12/22/2011

1

Medical Physics

12/22/2011 1

Thermal Physics

Dr. Mohamed Al-Fadhali

Dr. Mohamed Al- Fadhali

2

I feel hot

He is hot

Properties of materials change with temperature– Length– Volume– Resistance

12/22/2011

2

Temperature ScalesThermometers can be calibrated by placing them in thermal contact with

an environment that remains at constant temperature– Environment could be mixture of ice and water in thermal

equilibrium– Also commonly used is water and steam in thermal equilibrium

Temperature of an ice-water mixture is defined as 0º CThis is the freezing point of water

Temperature of a water-steam mixture is defined as 100º CThis is the boiling point of water

Distance between these points is divided into 100 segments

Celsius Scale

Kelvin Scale

§ When the pressure of a gas goes to zero, its temperature is –273.15º C§ This temperature is called absolute zero§ This is the zero point of the Kelvin scale (–273.15º C = 0 K)

§ To convert: TC = TK – 273.15

Fahrenheit Scales§ Most common scale used in the US§ Temperature of the freezing point is 32º§ Temperature of the boiling point is 212º§ 180 divisions between the points

12/22/2011

3

Comparing Temperature Scales

273.159 325

95

C K

F C

F C

T T

T T

T T

= −

= +

∆ = ∆

Type of Thermometer

vChange in electrical resistance (convenient but not very linear)vChange in length of a bar (bimetallic strip)vChange in volume of a liquid vChange in volume of gas (very accurate but

slow and bulky)

12/22/2011

4

Thermal Expansion

• The thermal expansion of an object is a consequence of the change in the average separation between its constituent atoms or molecules

• At ordinary temperatures, molecules vibrate with a small amplitude

• As temperature increases, the amplitude increasesThis causes the overall object as a whole to expand

Rails expand and may buckle on a hot summer day

In most liquids or solids, when temperature risesmolecules have more kinetic energy

they are moving faster, on the averageconsequently, things tend to expand (works for a gas)

amount of expansion ∆L depends on…change in temperature ∆Toriginal length L0

coefficient of thermal expansionL0 + ∆L = L0 + α L0 ∆T∆L = α L0 ∆T (linear expansion)∆V = β V0 ∆T (volume expansion)β ≅ 3α

L0 ∆L

V

V + ∆V

12/22/2011

5

Linear (area, volume) Expansion

v For small changes in temperature

v The coefficient of linear expansion, depends on the material

v Similar in two dimensions (area expansion)

v … and in three dimensions (volume expansion)

tLL o ∆=∆ αα

αγγ 2, =∆=∆ tAA o

αββ 3,solidsfor =∆=∆ tVV o

Thermal Expansion and Teeth

Crazing:

Thermal expansion matching is important in choosing materials for fillings.

12/22/2011

6

Thermal Expansion and Teeth

Coefficients of linear expansion:

Enamel: 11.4 x 10-6 °CDentin: 8.3 x 10-6 °C

If you quickly switch from eating/drinkingsomething hot to somethingcold, the brittle enamel will contract morethan the dentin, and develop small cracks called crazes.

Example

A copper telephone wire has essentially no sag between poles 35.0 m apart on a winter day when the temperature is –20.0°C. How much longer is the wire on a summer day when TC = 35.0°C? Assume that the thermal coefficient of copper is constant throughout this range at its room temperature value.

12/22/2011

7

Applications of Thermal Expansion

1. Thermostats– Use a bimetallic strip– Two metals expand differently

2. Pyrex Glass– Thermal stresses are smaller than for ordinary glass

3. Train rails– Keeping space between joints for possible expansion and

contraction

Ideal Gas• Properties of gases

o A gas does not have a fixed volume or pressureo In a container, the gas expands to fill the container

• Ideal gas:o Collection of atoms or molecules that move randomlyo Molecules exert no long-range force on one anothero Molecules occupy a negligible fraction of the volume of their container

• Most gases at room temperature and pressure behave approximately as an ideal gas

Molesv It’s convenient to express the amount of gas in a given volume in terms of the

number of moles, n

v One mole is the amount of the substance that contains as many particles as there are atoms in 12 g of carbon-12

massmolarmassn =

12/22/2011

8

Avogadro’s Hypothesis

§ Equal volumes of gas at the same temperature and pressure contain the same numbers of molecules– Consequences: At standard temperature and pressure, one

mole quantities of all gases contain the same number of molecules

– This number is called Avogadro’s Number (NA =6.02 x 1023 particles / mole)

– Can also look at the total number of particles: N = n NA

The mass of an individual atom can be calculated as follows:

Aatom N

massmolarm =

What is the volume of 1 mol of gas at STP ?T = 0 °C = 273 Kp = 1 atm = 1.01 x 105 Pa

nRTpV =

( )

l4.22m0224.0

Pa10x01.1K273Kmol/J31.8

PRT

nV

3

5

==

⋅=

=

The Ideal Gas Law

12/22/2011

9

Equation of State for an Ideal Gas

v Boyle’s LawAt a constant temperature, pressure is

inversely proportional to the volumev Charles’ Law

At a constant pressure, the temperature is directly proportional to the volume

v Gay-Lussac’s LawAt a constant volume, the pressure is

directly proportional to the temperature

Ideal Gas Law

vSummarizes Boyle’s Law, Charles’ Law, and Guy-Lussac’s Law

PV = n R TR is the Universal Gas ConstantR = 8.31 J / mole KR = 0.0821 L atm / mole K

P V = N kB TkB is Boltzmann’s ConstantkB = R / NA = 1.38 x 10-23 J/ K

12/22/2011

10

Questions1. An ideal gas is confined to a container with constant volume. The number of moles is constant. By what factor will the pressure change if the absolute temperature triples?

a. 1/9b. 1/3c. 3.0d. 9.0

2. An ideal gas is confined to a container with adjustable volume. The number of moles and temperature are constant. By what factor will the volume change if pressure triples?

a. 1/9b. 1/3c. 3.0d. 9.0

Volume and Pressure of a Gas

In the kelvin scale, the lowest possible temperature is 0 K. (zero volume and zero pressure)

Any two temperatures defined by the ratiop1 T2 = p2 T1 or V1 T2 = V2 T1

The zero point is fixed -Absolution Zero (≈-273.15°C)

A bottle of hair spray is filled to a pressure of 1atm at 20°C

What is the canister pressure if it is placed into boiling water?

p1 T2 = p2 T1

1 x 373 = p2 x 293p2 = 373/293p2 = 1.27 atm

Example

12/22/2011

11

Pressure of an Ideal Gas

§ The pressure is proportional to the number of molecules per unit volume and to the average translational kinetic energy of a molecule

= 2mv

21

VN

23P

Molecular Interpretation of TemperatureØ Temperature is proportional to the average kinetic energy of the

molecules

Ø The total kinetic energy is proportional to the absolute temperature

Ø In a monatomic gas, the KE is the only type of energy the molecules can have

Ø U is the internal energy of the gasØ In a polyatomic gas, additional possibilities for contributions to the

internal energy are rotational and vibrational energy in the molecules

Tkmv B23

21 2 =

nRTKE total 23

=

nRTU23

=

12/22/2011

12

Speed of the Molecules

• Expressed as the root-mean-square (rms) speed

• At a given temperature, lighter molecules move faster, on average, than heavier ones• Lighter molecules can more easily reach escape speed

from the earth

MTR

mTkv B

rms33

==

Internal Energy vs. Heat

v Internal Energy, U, is the energy associated with the microscopic components of the system– Includes kinetic and potential energy associated with the random

translational, rotational and vibrational motion of the atoms or molecules

– Also includes the intermolecular potential energy

v Heat is energy transferred between a system and its environment because of a temperature difference between them– The system Q is used to represent the amount of energy transferred

by heat between a system and its environment

12/22/2011

13

Units of Heatv 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

1 cal = 4.186 J– This is called the Mechanical Equivalent of Heat

v BTU (US Customary Unit)– 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

UnitsUnitsSISI Joule (J)Joule (J)CGSCGS Calorie (cal)Calorie (cal)

Specific Heatv Every substance requires a unique amount of energy per

unit mass to change the temperature of that substance by 1° C– directly proportional to mass (thus, per unit mass)

v The specific heat, c, of a substance is a measure of this amount

TmQc∆

=

12/22/2011

14

Example1:How much heat is needed to raise temperature of aluminum by 5°C?

Given:

Mass: m=0.5 kgTemp. ∆T= 5°Specific heat: cAl =900 J/kg°C

Find:

Q=?

( )( )( ) JoulesCCkgJkg

TmcQ Al

225059005.0 +=+=

∆=oo

ü

Heat is related to mass and temperature by

Thus, energy is flowing into the system!

Calorimeter§ A technique for determining the specific heat of a

substance is called calorimetry § A calorimeter is a vessel that is a good insulator that allows

a thermal equilibrium to be achieved between substances without any energy loss to the environment

CalorimetryAnalysis performed using a calorimeterConservation of energy applies to the isolated systemThe energy that leaves the warmer substance equals the

energy that enters the waterQcold = -Qhot

Negative sign keeps consistency in the sign convention of ΔT

12/22/2011

15

A 0.010-kg piece of unknown metal heated to 100°C and dropped into the bucket containing 0.5 kg of water at 20°C. Determine specific heat of metal if the final temperature of the system is 50°C

Given:

Mass: m1=0.010 kgm2=0.5 kg

Specific heat (water): cW =4186 J/kg°CTemperatures:

T1=100 °CT2=20 °CTf=50 °C

Find:

Specific heat =?

( ) ( ) ( )( )( )( ) 0627905.0

205041865.01005001.0

0222

=+−=−+−=

∆+∆==+

JcCCCkgJkgCCckg

TcmTcmQQ

metal

metal

OHOHOHmetalmetalmetalmetalwater

ooooo

ü

Conservation of energy: heat lost by metal is the same as heat acquired by water:

Solve this equation:0=+ metalwater QQ

CkgJcmetalo51025.1 ×=

iron

Example:

Latent Heat

§ During a phase change, the amount of heat is given asQ = m L

§ L is the latent heat of the substanceLatent means hidden or concealed

§ Choose a positive sign if you are adding energy to the system and a negative sign if energy is being removed from the system

§ Latent heat of fusion is used for melting or freezing§ Latent heat of vaporization is used for boiling or

condensing

12/22/2011

16

Graph of Ice to Steam

Methods of Heat Transfer

vMethods includeØConductionØConvectionØRadiation

1. ConductionThe 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 from more energetic particles

Rate of conduction depends upon the characteristics of the substance

12/22/2011

17

Conduction example

§ The molecules vibrate about their equilibrium positions

§ Particles near the flame vibrate with larger amplitudes

§ These collide with adjacent molecules and transfer some energy

§ Eventually, the energy travels entirely through the rod

Conduction can occur only if there is a difference in temperature between two parts of the conducting medium

Conduction§ The slab allows energy to

transfer from the region of higher temperature to the region of lower temperature

LTTkA

tQP ch −

==

Heat flow Thermal conductivity

12/22/2011

18

• A is the cross-sectional area• L = Δx is the thickness of the slab or the

length of a rod• P is in Watts when Q is in Joules and t is in

seconds• k is the thermal conductivity of the material

• Good conductors have high k values and good insulators have low k values

LTTkA

tQP ch −

==

2. ConvectionvEnergy transferred by the

movement of a substance• When the movement

results from differences in density, it is called natural conduction

• When the movement is forced by a fan or a pump, it is called forced convection

12/22/2011

19

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

§ Applications:§ Radiators§ Cooling automobile

engines

3. Radiation

vRadiation does not require physical contactvAll objects radiate energy continuously in

the form of electromagnetic waves due to thermal vibrations of the moleculesvRate of radiation is given by Stefan’s Law

12/22/2011

20

Radiation example

• The electromagnetic waves carry the energy from the fire to the hands

• No physical contact is necessary

Radiation equation

P = σAeT4

– P is the rate of energy transfer, in Watts– σ = 5.6696 x 10-8 W/m2 K4

– 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

12/22/2011

21

Energy Absorption and Emission by Radiation

• With its surroundings, the rate at which the object at temperature T with surroundings at To radiates is

Pnet = σAe(T4 – T4o)

• When an object is in equilibrium with its surroundings, it radiates and absorbs at the same rate• Its temperature will not change

Example: Determine solar energy over the area of 1 m2. Temperature of Sun’s surface is 6000 K and temperature of surroundings is 300 K.

Given:

Area: A= 1 m2

Temperatures: T1=6000 KT2=300 K

Find:

Power =?ü

Use Stefan’s law:

( )40

4 TTAPower −= εσ

( )( )( )( )( )

sJKm

KKAPower

7

41528

44

103.7103.1111067.5

3006000

×=

××=

−=−

εσ

Temperature of Sun’s surface Temperature on the Earth

12/22/2011

22

Ideal Absorbers and Reflectors

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 An ideal reflector absorbs none of the energy

incident on ite = 0

Applications of Radiation

Ø Clothing• Black fabric acts as a good absorber• White fabric is a better reflector

Ø Thermography• The amount of energy radiated by an object can be

measured with a thermographØ Body temperature

• Radiation thermometer measures the intensity of the infrared radiation from the eardrum

12/22/2011

23

Question

The use of fiberglass insulation in the outer walls of a building is intended to minimize heat transfer through what process?

a. conductionb. radiationc. convectiond. vaporization

Work in Thermodynamic Processes

vState of a system– Description of the system in terms of state

variables• Pressure• Volume• Temperature• Internal Energy

– A macroscopic state of an isolated system can be specified only if the system is in internal thermal equilibrium

12/22/2011

24

Work

§ Work is an important energy transfer mechanism in thermodynamic systems

§ Heat is another energy transfer mechanism

Ø Example: gas cylinder with pistono The gas is contained in a cylinder with a

moveable pistono The gas occupies a volume V and exerts

pressure P on the walls of the cylinder and on the piston

§ A force is applied to slowly compress the gas§ The compression is

slow enough for all the system to remain essentially in thermal equilibriumW = - P ΔV

§ This is the work done on the gas

12/22/2011

25

Work on a Gas Cylinder

ØWhen the gas is compressedΔV is negative The work done on the gas is positive

ØWhen the gas is allowed to expandΔV is positiveThe work done on the gas is negative

ØWhen the volume remains constantNo work is done on the gas

W = W = -- P ΔVP ΔV

Notes about the Work Equation

If the pressure remains constant during the expansion or compression, the process is called an isobaric process

If the pressure changes, the average pressure may be used to estimate the work done

W = W = -- P ΔVP ΔV

Work done on the gas

Work=Area under the curve

W = W = -- P ΔVP ΔV

12/22/2011

26

QuestionFind work done by the gas in this cycle.

P2

P1

V1 V2

Note: work is equal to the area:( )( )1212 VVppW −−=

Other ProcessesØ Isovolumetric

Volume stays constantVertical line on the PV diagram

Ø IsothermalTemperature stays the same

ØAdiabaticNo heat is exchanged with the surroundings

12/22/2011

27

Example:

Given:

n = 1 moleTi = 96.2 KTf = 144.3 KVi = 0.2 m3

Vf = 0.3 m3

P = const

Find:

W=?

( ) ( )J

mVVPVPW if

4002.00.3mPa4000 33

=

−=−=∆=

ü

1. Isobaric expansion:

Calculate work done by expanding gas of 1 mole if initial pressure is 4000 Pa, initial volume is 0.2 m3, and initial temperature is 96.2 K. Assume a two processes: (1) isobaric expansion to 0.3 m3, Tf=144.3 K (2) isothermal expansion to 0.3 m3.

Also:

5.12.03.0

3

3

====mm

VV

nRVP

nRVP

TT

i

f

ii

ff

i

f

A 50% increase in temperature!

Example:

Given:

n = 1 moleTi = 96.2 KVi = 0.2 m3

Vf = 0.3 m3

T = const

Find:

W=?

( )( ) Jmmm

VV

VPVV

nRTWi

fii

i

f

3242.03.0ln2.0Pa4000

lnln

3

33 ==

=

=

ü

2. Isothermal expansion:

Calculate work done by expanding gas of 1 mole if initial pressure is 4000 Pa, initial volume is 0.2 m3, and initial temperature is 96.2 K. Assume a two processes: (1) isobaric expansion to 0.3 m3, Tf=144.3 K (2) isothermalexpansion to 0.3 m3.

Also:

PammPa

VVPP

f

iif 2667

3.02.04000 3

3

===

A ~67% decrease in pressure!

12/22/2011

28

Processes for Transferring Energy

vBy doing work– Requires a macroscopic displacement of the point of

application of a forcevBy heat

– Occurs by random molecular collisionsvResults of both

– Change in internal energy of the system– Generally accompanied by measurable macroscopic

variables• Pressure• Temperature• Volume

First Law of Thermodynamics

v Consider energy conservation in thermal processes. Must include:– Q (Heat)

Positive if energy is transferred to the system– W (Work)

Positive if done on the system– U(Internal energy )

• Positive if the temperature increases

The relationship among U, W, and Q can be expressed as

ΔU = Uf – Ui = Q + WThis means that the change in internal energy of a system is equal to the sum of the energy transferred across the system boundary by heat and the energy transferred by work

12/22/2011

29

Example:

Given:

n = 1 moleVi = 0.2 m3

Vf = 0.3 m3

P = constQ=500 J

Find:

∆U=?

( ) ( )J

mVVPVPW if

4002.00.3mPa4000 33

=

−=−=∆=

ü

1. Isobaric expansion:

If 500 J of heat added to ideal gas that is expanding from 0.2 m3 to 0.3 m3 at a constant pressure of 4000 Pa, what is the change in its internal energy?

Use 1st law of thermodynamics:

JJJWQUWUQ

100400500 =−=−=∆+∆=

What if volume is kept constant?

The First Law and Human Metabolism

vThe First Law can be applied to living organismsvThe internal energy stored in humans goes into

other forms needed by the organs and into work and heat

vThe metabolic rate (ΔU / ΔT) is directly proportional to the rate of oxygen consumption by volume

Basal metabolic rate (to maintain and run organs, etc.) is about 80 W

12/22/2011

30

Various Metabolic Rates

Fig. T12.1, p. 369

Slide 11