Thermal Physics

Chapter 11

• Heat transfer mechanisms• Thermal equilibrium• Temperature• Newton’s law of cooling

Recap of GCSE ideas

Conduction

• The molecules in the pan vibrate about their equilibrium positions

• Particles in the pan near the stove coil vibrate with larger amplitudes

• These collide with adjacent molecules and transfer some energy

• Eventually, the energy travels entirely through the pan and its handle

Convection

• Fluid heats up and expands• Expanded fluid is less dense, so rises• Cooler fluid replaces it, is warmed…

Radiation

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

• No physical contact is necessary• Cannot be accounted for by conduction or

convection

Thermal equilibrium

• If you leave a spoon in a cup of coffee what happens?

• Coffee cools, spoon heats up until they are in THERMAL EQUILIBRIUM

• No further flow of heat

• If two systems are at the same temperature they are said to be in THERMAL EQUILIBRIUM and there is no resultant flow of heat

• Temperature can be thought of as a measure of the average kinetic energy of the molecules.

• Heat flows from higher temperature objects to lower temperature objects.

So what is “temperature”?

• Rate of change of temperature is proportional to temperature difference

•

• Solve for T: exponential drop

Newton’s Law of Cooling

)( STTkdt

dT

• What type of energy do the coffee molecules have?

Internal Energy

Internal energy• The INTERNAL ENERGY (U) of an object is the sum of

the random distribution of the KINETIC and POTENTIAL energies of its molecules– Kinetic energy – molecules moving (in a disorganised way)– Potential energy – due to bond with other molecules

(changes as the molecules get closer and further apart)

• On average the molecules in a hot cup of coffee have more energy than a cold cup of coffee.

Internal energy and temperature

• Which has the higher temperature?• Which has the greater internal energy?

Temperature Scales

Temperature Scales

• A temperature scale is constructed by choosing two fixed points and then dividing the resulting range into some number of degrees.Scale Lower fixed pt Upper fixed pt Degrees

between pts

Celsius Melting pt of ice Steam above boiling water

100

Fahrenheit

Frigorific mixture of brine

Mrs Fahrenheit’s armpit (really!)

100

Kelvin Absolute zero Triple point of water

273.16

Thermometers

• Thermometers are used to measure temperature. They can use a number of different technologies:– Mercury in glass• uses expansion

– Thermocouple• voltage at junction between two metals

– Resistance• temperature-dependent resistance of Pt wire

Specific Heat Capacity

• The Specific Heat Capacity of a substance is the heat required to produce a 1 K rise in 1 kg.– units: Jkg-1K-1

• The total amount of heat (Q) needed to raise the temperature of a sample depends on:– The Specific Heat Capacity (c) of the material

– The mass of the sample (m)– The temperature rise (DT)

TmcQ

Questions

• 1. How much heat is needed to raise the temperature by 10 °C of 5 kg of a substance of specific heat capacity 300 J/kg°C?

• 2. The same quantity of heat was given to different masses of three substances. From the table, calculate the specific heat capacities of A, B, and C.

Material Mass (kg)

Heat in (J)

Temp rise (°C)

A 1.0 2000 1.0

B 2.0 2000 5.0

C 0.5 2000 4.0

Some Specific Heat Capacities

• SHC of Aluminium is 900 J/kg°C• SHC of water is 4200 J/kg°C• SHC of milk is 3770 J/kg°C• SHC of ice is 2060 J/kg°C

• So SHC depends on state!– Why?– Degrees of freedom (complicated, not examinable)

Specific Heat Capacity of Water

• The Specific Heat Capacity of water is particularly high at 4200 J/kg°C.

• This is due to its polarized molecule which gives rise to hydrogen bonding.

• This is why water is a good substance to use in cooling or heating systems – it can absorb and transport a lot of heat.

• It also explains why we get sea breezes and why an island like Britain has a much milder climate than mainland countries at the same latitudes.

• Energy supplied by electric heater Q= ItV• Measure mass, I, V, t and temp. change• Not hugely accurate – why?

Measuring c

solid liquid

• GPE to heat– waterfalls, inversion tube

• friction• ...

Other ways of heating objects

1 .2 m

lead

Practice questions

• Do questions 3 & 4 on page 204.

Cooling Curves

• As a liquid cools, its temperature drops. • When the temperature reaches the freezing (melting) point the

liquid begins to solidify.• While it solidifies it continues to give out heat but the temperature

remains constant until all of the substance has solidified.• The solid then continues to cool.

Cooling curve for Benzene

Freezing point

Note exponential cooling

Heating curves

• The same pattern is observed when heating a substance at a constant rate.

• As the substance changes state, from solid to liquid or from liquid to gas, heat must be put in but the temperature remains constant until the state change is complete.

Heating curve for ethanoic acid

Note constant gradient, =P/mc

Energy changes during state change

• Liquefaction and solidification are processes of bond formation; melting and vaporisation are processes of bond-breaking.

• Energy must be put into a substance to break bonds, it is released when bonds are made.

• During a change of state, the temperature remains constant although heat is being put in or given out.

• This is because the heat change is not causing a change in the average kinetic energy of the molecules, but rather a change in their potential energy.

Specific Latent Heat• Heat that is absorbed by a solid during melting or given out by a

liquid when solidifying is called LATENT HEAT OF FUSION.• The SPECIFIC LATENT HEAT OF FUSION (lf) of a substance is the

quantity of heat needed to change one kg from solid to liquid without temperature change.

• Units of lf are J/kg. • Heat that is absorbed by a liquid during vapourisation or given out by

a vapour when liquefying is called LATENT HEAT OF VAPOURISATION.• The SPECIFIC LATENT HEAT OF VAPOURISATION (lv) of a substance is

the quantity of heat needed to change one kg from liquid to gas without temperature change.

•

flmQ

vlmQ

Question

• How much heat is needed to change 20 g of ice at 0 °C to steam at 100 °C? 0..(specific heat capacity of water is 4200 J/kg °C, specific latent heat of ice is 340,000 J/kg, specific latent heat of water is 2,300,000 J/kg)

• Ice at 0 °C to water at 0 °C:• E=m lf , =0.02 x 340000=6800 J.• Water at 0 °C to water at 100 °C:• E=mcDT, =0.02 x 4200 x 100 = 8400 J• Water at 100 °C to steam at 100 °C:• E=m lv , =0.02 x 2300000=46000 J.• Total E = 6800+8400+46000=61200 J

Evaporation• Liquid particles have a distribution of energies.• Some molecules close to the surface of a liquid have

enough energy to escape as vapour – EVAPORATION.• The rate of evaporation depends on:• temperature• Surface area• any draughts

• Latent Heat is required for evaporation, which it acquires from its surroundings.

• Evaporation removes the most energetic molecules from the liquid. The average KE of the remaining molecules is lower, so the temperature falls.

Examples of evaporative cooling

• Sweating

• Hot coffee “steams”

• Refrigerator• (Removes energy via the latent• heat of vaporisation of coolant)

• Now make sure you can do the ESQs on p. 208-9

Gas Laws and Kinetic theory

Chapter 12

Experimental Gas Laws

• 3 experimentally derived laws which relate pressure, volume and temperature for gases

• Boyle’s Law

• Charles’ (Gay-Lussac) Law

• Pressure (Amonton’s) Law

Vp

1

TV

Tp

Boyle’s Law

• At constant temperature:

– The lower the temperature, the closer the curve is to the axes

• Marshmallows video

constant

1

pVV

p

Charles’ Law

• At constant pressure:

• What does this predict at absolute zero?

constant

T

V

TV

• Measuring variation of pressure with temperature for a fixed volume of gas

• For all gases, lines extrapolate back to the same temperature: absolute zero– The lowest possible temperature

• An object at absolute zero has minimum internal energy.

Absolute zero

Pressure Law

• At constant volume:

constant

T

p

Tp

Counting particles and moles

• 1 mole contains NA particles. NA is the number of atoms in 12g of C-12. NA=6.023x1023.• So 1 mole of a substance is ≈ no. of particles in its atomic mass

in g• The Molar mass is the mass of 1 mole of a substance.

• Avogadro proposed that equal volumes of different gases at the same T and p contain equal numbers of molecules.– But they will have different masses, and hence

densities.• (1 mole of any gas occupies 22.4 litres at STP)

The Ideal Gas Law

• Combining these three laws we get:

• We can define R, the molar gas constant:

2

22

1

11or constant,T

Vp

T

Vp

T

pV

constant sBoltzmann'k whereor so

KJmol 8.31 1-1-

NkTpVnRTpV

RT

pVm Vm is the volume occupied by a mole of any gas at p and T

Now buy the T shirt!

This is a good approximation to reality for gases at low pressure and high temperatures.

(k=1.38x10-23 JK-1)

• Now practise the summary questions on p. 214

Kinetic Theory of gases

• Kinetic theory attempts to explain the microscopic scale causes of macroscopic phenomena– Pressure is caused by molecules

colliding with the walls of the container.

– Temperature is a measure of the average speed at which molecules are moving

Molecular speeds• There is a continuous distribution of speeds of

molecules in a gas sample.• Individual molecules change speed when they

collide, but the distribution stays the same (provided T is const)

• As on average there are always equal numbers of molecules travelling in opposite directions, the mean velocity is zero.

• A more useful quantity is the root mean square (rms) speed.

• If the temperature is raised the rms speed increases

N

cccc Nrms

222

21 ...

Ideal Gas Assumptions

1. The gas sample contains a large number of molecules, so statistics are reliable.

2. Molecules constantly move randomly and rapidly.3. There are no attractive forces between molecules (i.e. all internal

energy is KE).4. All collisions between particles and other particles or container

walls are perfectly elastic (i.e. no KE lost).5. All collisions are effectively instantaneous (not too dense).6. Molecules have negligible volume compared with volume of gas7. Gravitational, quantum and relativistic effects can be neglected.

• All reasonable under most circumstances – learn them!

Pressure of an Ideal Gas (i)• Consider 1 molecule of mass m travelling with

velocity c1 in a box of sides lx, ly, lz.

• u1, v1, w1 are the x, y, z components of c1.• Each collision with the shaded face of the box

changes the molecule’s momentum:

• The time between successive collisions with the shaded face (assuming no other collisions) is:

• The force on the molecule is therefore:

• By N3, the force on the wall must be the same, so the pressure due to this molecule on the shaded wall is:

111 2mumumumom

1

2

u

lt x

xx l

mu

ul

mu

t

momF

21

1

1

/2

2

V

mu

lll

mu

A

Fp

zyx

21

21

)(

Pressure of an Ideal Gas (ii)• For N molecules in the box, all moving in

different directions, the total pressure is:

• There is no preferred direction of motion, though, so we can write similar formulae for the y and z directions.

• So

• or

velocity theofcomponent - xsquaremean theis 2 where

222

21

22

21 ......

u

V

uNmuu

V

m

V

mu

V

mup

V

wNmp

V

vNmp

22

,

2222

33 rmscV

Nmwvu

V

Nmp

2

3

1rmsNmcpV You need to be able to reproduce

this derivation in the exam!

22222

2222

3 so ucwvu

wvuc

rms

rms

Kinetic Energy and Temperature

•

nRT

RTkTN

kTmc

nRNkN

nRTmc

nRTNmc

NmcpVnRTpV

A

rms

rms

rms

rms

2

3 molesn of KE Total

2

3

2

3 gas of mole 1 of KE Total

2

3

2

1 KEmean so

remember but ,2

3

2

1

:molecule a of KEmean thegives grearrangin and 3/2by gmultiplyin3

13

1 and have weSo

2

2

2

2

• Now practise the summary questions on p. 218...

• ...and the ESQs that follow.