/*! 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} Q28P In the first stage of a two-stag... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1at temperature T1, work W1is done, and energy is expelled as heat Q2at a lower temperature T2. The second stage absorbs that energy as heat Q2 does workW2, and expels energy as heat Q3at a still lower temperature T3. Prove that the efficiency of the engine is(T1-T3)/T1.

Short Answer

Expert verified

The efficiency of the Carnot engine is T1-T3T1.

Step by step solution

01

The given data

Absorbed energy in the first stage of the Carnot engine is Q1.

The initial temperature of the first stage of the Carnot engine is T1.

The energy that is expelled out by the first stage of the Carnot engine is Q2.

The amount of work done by the first stage of the Carnot engine is W1.

Absorbed energy in the second stage of the Carnot engine is Q2.

The initial temperature of the second stage of the Carnot engine is T2.

The energy that is expelled out by the second t stage of the Carnot engine is Q3.

The amount of work done by the second stage of the Carnot engine is W1.

The final temperature of the Carnot engine is T3.

02

Understanding the concept of the Carnot engine’s efficiency

We can write the efficiency of the Carnot engine in terms of total work done. Then using the relation between heat energy absorbed and expelled out with work we can write it in terms of heat energy. Then using the relation between heat energy and temperature at higher and lower temperature reservoirs we can find the efficiency in terms of those temperatures.

Formulae:

The efficiency of the Carnot engine,

ε=WQH (1)

The work done per cycle of the gas,

W=QH-QL (2)

The heat and temperature relation of a Carnot engine,

QHQL=THTL (3)

03

Calculation of the efficiency of the engine

In the first case,W=W1+W2andQH=Q1

The efficiency of the Carnot cycle using equation (1) of the first stage can be given as:

ε=W1+W2Q1

Using equation (2), the works per cycle of the first and second stages are

W1=Q2-Q1,andW2=Q2-Q3respectively.

So, the efficiency becomes

ε=Q2-Q1+Q2-Q3Q1=Q1-Q3Q1=1-Q3Q1

Now, using equation (3), the efficiency of the engine is given usingQ3Q1=T3T1 as,

ε=1-T3T1=T1-T3T1

Hence, the efficiency of the engine is proved asT1-T3T1

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An ideal gas ( 1.0 mol) is the working substance in an engine that operates on the cycle shown in Figure 20-30. Processes BC andDA are reversible and adiabatic. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is the engine efficiency?

A box contains 100 atoms in a configuration that has 50atoms in each half of the box. Suppose that you could count the different microstates associated with this configuration at the rate of 100billion states per second, using a supercomputer. Without written calculation, guess how much computing time you would need: a day, a year, or much more than a year.

A 50.0 kgblock of copper whose temperature is 400kis placed in an insulating box with a 100gblock of lead whose temperature is 200k. (a) What is the equilibrium temperature of the two-block system? (b) What is the change in the internal energy of the system between the initial state and the equilibrium state? (c) What is the change in the entropy of the system? (See Table 18-3.)

At very low temperatures, the molar specific heat CVof many solids is approximately Cv=AT3, where depends on the particular substance. For aluminum,A=3.15×10-5J/mol.K4. Find the entropy change for 4.00 mlof aluminum when its temperature is raised from 5.00 kto 10.0 k.

A box contains Nmolecules. Consider two configurations: Configuration Awith an equal division of the molecules between the two halves of the box, and configuration Bwith 60.0%of the molecules in the left half of the box and 40.0%in the right half. For N =50, what are (a) the multiplicity WAof configuration A, (b) the multiplicityWb of configuration B, and (c) the ratiofB/Aof the time the system spends in configuration Bto the time it spends in configuration A? For N =100, what are (d)WA, (e)WB, and (f)fA/B? ForN =200, what are (g)WA, (h)WB, and (i)fA/B? ( j) With increasingN, doesincrease, decrease, or remain the same?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.