm{~K} \). Thus,\[ \eta_1 = 1 - \frac{670}{890} = 1 - 0.7528 = 0.2472 \] Hence, the efficiency of the first engine is 0.2472 or 24.72%.
02

Calculate Work of First Engine

The work \( W \) done by the first Carnot engine is given by the product of its efficiency and the heat input \( Q_{in} \): \[ W_1 = \eta_1 \times Q_{in} \]Given \( Q_{in} = 4800 \mathrm{~J} \) and \( \eta_1 = 0.2472 \), we have:\[ W_1 = 0.2472 \times 4800 = 1186.56 \mathrm{~J} \]Therefore, the first engine delivers 1186.56 J of work.
03

Determine Rejected Heat from First Engine

The heat rejected by the first engine to its cold reservoir is calculated by subtracting the work done from the heat input:\[ Q_{rejected,1} = Q_{in} - W_1 = 4800 - 1186.56 = 3613.44 \mathrm{~J} \]Thus, 3613.44 J is rejected to the second engine.
04

Determine Efficiency of Second Carnot Engine

The second engine uses the 670 K reservoir as the hot reservoir and the 420 K reservoir as the cold reservoir. Its efficiency is:\[ \eta_2 = 1 - \frac{420}{670} = 1 - 0.6269 = 0.3731 \]Therefore, the efficiency of the second engine is 0.3731 or 37.31%.
05

Calculate Work of Second Engine

With the rejected heat from the first engine \( Q_{rejected,1} = 3613.44 \mathrm{~J} \) as input, the work done by the second engine is:\[ W_2 = \eta_2 \times Q_{rejected,1} = 0.3731 \times 3613.44 = 1348.77 \mathrm{~J} \]Hence, the second engine delivers 1348.77 J of work.
06

Calculate Total Work Delivered by the Two Engines

The total work delivered by both engines is the sum of their individual works:\[ W_{total} = W_1 + W_2 = 1186.56 + 1348.77 = 2535.33 \mathrm{~J} \]Thus, the combined work from both engines is 2535.33 J.

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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 of how well an engine converts heat into work. It's crucial for understanding systems like the Carnot engine. The Carnot engine, a theoretical construct, sets the upper benchmark for efficiency among heat engines because it operates between two thermal reservoirs. This makes it instrumental in understanding real-world energy systems.

Thermodynamic efficiency is determined by the temperatures of the engine's hot and cold reservoirs. The efficiency \( \eta \) is calculated using the formula:
  • \( \eta = 1 - \frac{T_c}{T_h} \)
where \( T_h \) is the temperature of the hot reservoir and \( T_c \) is the temperature of the cold reservoir.

In our exercise, the efficiency of the first Carnot engine is 24.72%. This means that only about 25% of the heat energy put in is converted into useful work, indicating the significant role of reservoir temperatures in determining efficiency. The limitation set by these temperatures also highlights why managing heat sinks and sources is vital for energy systems.
Heat Transfer
Heat transfer is the movement of thermal energy from one object or place to another. In a Carnot engine, this transfer process plays a crucial role in its operation. The engine utilizes heat from the hot reservoir, performs work, and then expels leftover heat to the cold reservoir. Understanding how this flow of heat occurs helps clarify why not all input heat can be converted to work.

In the given problem, we see an example of heat transfer at work. Initially, the first engine receives 4800 J. After doing useful work, not all of this energy remains in work form:
  • 1186.56 J is converted into work.
  • 3613.44 J is rejected as waste heat to the next reservoir.
The concept of rejected heat is fundamental in thermodynamics. This unutilized portion of heat energy can often be repurposed or has to be managed, especially in setups like the consecutive Carnot engines shown. Here, the rejected heat becomes the input energy for the second engine which further extracts work from it.
Thermodynamic Cycle
A thermodynamic cycle is a series of processes that a system undergoes where it returns to its initial state, completing a circuit path in terms of energy and matter. This cycle is essential to heat engines, like Carnot engines, which convert heat into useful work through cyclical operations.

In our problem, the thermodynamic cycle is reflected in the two-step operation involving two linked Carnot engines:
  • The first engine operates a cycle from a high-temperature reservoir to a medium-temperature reservoir, converting portion of the energy into work.
  • The second engine continues the cycle from the medium-temperature reservoir to a low-temperature one, extracting additional work.
By completing these cycles, the two-engine system maximizes the total work output possible ( 2535.33 J in total).

The interesting part of thermodynamic cycles is that they illustrate energy transformation efficiency and waste, ultimately aiming to bring the system back to its starting point. This continuous flow and cycle allow the engine to do work persistently, showcasing how energy is conserved and utilized within heat engines.

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Most popular questions from this chapter

The pressure and volume of an ideal monatomic gas change from \(A\) to \(B\) to \(C,\) as the drawing shows. The curved line between \(A\) and \(C\) is an isotherm. (a) Determine the total heat for the process and (b) state whether the flow of heat is into or out of the gas.

The temperature of 2.5 mol of a monatomic ideal gas is \(350 \mathrm{~K}\). The internal energy of this gas is doubled by the addition of heat. How much heat is needed when it is added at (a) constant volume and (b) constant pressure?

An air conditioner keeps the inside of a house at a temperature of \(19.0^{\circ} \mathrm{C}\) when the outdoor temperature is \(33.0^{\circ} \mathrm{C}\). Heat, leaking into the house at the rate of 10500 joules per second, is removed by the air conditioner. Assuming that the air conditioner is a Carnot air conditioner, what is the work per second that must be done by the electrical energy in order to keep the inside temperature constant?

How long would a \(3.00-\mathrm{kW}\) space heater have to run to put into a kitchen the same amount of heat as a refrigerator ( (coefficient of performance \(=3.00\) ) ) does when it freezes \(1.50 \mathrm{~kg}\) of water at \(20.0^{\circ} \mathrm{C}\) into ice at \(0.0^{\circ} \mathrm{C} ?\)

Concept Questions (a) A real (irreversible) engine operates between hot and cold reservoirs whose temperatures are \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}}\). The engine absorbs heat of magnitude \(Q_{\mathrm{H}} \mid\) from the hot reservoir and performs work of magnitude \(|W| .\) What is the expression that gives the change in entropy \(\Delta S_{\text {universe }}\) of the universe associated with the operation of this engine? Express your answer in terms of \(T_{\mathrm{H}}, T_{\mathrm{C}},\left|Q_{\mathrm{H}}\right|,\) and \(\left|Q_{\mathrm{C}}\right|\) Would you expect the change in the entropy of the universe to be greater than, less than, or equal to zero? Provide a reason for your answer. (c) Suppose that a reversible engine (a Carnot engine) operates between the same hot and cold temperatures as the irreversible engine described above. For the same input heat \(\left|Q_{\mathrm{H}}\right|,\) would the reversible engine produce more, less, or the same work as the irreversible engine? Justify your answer. (d) In terms of \(\Delta\) S universe and \(T_{\mathrm{C}}\), how could one determine the difference, if any, between the work done by the reversible engine and that done by the irreversible engine? Problem (a) An irreversible engine operates between temperatures of 852 and \(314 \mathrm{~K}\). It absorbs \(1285 \mathrm{~J}\) of heat from the hot reservoir and does \(264 \mathrm{~J}\) of work. What is the change \(\Delta S_{\text {universe }}\) in the entropy of the universe associated with the operation of this engine? (b) If the engine were reversible, what is the magnitude \(|W|\) of the work it would have done, assuming it operated between the same temperatures and absorbed the same heat as the irreversible engine? (c) Using the results of parts (a) and (b), find the difference between the work produced by the reversible and irreversible engines.

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