Chapter Eleven

Raymond A. SerwayChris Vuille

Chapter ElevenEnergy in Thermal Processes

Read and take notes on pages: 352-353

Read and take notes on pages: 366-367

Energy Transfer• When two objects of different temperatures are

placed in thermal contact, the temperature of the warmer decreases and the temperature of the cooler increases

• The energy exchange ceases when the objects reach thermal equilibrium

• The concept of energy was broadened from just mechanical to include internal– Made Conservation of Energy a universal law of nature

Introduction

Heat Compared to Internal Energy

• Important to distinguish between them– They are not interchangeable

• Heat involves a transfer of energy

Section 11.1

Internal Energy

• Internal Energy, U, is the energy associated with the atoms and molecules of the system– Includes kinetic and potential energy associated with the

random translational, rotational and vibrational motion of the particles that make up the system

– Also includes any potential energy bonding the particles together

Section 11.1

Link toBright storm on

Heat

Heat

• Heat is the transfer of energy between a system and its environment because of a temperature difference between them– The symbol Q is used to represent the amount of

energy transferred by heat between a system and its environment

Section 11.1

Units of Heat

• Calorie– An historical unit, before the connection between

thermodynamics and mechanics was recognized– A calorie is the amount of energy necessary to

raise the temperature of 1 g of water from 14.5° C to 15.5° C .• A Calorie (food calorie) is 1000 cal

Section 11.1

Units of Heat, cont.

• US Customary Unit – BTU• BTU stands for British Thermal Unit– A BTU is the amount of energy necessary to raise

the temperature of 1 lb of water from 63° F to 64° F • 1 cal = 4.186 J– This is called the Mechanical Equivalent of Heat

Section 11.1

James Prescott Joule

• 1818 – 1889• British physicist• Conservation of Energy• Relationship between

heat and other forms of energy transfer

Section 11.1

Link toBright storm on

Heat Transfer

Methods of Heat Transfer

• Need to know the rate at which energy is transferred

• Need to know the mechanisms responsible for the transfer

• Methods include– Conduction– Convection– Radiation

Section 11.5

Conduction

• The transfer can be viewed on an atomic scale– It is an exchange of energy between microscopic particles

by collisions– Less energetic particles gain energy during collisions with

more energetic particles• Rate of conduction depends upon the characteristics

of the substance

Section 11.5

Conduction example• The molecules vibrate

about their equilibrium positions

• Particles near the stove coil vibrate with larger amplitudes

• These collide with adjacent molecules and transfer some energy

• Eventually, the energy travels entirely through the pan and its handle

Section 11.5

Conduction, cont.

• The rate of conduction depends on the properties of the substance

• In general, metals are good conductors– They contain large numbers of electrons that are relatively

free to move through the metal– They can transport energy from one region to another

• Conduction can occur only if there is a difference in temperature between two parts of the conducting medium

Section 11.5

Conduction, equation

• The slab of material allows energy to transfer from the region of higher temperature to the region of lower temperature

• A is the cross-sectional area

Section 11.5

Conduction, equation explanation• A is the cross-sectional area• Through a rod, Δx = L• P is in Watts when Q is in

Joules and t is in seconds• k is the thermal

conductivity of the material– See table 11.3 for some

conductivities– Good conductors have high

k values and good insulators have low k values

Section 11.5

EXAMPLE 11.9 Energy Transfer Through a Concrete Wall

Goal Apply the equation of heat conduction. Problem Find the energy transferred in 1.00 h by conduction through a concrete wall 2.0 m high, 3.65 m long, and 0.20 m thick if one side of the wall is held at 20°C and the other side is at 5°C. Strategy = kA(Th - Tc)/L gives the rate of energy transfer by conduction in joules per second. Multiply by the time and substitute given values to find the total thermal energy transferred.

SOLUTION

Multiply the energy transfer equation by Δt to find an expression for the total energy Q transferred through the wall. Q = Δt = kA(Th - Tc /L)Δt Substitute the numerical values to obtain Q, consulting the table for k. Q = (1.3 J/s · m °C)(7.3 m2)(15°C/0.20 m)(3600 s) = 2.6 106 J

LEARN MORE

Remarks Early houses were insulated with thick masonry walls, which restrict energy loss by conduction because k is relatively low. The large thickness L also decreases energy loss by conduction, as shown by energy transfer equation. There are much better insulating materials, however, and layering is also helpful. Despite the low thermal conductivity of masonry, the amount of energy lost is still rather large, enough to raise the temperature of 600 kg of water by more than 1°C. Question True or False: Materials having high thermal conductivities provide better insulation than materials having low thermal conductivities.

True. Thermal conductivity measures a material's usefulness as heat insulation. False. A

good insulator transfers energy by heat effectively, and has low thermal conductivity. False. A good insulator does not transfer energy by heat readily through it, and has low thermal

conductivity. True. The rate of heat flow in a material is proportional to its thermal conductivity

Read and take notes on page: 368

Read and take notes on pages: 371-374

Convection

• Energy transferred by the movement of a substance– When the movement results from differences in

density, it is called natural convection– When the movement is forced by a fan or a pump,

it is called forced convection

Section 11.5

Convection example• Air directly above the

flame is warmed and expands

• The density of the air decreases, and it rises

• The mass of air warms the hand as it moves by

Section 11.5

Convection applications

• Boiling water• Radiators• Upwelling• Cooling automobile engines• Algal blooms in ponds and lakes

Section 11.5

Convection Current Example• The radiator warms the air

in the lower region of the room

• The warm air is less dense, so it rises to the ceiling

• The denser, cooler air sinks• A continuous air current

pattern is set up as shown

Section 11.5

Radiation

• Radiation does not require physical contact• All objects radiate energy continuously in the

form of electromagnetic waves due to thermal vibrations of the molecules

• Rate of radiation is given by Stefan’s Law

Section 11.5

Radiation example

• The electromagnetic waves carry the energy from the fire to the hands

• No physical contact is necessary• Cannot be accounted for by conduction or convection

Section 11.5

Radiation equation

• P = σ A e T4

– The power is the rate of energy transfer, in Watts– σ = 5.669 6 x 10-8 W/m2.K4

• Called the Stefan-Boltzmann constant– A is the surface area of the object– e is a constant called the emissivity• e varies from 0 to 1

– T is the temperature in Kelvins

Section 11.5

Energy Absorption and Emission by Radiation

• The rate at which the object at temperature T with surroundings at To radiates is– Pnet = σ A e (T4 - To

4)– When an object is in equilibrium with its

surroundings, it radiates and absorbs at the same rate• Its temperature will not change

Section 11.5

Ideal Absorbers

• An ideal absorber is defined as an object that absorbs all of the energy incident on it– e = 1

• This type of object is called a black body• An ideal absorber is also an ideal radiator of

energy

Section 11.5

Ideal Reflector

• An ideal reflector absorbs none of the energy incident on it– e = 0

Section 11.5

Applications of Radiation• Clothing– Black fabric acts as a good absorber– White fabric is a better reflector

• Thermography– The image of the pattern formed by varying radiation

levels is called a thermogram• Body temperature– Radiation thermometer measures the intensity of the

infrared radiation from the eardrum

Section 11.5

Resisting Energy Transfer• Dewar flask/thermos bottle• Designed to minimize

energy transfer to surroundings

• Space between walls is evacuated to minimize conduction and convection

• Silvered surface minimizes radiation

• Neck size is reduced

Section 11.5

Global Warming

• Greenhouse example– Visible light is absorbed and re-emitted as infrared

radiation– Convection currents are inhibited by the glass

• Earth’s atmosphere is also a good transmitter of visible light and a good absorber of infrared radiation

Section 11.6

Raymond A. SerwayChris Vuille

Chapter TwelveThe Laws of Thermodynamics

Read and take notes on pages: 385-387

First Law of Thermodynamics

• The First Law of Thermodynamics tells us that the internal energy of a system can be increased by– Adding energy to the system– Doing work on the system

• There are many processes through which these could be accomplished– As long as energy is conserved

Introduction

Video # 1 First Law of Thermodynamics & Internal Energy

Khan Academy The next four videos are long but a

MUST WATCH!!!!!!The concepts being covered I find to be very difficult, these videos when

watched together, do a great job explaining the first law of

Thermodynamics

Video # 2More on I

nternal Energy (Khan Academy)

Video # 3 Work from Expansion

(Khan Academy)

Video # 4 PV Diagrams and Expansion Work

(Khan Academy)

Second Law of Thermodynamics

• Constrains the First Law• Establishes which processes actually occur• Heat engines are an important application

Introduction

Work in Thermodynamic Processes – Assumptions

• Dealing with a gas• Assumed to be in thermodynamic equilibrium– Every part of the gas is at the same temperature– Every part of the gas is at the same pressure

• Ideal gas law applies

Section 12.1

Work in a Gas Cylinder• The gas is contained in a

cylinder with a moveable piston

• The gas occupies a volume V and exerts pressure P on the walls of the cylinder and on the piston

Section 12.1

Work in a Gas Cylinder, cont.• A force is applied to

slowly compress the gas– The compression is slow

enough for all the system to remain essentially in thermal equilibrium

• W = - P ΔV– This is the work done on

the gas where P is the pressure throughout the gas

Section 12.1

More about Work on a Gas Cylinder

• When the gas is compressed– ΔV is negative – The work done on the gas is positive

• When the gas is allowed to expand– ΔV is positive– The work done on the gas is negative

• When the volume remains constant– No work is done on the gas

Section 12.1

Work By vs. Work On

• The definition of work, W, specifies the work done on the gas– This definition focuses on the internal energy of

the system

• Wenv is used to denote the work done by the gas– The focus would be on harnessing a system’s

internal energy to do work on something external to the gas

• W = - Wenv Section 12.1

Notes about the Work Equation

• The pressure remains constant during the expansion or compression– This is called an isobaric process

• The previous work equation can be used only for an isobaric process

Section 12.1

EXAMPLE 12.1 Work Done by an Expanding Gas

(a) A gas in a cylinder occupying a volume V at a pressure P. (b) Pushing the piston down compresses the gas. Goal Apply the definition of work at constant pressure. Problem In a system similar to that shown in the figure, the gas in the cylinder is at a pressure equal to 1.01 105 Pa and the piston has an area of 0.100 m2. As energy is slowly added to the gas by heat, the piston is pushed up a distance of 4.00 cm. Calculate

the work done by the expanding gas on the surroundings, Wenv, assuming the pressure remains constant. Strategy The work done on the environment is the negative of the work done on the gas. Compute the change in volume and multiply by the pressure.

SOLUTION

Find the change in volume of the gas, ΔV, which is the cross-sectional area times the displacement. ΔV = AΔy = (0.100 m2)(4.00 10-2 m) = 4.00 10-3 m3

Multiply this result by the pressure, getting the work the gas does on the environment, Wenv. Wenv = PΔV = (1.01 105 Pa)(4.00 10-3 m3) = 404 J

LEARN MORE

Remarks The volume of the gas increases, so the work done on the environment is positive. The work done on the system during this process is W = −404 J. The energy required to perform positive work on the environment must come from the energy of the gas. Question Which of the following is/are true if no energy is added as heat during the expansion of an ideal gas? (Select all that apply.)

The temperature of the gas must change because of changes in

internal energy. The pressure cannot remain constant. The

expanding gas does work on the walls of its container. The pressure

need not change. The temperature of the gas need not change.

PV Diagrams• Used when the pressure

and volume are known at each step of the process

• The work done on a gas that takes it from some initial state to some final state is equal in magnitude to the area under the curve on the PV diagram– This is true whether or not the

pressure stays constant

Section 12.1

PV Diagrams, cont.

• The curve on the diagram is called the path taken between the initial and final states

• The work done depends on the particular path– Same initial and final states, but different amounts of work are done

Section 12.1

First Law of Thermodynamics

• Energy conservation law• Relates changes in internal energy to energy

transfers due to heat and work• Applicable to all types of processes• Provides a connection between microscopic

and macroscopic worlds

Section 12.2

First Law, cont.

• Energy transfers occur– By doing work• Requires a macroscopic displacement of an object

through the application of a force– By heat• Due to a temperature difference• Usually occurs by radiation, conduction and/or

convection– Other methods are possible

• All result in a change in the internal energy, DU, of the system

Section 12.2

First Law, Equation

• If a system undergoes a change from an initial state to a final state, then DU = Uf – Ui = Q + W– Q is the energy transferred between the system

and the environment– W is the work done on the system– DU is the change in internal energy

Section 12.2

First Law – Signs • Signs of the terms in the equation– Q

• Positive if energy is transferred into the system• Negative if energy is removed from the system

– W• Positive if work is done on the system• Negative if work is done by the system

– DU• Positive if the temperature increases• Negative if the temperature decreases

Section 12.2

Notes About Work

• Positive work increases the internal energy of the system

• Negative work decreases the internal energy of the system

• This is consistent with the definition of mechanical work

Section 12.2

EXAMPLE 12.2 Work and PV Diagrams

Goal Calculate work from a PV diagram. Problem Find the numeric value of the work done on the gas in (a) Figure a, and (b) Figure b. Strategy The regions in question are composed of rectangles and triangles. Use basic geometric formulas to find the area underneath

each curve. Check the direction of the arrow to determine signs.

SOLUTION

(a) Find the work done on the gas in Figure a. Compute the areas A1 and A2 in Figure a. A1 is a rectangle and A2 is a triangle. A1 = height width

A1 = (1.00 105 Pa)(2.00 m3) = 2.00 105 J

A2 = ½ base height

A2 = ½(2.00 m3)(2.00 105 Pa) = 2.00 105 J

Sum the areas (the arrows point to increasing volume, so the work done on the gas is negative). Area = A1 + A2 = 4.00 105 J

W = −4.00 105 J

(b) Find the work done on the gas in Figure b.

Compute the areas of the two rectangular regions.

A1 = height width

A1 = (1.00 105 Pa)(1.00 m3) = 1.00 105 J

A2 = height width

A2 = (2.00 105 Pa)(1.00 m3) = 2.00 105 J

Sum the areas (the arrows point to decreasing volume, so the work done on the gas is positive).

Area = A1 + A2 = 3.00 105 J

W = +3.00 105 J

LEARN MORE

Remarks Notice that in both cases the paths in the PV diagrams start and end at the same points, but the answers are different. Question Is work done on a system during a process in which its volume remains constant? Explain. (Select all that apply.)

Work requires not only the force from pressure acting over an area,

but also a displacement. Heating a gas at constant volume does work.

No work is done during such a process. Work is done during

such a process. Changing the pressure of a gas at constant volume does work.

Read and take notes on pages: 388-389

EXAMPLE 12.3 Heating a Gas

Goal Combine the first law of thermodynamics with work done during a constant pressure process. Problem An ideal gas absorbs 5.00 103 J of energy while doing 2.00 103 J of work on the environment during a constant pressure process. (a) Compute the change in the internal energy of the gas. (b) If the internal energy now drops by 4.50 103 J and 7.50 103 J is expelled from the system, find the change in volume, assuming a constant pressure process at 1.01 105 Pa. Strategy Part (a) requires substitution of the given information into the first law. Notice, however, that the given work is done on the environment. The negative of this amount is the work done on the system, representing a loss of internal energy. Part (b) is a matter of substituting the equation for work at constant pressure into the first law and solving for the change in volume.

SOLUTION

(a) Compute the change in internal energy of the gas. Substitute values into the first law, noting that the work done on the gas is negative. ΔU = Q + W = 5.00 103 J - 2.00 103 J = 3.00 103 J

(b) Find the change in volume, noting that ΔU and Q are both negative in this case.

Substitute the equation for work done at constant pressure into the first law.

ΔU = Q + W = Q - PΔV

-4.50 103 J = -7.50 103 J - (1.01 105 Pa)ΔV

Solve for the change in volume, ΔV.

ΔV = -2.97 10-2 m3

LEARN MORE

Remarks The change in volume is negative, so the system contracts, doing negative work on the environment, whereas the work W on the system is positive. Question When an ideal gas expands at constant pressure, is the change in the internal energy positive or negative? mc

It can be either positive or negative depending on other details.

It is positive. It is negative.

Read and take notes on pages: 390-391

EXAMPLE 12.4 Expanding Gas

Problem Suppose a system of monatomic ideal gas at 2.00 105 Pa and an initial temperature of 293 K slowly expands at constant pressure from a volume of 1.00 L to 2.50 L. (a) Find the work done on the environment. (b) Find the change in internal energy of the gas. Strategy This problem mainly involves substituting values into the appropriate equations. Substitute into the equation for work at constant pressure to obtain the answer to part (a). In part (b) use the ideal gas law twice: to find the temperature when V = 2.00 L and to find the number of moles of the gas. These quantities can then be used to obtain the change in internal energy, ΔU.

SOLUTION

(a) Find the work done on the environment. Apply the definition of work at constant pressure. Wenv = PΔV = (2.00 105 Pa)(2.50 10-3 m3 - 1.00 10-3 m3)

Wenv = 3.00 102 J

(b) Find the change in the internal energy of the gas.

First, obtain the final temperature, using the ideal gas law, noting that Pi = Pf.

PfVf = Tf → Tf = Ti

Vf = (293 K) (2.50 10-3 m3)

PiVi Ti Vi (1.00 10-3m3)

Tf = 733 K

Again using the ideal gas law, obtain the number of moles of gas.

n = PiVi =

(2.00 105 Pa)(1.00 10-3 m3) = 8.21 10-2 mol

RTi (8.31 J/K · mol)(293 K) Use these results and given quantities to calculate the change in internal energy, ΔU.

ΔU = nCvΔT = 3/2nRΔT

= 3/2(8.21 10-2 mol)(8.31 J/K · mol)(733 K - 293 K)

ΔU = 4.50 102 J

LEARN MORE

Remarks Notice that problems involving diatomic gases are no harder than those with monatomic gases. It's just a matter of adjusting the molar specific heats. Question Which of the following is true about a constant pressure compression of an ideal gas? (Select all that apply.)

The internal energy of the ideal gas must decrease The temperature of an ideal

gas must always decrease during the process The gas must always exhaust thermal

energy (Q < 0) The internal energy can increase because of the work being done