Download Heat

Raymond A. Serway
Chris Vuille
Chapter Eleven
Energy 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 to
Bright 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 to
Bright 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.
conductivity
True. The rate of heat flow in a material is proportional to its thermal
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 - To4)
– 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. Serway
Chris Vuille
Chapter Twelve
The 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 # 2
More on Internal 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.
expanding gas does work on the walls of its container.
need not change.
The
The pressure
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.
such a process.
work.
Work is done during
Changing the pressure of a gas at constant volume does
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.
(2.50
PfVf Tf
Vf
=
→ Tf = Ti = (293 K) (1.00
PiVi Ti
Vi
10-3 m3)
10-3m3)
Tf = 733 K
Again using the ideal gas law, obtain the number of moles of gas.
PiVi (2.00 105 Pa)(1.00 10-3 m3)
=
= 8.21
n=
(8.31 J/K · mol)(293 K)
RTi
10-2 mol
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
gas must always decrease during the process
energy (Q < 0)
The temperature of an ideal
The gas must always exhaust thermal
The internal energy can increase because of the work being done