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Thermodynamics
Thermodynamics
Thermodynamics – What is it?
The branch of physics that is built upon the fundamental laws that heat and work obey.
Heat & Internal Energy
Heat is energy that flows from a higher-temperature object to a lower temperature object due to the differences in temperature.
Heat is positive when a system gains heat and negative when it loses heat.
Internal energy is a measure of the temperature of a substance due to the kinetic energy of the individual molecules or atoms as well as potential energy of these particles .
Zeroth Law of Thermodynamics
Thermal Equilibrium:• If two object have the same temperature, then there
will be no flow of heat between the two systems.
• If two objects are in thermal equilibrium with a third object, then they must be in thermal equilibrium with each other.
• When two objects are brought in contact with one another, heat will flow from the warmer object to the cooler one.
Temperature Scales
A common scale used for measuring and recording temperature is Celsius.
• Water freezes at 0°C and boils at 100°C. To convert from °F to °C:
• T (°C) = 5/9 (T(°F) – 32) The Kelvin temperature scale is of more significances than
the Celsius scale.
• It is measured in the same increments as the Celsius scale, but is measured from absolute 0.
• The temperature at which water freezes (ice point) is 273.15 K / 0°C, and the temperature at which water boils (steam point) is 373.15 K / 100°C.
Heat and Energy
What is Heat? What is Energy? Is there a difference?
• Yes
• Heat is a measure of the transfer of energy from one object that is at a higher temperature to another that is at a lower energy level.
• Energy is a measure of the average kinetic energy of all the particles in a gas sample, the potential energy that exists between the atoms and molecules, etc.
• Objects do not contain heat, they contain energy.
How can heat be transferred?
Convection: The process of transferring heat by the bulk movement of a gas or liquid.
Conduction: The process of transferring heat through a body whereby the bulk motion of the material plays no role in the transfer.
Radiation: Process by which energy is transferred by means of electromagnetic radiation.
Phases of Matter
Matter comes in three phases – gas, liquid and solid.• When a substance undergoes a phase change, the
temperature does not change.
• Solids melt to become liquid or sublimate to become a vapor.
• Liquids evaporate to become a vapor or freeze to become a solid.
• Gases condense to become a liquid, or go through deposition to become a solid.
Heat Transfer – Change in T
Heat is measure of the amount of energy that flows to or from a substance is dependent on the temperature, mass of the object and the specific heat (c).
Q = mcΔT• c is called the specific heat and is unique for
every substance.
• If ΔT is positive, then Q is positive and heat flows into the substance.
Phase ChangesHeat Transfer – No Change in T
When a substance goes through a phase change, the temperature does not change, so the formula Q = mcΔT will not hold true. Instead:
Q = mL• Where
• L = Latent Heat of Transformation
• For a solid to liquid L is called the heat of fusion
• For a liquid to gas L is called the heat of vaporization
Linear Thermal Expansion
Most materials will expand when heated and shrink when cooled.• Railroad tracks, bridges and pipelines are a few
examples of things that expand when heated.
L = Lf – Li = Li (Tf – Ti)Where:
L is the length.T is the temperature. is the coefficient of linear expansion.
Example 1: Linear Expansion
How much will the length of a 30 m beam change as the temperature rises from 15°C to 35°C? The coefficient of linear expansion of the steel beam is 1.2 x 10-5/°C.
L = Li (Tf – Ti)
L = (1.2 x 10-5/°C)(30m)(35°C - 15°C)
L = 7.2 x 10-3 m = 7.2 mm
Volume Thermal Expansion
Since expansion should not be limited to one dimension, a similar expression to linear expansion exists for volumetric expansion.
V = Vi (Tf – Ti)
• For many solids, ~ 3
14.2 The Ideal Gas Law
An ideal gas is an idealized model for real gases that have sufficiently low densities.
The condition of low density means that the molecules are so far apart that they do not interact except during collisions, which are effectively elastic.
TP
At constant volume the pressureis proportional to the temperature.
At constant temperature, the pressure is inversely proportional to the volume.
VP 1
The pressure is also proportional to the amount of gas.
nP
14.2 The Ideal Gas Law
The absolute pressure of an ideal gas is directly proportional to the Kelvin temperature and the number of moles of the gas and is inversely proportional to the volume of the gas.
V
nRTP
nRTPV
14.2 The Ideal Gas Law
: AA
RSince N nN and k
N
UniversalGas Constant 8.314 /( )R J mol K
For an alternate representation:
PV NkTk = Boltsmann’s constant = 1.38 x 10-23 J/K:
Consider a sample of an ideal gas that is taken from an initial to a final state, with the amount of the gas remaining constant.
nRTPV
i
ii
f
ff
T
VP
T
VP
constant nRT
PV
14.2 The Ideal Gas Law
i
ii
f
ff
T
VP
T
VP
14.2 The Ideal Gas Law
Constant P, constant n:
i
i
f
f
T
V
T
V
Charles’ law
Constant T, constant n:iiff VPVP Boyle’s law
•The particles are in constant, random motion, colliding with each other and with the walls of the container.
•Each collision changes the particle’s speed.
•As a result, the atoms and molecules have different speeds.
14.3 Kinetic Theory of Gases
14.3 Kinetic Theory of Gases
Using the Impulse-Momentum Theorem:
L
mv
vL
mvmv 2
2
collisions successivebetween Time
momentum Initial-momentum Final force Average
t
mv
t
vmmaF
14.3 Kinetic Theory of Gases
L
mvF
2
For a single molecule, the average force is:
For N molecules moving randomly in three dimensions, the average force is:
L
vmNF
2
3 root-mean-squarespeed
3
2
2 3 L
vmN
L
F
A
FP
volume
14.3 Kinetic Theory of Gases
V
vmNP
2
3
221
322
31
rmsrms mvNmvNPV
NkT KE
kTmvrms 232
21KE
• This formula shows that the kinetic energy associated with the translational motion of a particle can be determined if we know the temperature of the system.
Average Speed of a Particle
Since KEavg = ½ mv2, we can find the average speed of a particle.
3 B
rms
k Tv
m
avg B
3KE = k T
2
Internal Energy of a Monatomic Ideal Gas
Internal Energy: The sum of the molecular kinetic energy, the molecular potential energy, and other kinds of energy.
U = 3/2 nRT
Where:
n = number of moles.
R = gas constant
T = Temperature (K)
Note: A substance has internal energy, not “Heat”.
Note: Internal energy depends only on the state of a system, and not on the method by which the system got their.
Note: Formula applies to monatomic ideal gases.
The 1st Law of Thermodynamics
Conservation of energy says that energy cannot be created nor destroyed, but may be converted from one form to another.
The 1st Law of Thermodynamics (cont.)
If a system only gains heat:
U = Uf – Ui = Q
Heat is positive when a system gains heat and negative when it loses heat.
The 1st Law of Thermodynamics (cont.)
If a system only does work on its surroundings and no heat is transferred:
U = Uf – Ui = -W
• Work is positive when it is done by the system on the surroundings and negative when it is done on the system by the surroundings.
W = PV where:
P = pressureV = volume
The 1st Law of Thermodynamics (cont.)
A system that consists of changes to both heat (internal energy) and work simultaneously is represented by:
ΔU = Uf – Ui = Q - W
First Law of Thermodynamics = Law of Conservation of energy.
Example 2: 1st Law of Thermo. The temperature of 3 moles of a monatomic ideal gas is
reduced from Ti = 540K to Tf = 350K when 5500 J of heat flows into the gas. What is the change in internal energy (ΔU) and work (W) done by the gas?
Using 1st Law of Thermodynamics: ΔU = Q – W
• W = Q – ΔU = 5500J – (-7100J) = 12,600J
3( )
2f iU nR T T
3(3.0 )(8.314 / )(350 540 )
27100
U moles J mole K K K
U J
Thermal Processes – Constant Pressure
An isobaric process is one that occurs at constant pressure.
ΔU = Q - W
Since the pressure is constant, the amount of work done is:
W = PV = P(Vf – Vi)
ΔU = Q – PΔV
Example 3: Isobaric Process 1 Gram of water is placed inside a cylinder where the pressure
is maintained at 2.0 x 105 Pa. The temperature of the water is raised by 31 C°. The water is in the gaseous phase and expands by 7.1 x 10-5 m3, and has a specific heat capacity of 2020J/(kg·C°). What is the change in internal energy of the water?
For an isobaric process we will use:
• Q = cmΔT for heat added.
• W = PΔV for work done by the gas. Using 1st Law of Thermodynamics:
• ΔU = Q - W
• ΔU = cmΔT - PΔV
ΔU = cmΔT – PΔV
ΔU = (2020 J/(kg·C°)(0.0010 kg)(31 C°) – (2.0 x 105 Pa)(7.1 x 10-5 m3)
ΔU = 63 J – 14 J
ΔU = 49 J
Table of Known & Unknowns
Mass of water m 0.0010 kg
Pressure on water P 2.0 x 105 Pa
Increase in T ΔT 31C°
Increase in V ΔV 7.1 x 10-5 m3
Specific Heat of H2O gas cgas 2020 J/(kg·C°)
ΔU of gas ΔUgas ?
Thermal Processes – Constant Volume
An isochoric process is one that occurs at constant volume.
Since the volume (ΔV) is 0, the work is zero.
U = Q - W = Q
U = 3/2nR(Tf – Ti)
Thermal Processes – Constant Temperature
An isothermal process is one that occurs at constant temperature.
U = Q - W Where W = nRT*ln (Vf/Vi)
U = Q - nRT*ln (Vf/Vi) For an ideal gas:
U = 3/2nR(Tf – Ti) Since the temperature does
not change, U = 0. Therefore:
W = Q = nRT*ln (Vf/Vi)
Example 4: Isothermal Expansion 2 moles of the monatomic gas argon expand isothermally at
298 K, from the initial volume of Vi = 0.025 m3 to a final volume of Vf = 0.050 m3. Assuming that argon is an ideal gas, find the work done by the gas, the change in internal energy of the gas and the heat supplied by the gas.
For an isothermal process we will use:
Since the temperature does not change,
the change in internal energy will be 0.
lnf
i
VW nRT
V
3( ) 0
2f iU nR T T
Using the 1st Law of Thermodynamics: ΔU = Q - W
• Since ΔU = 0:
Q = W
Table of Known & Unknowns
Initial Volume Vi 0.025 m3
Final Volume Vf 0.050 m3
Temperature T 298 K
Gas Constant R 8.314 J/(kg·K)
Heat supplied to gas Q ?
lnf
i
VQ W nRT
V
3
3
0.050(2 )(8.314 /( )(298 ) ln 3400
0.025
mQ moles J mole K K J
m
Thermal Processes – Adiabatic
An adiabatic process is one that occurs without the transfer of heat (Q = 0).
U = Q - W = -W
and since
U = 3/2nRT
W = -U = 3/2nR(Ti – Tf)
If a gas expands adiabatically, then the final temperature will be lower than the initial temperature.
1st Law Relationships
Specific Heat Capacities
Specific Heat: A measure of the amount of heat (Q) that must be supplied to change the temperature of a substance by an amount T.
• First Law: U = Q – W
• Solving for Heat: Q = U + W (1)
• Using Specific Heat:Q = nCT (2)• Where mc = nC
Specific Heat Capacities (const. P)
• From 1st Law: U = Q – W
Q = U + PV
(3)
• Substituting (2) for Q in (3) and solving for C gives:
3( ) ( )
2f i f iQ nR T T nR T T
5( )
2f iQ nR T T
5( ) ( )
2p f i f inC T T nR T T
5
2pC R
Note: This relationship holds true for monatomic ideal gases.
Specific Heat Capacities (const. V)
• From 1st Law: U = Q – W
Q = U + PV (Since V = 0, W = 0)
(3)
• Substituting (2) for Q in (3) and solving for C gives:
3( ) 0
2f iQ nR T T
3( )
2f iQ nR T T
3( ) ( )
2v f i f inC T T nR T T
3
2vC R
Note: This relationship holds true for monatomic ideal gases.
Specific Heat
An interesting outcome results when you evaluate the difference between Cp and Cv:
• The reason is due to the fact that work is done under constant pressure where no work is done when the volume is constant.
5 3
2 2R R R
p vC C R
2nd Law of Thermodynamics
Heat Flows Spontaneously from a substance at higher temperature to a substance at a lower temperature, while the reverse is not true.
In other words, natural processes go in a direction that maintains or increases the entropy of the universe. Entropy: A measure of disorder or
randomness of a system.
Note: Reversible processes do not change the entropy of a system,….or universe.
Entropy
15.8 Heat Engines
A heat engine is any device that uses heat to perform work. It has three essential features.
1. Heat is supplied to the engine at a relatively high temperature from a place called the hot reservoir.
2. Part of the input heat is used to perform work by the working substance of the engine (gasoline, steam, etc).
3. The remainder of the input heat is rejected to a place called the cold reservoir.
heatinput of magnitude HQ
heat rejected of magnitude CQ
done work theof magnitude W
15.8 Heat Engines
The efficiency of a heat engine is defined asthe ratio of the work done to the input heat:
HQ
We
If there are no other losses, then
CH QWQ
H
C
Q
Qe 1
15.8 Heat Engines
Example 5 An Automobile Engine
An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine?
HQ
We
CH QWQ
e
WQH
15.8 Heat Engines
CH QWQ
e
WQH
J 89001220.0
1J 2510
11
eWW
e
WQC
WQQ HC
15.9 Carnot’s Principle and the Carnot Engine
A reversible process is one in which both the system and the environment can be returned to exactly the states they were in before the process occurred.
CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECONDLAW OF THERMODYNAMICS
• No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures.
• Furthermore, all reversible engines operating between the same temperatures have the same efficiency.
15.9 Carnot’s Principle and the Carnot Engine
The Carnot engine is useful as an idealizedmodel.
All of the heat input originates from a singletemperature, and all the rejected heat goesinto a cold reservoir at a single temperature.
Since the efficiency can only depend onthe reservoir temperatures, the ratio of heats can only depend on those temperatures.
H
C
H
C
T
T
Q
Qe 11
H
C
H
C
T
T
Q
Q
15.9 Carnot’s Principle and the Carnot Engine
Example 6 A Tropical Ocean as a Heat Engine
Water near the surface of a tropical ocean has a temperature of 298.2 K, whereasthe water 700 meters beneath the surface has a temperature of 280.2 K. It hasbeen proposed that the warm water be used as the hot reservoir and the cool wateras the cold reservoir of a heat engine. Find the maximum possible efficiency forsuch an engine.
H
C
T
Te 1carnot
060.0K 298.2
K 2.28011carnot
H
C
T
Te
15.9 Carnot’s Principle and the Carnot Engine
Conceptual Example 7 Natural Limits on the Efficiency of a Heat Engine
Consider a hypothetical engine that receives 1000 J of heat as input from a hot reservoir and delivers 1000J of work, rejecting no heat to a cold reservoirwhose temperature is above 0 K. Decide whether this engine violates the firstor second law of thermodynamics.
H
C
T
Te 1carnot
• As far as the first law is concerned, it is not violated since in theory, all the input heat can contribute to work being done by the system on the environment.
• However, the second law is violated since you can never lower the temperature to true absolute 0. Hence the efficiency cannot be truly equal to 1.
Key Ideas Thermal equilibrium exists between two bodies when they
both have the same temperature. Heat can be transferred through convection, conduction and
radiation. The 1st Law of Thermodynamics is based upon the principle
of Conservation of Energy. The 2nd Law of Thermodynamics says that heat will flow
spontaneously from a substance at a higher temperature to a substance at a lower temperature.
Irreversible processes lead to an increase in entropy of the universe.
