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