Download Ch. 15 - UCSB Physics

Chapter 15
Temperature and Heat
Mechanics vs. Thermodynamics
• Mechanics:
• obeys Newton’s Laws
• key concepts:
force
kinetic energy
static equilibrium
• Thermodynamics:
• will find new ‘laws’
• key concepts:
temperature, heat
internal energy
thermal equilibrium
Newton’s 2nd Law
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 Thermometers
Add 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)
Demonstration
• 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)?
Demonstration
• In (a), are A and B in thermal equilibrium?
• Yes, but it’s not obvious!
• It must be proved by experiment!
Demonstration
• Experimentally, consider going from (a) to (b):
• Thermally couple A to B and thermally decouple C.
• Experiments reveal no macroscopic changes in A, B!
Demonstration
• 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.
Demonstration
• This means: If two systems A and B are in thermal
equilibrium, they must have the same temperature
(TA=TB), and vice versa
Demonstration
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
9
TF  TC  32
5
• Celsius temperature
5
TC  (TF  32 )
9
• Kelvin temperature
TK  TC  273.15
Temperature Scales:
Kelvin vs. Celsius
• triple point of water:
• we measure TC, triple = 0.01oC
• we define
TK, triple = 273.16 K
• (DT)K = (DT)C so the unit of DT 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:
T (in kelvins)  (constant)  p
 273.16 K 
p
T (in kelvins)  
 p

triple


• At low density, see same graph for all gases
• Extrapolate to p=0 (at T = absolute zero K)
Demonstration
Thermal Expansion
Thermal Expansion
DL
 a DT
L0
• Empirical law for solids, valid for small DT
• (simple case: all directions expand equally)
For a > 0:
• If DT > 0: DL > 0 , material expands
• If DT < 0: DL < 0 , material compresses
Thermal Expansion
DL
 a DT
L0
a = 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 (DT)K = (DT)C
Thermal Expansion
DL
 a DT
L0
• Example: two
different materials
have different DL
• They can be used to
build a thermometer or
a thermostat
• Atomic explanation of thermal expansion!
• Recall ‘spring’ model for diatomic molecule:
• Van der Waals potential energy, U
Demonstration
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
DV
 b DT
V0
• b = coefficient of volume expansion
• varies with material (and range of T)
• unit: 1/K, or 1/oC since (DT)K = (DT)C
Thermal Volume Expansion:
Solids
DV
 b DT
V0
 3a DT
• Find a simple relationship between linear
and volume expansion coefficients:
• b = 3a
Thermal Expansion of Water
DV
 3a DT
V0
• ‘unusual’ state:
• a < 0 if
0o C < T < 4 o C
• (it’s why lakes
freeze from the
top down)
Thermal Stress
• Thermal stress= stress required to
counteract (balance) thermal expansion
• Tensile thermal stress:
F
 Ya DT
A
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
• (DT)K = (DT)C so the unit of DT 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  mc DT
• Q = heat energy required to change the
temperature of material (mass m) by DT
• c = ‘specific heat capacity’ of the material
(treat as independent T)
unit: J/(kg ·K)
Absorption of Heat
Q  mc DT
dQ  mc dT
1 dQ
c
m dT
• If Q and DT positive: heat absorbed by m
• If Q and DT negative: heat leaves m
Do 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 15-51
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
TH  TC
dQ
H
 kA
dt
L
• k = thermal conductivity of material
unit: W/(m·K)
• A = cross sectional area of material
• L = length of material
Do Exercises 15-57, 15-58
Notes on a composite conducting rod
Conduction
k  thermal conductivi ty, unit : W/(m  K)
TH  TC
dQ
H
 kA
dt
L
Convection (usually complicated)
Radiation
(e.g. emitted by the sun)
Radiation =
Electromagnetic Waves
Emission of Radiation
dQ
H
 AeT 4
dt
  5.67 10-8 W/(m 2  K 4 )
•
•
•
•
all bodies emit electromagnetic radiation
A = surface area of body
T = surface temperature of body
e = emissivity of body (0 < e < 1)
Do Exercise 15-67
Example of net radiation and Problem 15-89
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
H net  AeT  AeTS
4
 Ae (T 4  TS )
4
4