thermal & kinetic lecture 21 heat capacity and thermal conductivity of solids;

Thermal & Kinetic Lecture 21

Heat capacity and thermal conductivity of solids;

Problems/ Exam Questions

LECTURE 21 OVERVIEW

The Einstein model revisited: heat capacity of solids

Thermal conductivity of solids

The most efficient process: the Carnot cycle

Carnot engine is an idealisation.

We’ll use an ideal gas as our working substance.

Carnot cycle may be constructed from a combination of adiabatic and isothermal compressions and expansions.

Adiabatic

Isotherm

P

V

A

B

CD

QH

QL

W

Animation

The Einstein model of a solid revisited

Back in Lecture 1, the model to the right was introduced. In Lecture 12, we simplified this model further and considered the Einstein model…….

Furthermore, replace each 3D oscillator (i.e. each atom) with three independent 1D oscillators (x, y, z).

Consider each atom in solid as moving independently of its neighbours.

The Einstein model of a solid revisited

Energies of the atomic simple harmonic oscillators comprising the solid are quantised.

0)2

1( nEn

U(x)

x

E0

E2

E1

E3

Just as we considered for the Planck model of blackbody radiation (Set 2b of the lecture notes), the energy difference between consecutive energy levels is:

0

According to the equipartition of energy theorem what is the average thermal energy per degree of freedom?

kT

½ k

T

3kT

/2

Don’t

know

3% 6%

26%

65%a) kT

b) ½ kT

c) 3kT/2

d) Don’t know

Which means that the average thermal energy of an ideal gas molecule is……?

3kT

/2 kT

3 k

T

Don’t

know

84%

3%6%6%

a) 3kT/2

b) kT

c) 3 kT

d) Don’t know

Which in turn means that the average thermal energy of 1 mole of an ideal gas is..?

RT

3RT/2

2RT

Don’t

know

6% 3%9%

82%1. RT

2. 3RT/2

3. 2RT

4. Don’t know

Specific heats and equipartition of energy: revision

In a solid there are no translational or rotational degrees of freedom – only vibrational degrees of freedom remain.

What is the average thermal energy of 1 mole of a solid according to the equipartition theorem?

3RT

RT

3kT

Don’t

know

28%

6%3%

64%a) 3RT

b) RT

c) 3kT

d) Don’t know

Specific heats and equipartition of energy

Specific heat capacity, C = = 3R

for a solid according to the equipartition of energy theorem.

How does this compare to the experimental results?

C (JK-1)

T (K)

3R

Only at high temperatures is a value of 3R for C observed.

?? Why do we measure a value of 3R only at high temperatures?

VT

U

Specific heats, equipartition, and quantisation

Remember our discussions of specific heats (for gases) and blackbody radiation…….?

Energy

E0

E2

E1

E3

Only if the spacing of energy levels is small relative to kT will we reach the classical limit where the equipartition theorem is a good approximation.

Thermal conductivity in solids

NB Note that the following equation, which we definedpreviously for gases also holds for solids:

dz

dTA

dt

dQ

dT/dz is the temperature gradient.Rate of heat flow (heat current)

is the coefficient of thermal conductivity and A is the cross-sectional area.

Thermal expansion

The final topic we’ll cover (briefly) in the module brings us back to Lecture 1 when we considered potential energy curves…..

U(r)

r

With increasing temperature the inter-atomic bond length increases (centre of oscillations shifts to larger r). Hence object expands at higher temperatures.

Linear expansion coefficient, : Tl

l

A worked example on thermal conductivity…

The cavity wall of a modern house consists of two 10 cm thick brick walls separated by a 15 cm air gap. If the temperature inside the house is 25°C and outside is 0°C, calculate the rate of heat loss by conduction per unit area. [For this temperature range, brick = 0.8 Wm-1K-1 and air = 0.023 Wm-1K-1]

NB Note that heat current must be conserved as we pass from brick to air to brick.

321

dt

dQ

dt

dQ

dt

dQ

T = 25°C T = 0°C10 cm,brick

10 cm,brick

15 cm,air

1 2 3

T = 25°C T = 0°C10 cm,brick

10 cm,brick

15 cm,air

1 2 3

T1T2

How do we determine T1 and T2???1.0

25 1

1

TA

dt

dQbrick

15.021

2

TTA

dt

dQair

1.0

02

3

T

Adt

dQbrick

321

dt

dQ

dt

dQ

dt

dQ Solve to get T1 and T2 and then use any expression for dQ/dt to get dQ/dt = 3.7 Wm-2

Alternatively…..

Consider: • I equivalent to dQ/dt, • V equivalent to temperature difference• R equivalent to thermal resistance

kA

xRT

T = 25°C T = 0°C10 cm,brick

10 cm,brick

15 cm,air

1 2 3

T1T2

R1 R2 R3

dQ/dt

RTOTAL =R1 + R2 + R3

Thermal ‘circuits’

In analogy with V = IR: dQ/dt = T/RTOTAL

Make sure that you can get the same answer as before!

Thermal & Kinetic paper, ’06/’07: Q5

Thermal resistance, R = x/(kA) {1}

RCu = 0.1/(401 x 4 x 10-4) = 0.623 K/W {+}

Using a similar approach, RAg = 0.583 K/W {+}

1/REq = 1/RCu + 1/RCu {1}

1/REq = 0.301 K/W {+}

Heat current = T/REq {1}

Heat current 332 W{+}

Thermal & Kinetic paper, ’06/’07: Q5 – Worked solution

Worked example on thermal expansion

Steel railway tracks are laid when the temperature is - 5°C. A standard section of rail is 12 m long. What gap should be left between rail sections so there is no compression when the temperature gets as high as 42°C? (The linear expansion coefficient of steel is 11 x 10-6 K-1)

T = ? Ans: 47°C

So, l = ? Ans: 6.5 mm

Tll

Thermal & Kinetic paper, ’04/’05: Q1

Is it possible for the temperature of an ideal gas to rise without heat flowing into the gas?

Yes N

o

Don’t

know

97%

0%3%

1. Yes

2. No

3. Don’t know

Must the temperature of an ideal gas necessarily change as a result of hear flow into or out of it?

Yes N

o

Don’t

know

0% 0%

100%1. Yes

2. No

3. Don’t know

Thermal & Kinetic paper, ’02/’03: Paper 2, Q2

…compare to CW Set 4

Q1. 1 mole of an ideal gas originally at a pressure of 1.0 x 104 Pa and occupying a volume of 0.2 m3 undergoes the following cyclic process:

(i) an adiabatic compression until the pressure is 3.0 x 104 Pa;(ii) an isobaric expansion to a volume of 0.4 m3;(iii) an isothermal expansion until the pressure reaches 1.0 x 104 Pa;(iv) an isobaric compression to the original volume of 0.2 m3.

Draw a PV diagram for this process [3].

For each of the stages (i) – (iv) calculate the heat transferred [5], the work done [5], and the change in internal energy [5].

Show that only internal energy is a function of state [2].

V (m3)

P (kPa)

A

B C

D

30

10

0.2 0.4 V (m3)

P (kPa)

A

B C

D

30

10

0.2 0.4

F31ST1 CW4

Stage 1 Stage2 Stage 3 Stage 4 Total

Work done +1.5 kJ -9 kJ -13.175 kJ + 10 kJ -10.675 kJ

Heat transferred

0 22.5 kJ 13.175 kJ -25 kJ +10.675 kJ

Change in internal energy

+1.5 kJ +13.5 kJ 0 -15 kJ 0

Thermal & Kinetic paper, ’06/’07: Q7

Q5. F31ST1 ’03 – ’04 Exam Paper