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Chapter 15 Temperature and Heat. Mechanics vs. Thermodynamics Mechanics: obeys Newton’s Laws key...
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Transcript of Chapter 15 Temperature and Heat. Mechanics vs. Thermodynamics Mechanics: obeys Newton’s Laws key...
Chapter 15
Temperature and Heat
Mechanics vs. Thermodynamics
• Mechanics:• obeys Newton’s Laws• key concepts:
force kinetic energy static equilibrium
Newton’s 2nd Law
• Thermodynamics:• will find new ‘laws’• key concepts:
temperature, heatinternal energy thermal equilibrium
2nd Law of Thermodynamics
Temperature (T)
• Temperature = a macroscopic quantity
• (see later: T is related to KE of particles)
• many properties of matter vary with T: (length, volume, pressure of confined gas)
Temperature (T)
• Human senses can be deceiving
• On a cold day: iron railings feel colder than wooden fences, but both have the same T
• How can we define T ?
• Look for macroscopic changes in a system when heat is added to it
Two ThermometersAdd heat to (a) and (b).
(a) liquid thermometer• liquid level rises• T is measured by L
(b) constant volume gas thermometer
• gas pressure p rises• T is measured by p
Using Thermometers• put the bulb of (a) in
contact with a body
• wait until the value of L (i.e. T) settles out
• the thermometer and the body have reached thermal equilibrium (they have the same T)
• Consider thermal interactions of systems in (a).• red slab = thermal conductor (transmits interactions)• blue slab = thermal insulator (blocks interactions)
DemonstrationDemonstration
• Let A and C reach thermal equilibrium (TA=TC).
• Let B and C reach thermal equilibrium (TB=TC).
• Then are A and B in thermal equilibrium (TA=TB)?
DemonstrationDemonstration
• In (a), are A and B in thermal equilibrium?• Yes, but it’s not obvious!• It must be proved by experiment!
DemonstrationDemonstration
• Experimentally, consider going from (a) to (b):• Thermally couple A to B and thermally decouple C.• Experiments reveal no macroscopic changes in A, B!
DemonstrationDemonstration
• This suggests the Zeroth Law of Thermodynamics: • If C is in thermal equilibrium with both A and B,
then A and B in thermal equilibrium with each other.
DemonstrationDemonstration
• This means: If two systems A and B are in thermal equilibrium, they must have the same temperature (TA=TB), and vice versa
DemonstrationDemonstration
Temperature Scales
Temperature Scales
• Three scales: Fahrenheit, Celsius, Kelvin
• To define a temperature scale, we need one or more thermodynamic fixed points
• fixed point = a convenient, reproducible thermodynamic environment
Temperature Scales
• Both Fahrenheit and Celsius scales are defined using two fixed points:
• freezing point and boiling point of water
• Kelvin scale defined using one fixed point:
• ‘triple point’ of water (all three phases coexist: ice, liquid, vapor)
Temperature Scales: Summary
• Relations among temperature scales:
• Fahrenheit temperature
• Celsius temperature
• Kelvin temperature 15.273
)32(9
5
325
9
CK
FC
CF
TT
TT
TT
Temperature Scales:Kelvin vs. Celsius
• triple point of water:
• we measure TC, triple = 0.01oC
• we define TK, triple = 273.16 K
• (T)K = (T)C so the unit of T is K or oC
• the scales differ only by an offset, so: TK = TC +273.15
Kelvin Temperature Scale• Fixed point = triple point of water: TK, triple
• p = pressure of ‘ideal’ (i.e. low density) gas (on a constant volume gas thermometer) (has value ptriple at TK, triple)
• We define:
pp
T
pT
triple
K 16.273kelvins)(in
(constant)kelvins)(in
• At low density, see same graph for all gases• Extrapolate to p=0 (at T = absolute zero K)
DemonstrationDemonstration
Thermal Expansion
Thermal Expansion
• Empirical law for solids, valid for small T
• (simple case: all directions expand equally)
For > 0:
• If T > 0: L > 0 , material expands
• If T < 0: L < 0 , material compresses
TL
L
0
Thermal Expansion
= coefficient of linear expansion > 0 (almost always)
• characterizes thermal properties of matter
• varies with material (and range of T)
• unit: 1/K, or 1/oC since (T)K = (T)C
TL
L
0
Thermal Expansion
• Example: two different materials have different L
• They can be used to build a thermometer or a thermostat
TL
L
0
• Atomic explanation of thermal expansion!• Recall ‘spring’ model for diatomic molecule:• Van der Waals potential energy, U
DemonstrationDemonstration
Thermal Expansion• Similar for a solid
made of many atoms
• Each pair of atoms has a potential energy U
• The asymmetry of U explains thermal linear expansion!
Thermal Volume Expansion:Solids and Liquids
• = coefficient of volume expansion
• varies with material (and range of T)
• unit: 1/K, or 1/oC since (T)K = (T)C
TV
V
0
Thermal Volume Expansion:Solids
• Find a simple relationship between linear and volume expansion coefficients:
• = 3
T
TV
V
3
0
Thermal Expansion of Water
• ‘unusual’ state:• < 0 if
0o C < T < 4o C
• (it’s why lakes freeze from the top down)
TV
V
3
0
Thermal Stress
• Thermal stress= stress required to counteract (balance) thermal expansion
• Tensile thermal stress:
TYA
F
Announcements
• Midterms:will probably be returned Monday
• Homework 5: is returned at front
• Homework Extra Credit: is on record (but not yet listed on classweb if it brings a score over the maximum)
Temperature Scales:Kelvin vs. Celsius
• triple point of water:
• we measure TC, triple = 0.01oC
• we define TK, triple = 273.16 K
• (T)K = (T)C so the unit of T is K or oC
• the scales differ only by an offset, so: TK = TC +273.15
Heat and Heat Transfer
Quantity of Heat (Q)
• Heat = energy absorbed or lost by a body due to a temperature difference
• Heat = energy ‘in transit’
• SI unit: J• other units: 1 cal = 4.186 J
1 kcal = ‘calorie’ on food labels
Quantity of Heat (Q)
• Q > 0: heat is absorbed by a body
• Q < 0: heat leaves a body
• (we will see several expressions for Q)
Quantity of Heat (Q)
• Conservation of energy (‘calorimetry’):
• For an isolated system, the algebraic sum of all heat exchanges add to zero
Q1 + Q2 + Q3 + ... = 0
Absorption of Heat
• Q = heat energy required to change the temperature of material (mass m) by T
• c = ‘specific heat capacity’ of the material (treat as independent T) unit: J/(kg ·K)
TmcQ
Absorption of Heat
• If Q and T positive: heat absorbed by m
• If Q and T negative: heat leaves m
dT
dQ
mc
dTmcdQ
TmcQ
1
Do Exercise 15-35Do Exercise 15-35
Phase Changes
• ‘phase’ = state of matter = solid, liquid, vapor
• energy is needed to change phase of matter
• under a phase transition of matter:only its phase changes, not its temperature!
Phase Changes in Water
Solid-Liquid Phase Change:
Q = ± mLf • ± mLf = heat needed for phase change
• Lf = ‘(latent) heat of fusion’ of the material = (heat/unit mass) needed for transition unit: J/kg
• + for melting (solid to liquid)– for freezing (liquid to solid)
Do Exercise 15-51Do Exercise 15-51
Liquid-Vapor Phase Change:
Q = ± mLv • ± mLv = heat needed for phase change
• Lv = ‘(latent) heat of vaporization’ = (heat/unit mass) needed for transition unit: J/kg
• + for evaporating (liquid to vapor)– for condensing (vapor to liquid)
Heat Transfer
Heat Transfer
dQ/dt = rate of heat flow = ‘heat current’
Three mechanisms for achieving heat transfer:
• Conduction
• Convection
• Radiation
Heat Transfer Mechanisms
• Conduction: Collisions of molecules, no bulk motion
• Convection:Bulk motion from one region to another
• Radiation:Emission of electromagnetic waves
Conduction
Conduction
• k = thermal conductivity of material unit: W/(m·K)
• A = cross sectional area of material
• L = length of material
L
TTkA
dt
dQH CH
Conduction
L
TTkA
dt
dQH
k
CH
K) W/(m:unit ty,conductivi thermal
Do Exercises 15-57, 15-58Do Exercises 15-57, 15-58 Notes on a composite conducting rodNotes on a composite conducting rod
Convection (usually complicated)
Radiation (e.g. emitted by the sun)
Radiation =Electromagnetic Waves
Emission of Radiation
• all bodies emit electromagnetic radiation
• A = surface area of body
• T = surface temperature of body
• e = emissivity of body (0 < e < 1)
)KW/(m10675 428
4
.
TAedt
dQH
-
Do Exercise 15-67Do Exercise 15-67
Absorption of Radiation
• In general, bodies emit radiation and also absorb radiation from their surroundings
• T = surface temperature of body
• TS = surface temperature of surroundings
)( 4S
4
4S
4net
TTAe
TAeTAeH
Example of net radiation and Problem 15-89Example of net radiation and Problem 15-89