thermal properties of solids Thermal & Kinetic Lecture 18 Phase transitions; thermal properties of solids LECTURE 18 OVERVIEW Recap A (very brief) introduction to phase transitions. The Einstein model revisted and the specific heat of a solid Normal modes and phonons Thermal expansion and thermal conductivity

? Last time…. The Carnot cycle: entropy and efficiency Maxwell’s demon Changes of phase. If I leave the ‘fridge door open will my kitchen: (i) cool down, (ii) stay at the same temperature, or (iii) heat up? ?

Phase diagrams Relative fractions of solid, liquid and gas phases depend upon pressure, volume and temperature. P TC Above the critical temperature, TC the liquid can no longer exist and the substance is a gas (or at higher pressures, a solid). Above TC the substance cannot be liquefied by any amount of pressure. (For example, N2 can’t be liquefied at any pressure when its temperature exceeds 126 K). Liquid Solid Gas and liquid coexist in equilibrium at different temperatures: curve of saturated vapour pressure vs T Gas TTP T2

Isotherms and changes of phase Thus far in the module we’ve only discussed isotherms for ideal gases and haven’t considered changes of phase. What does an isotherm look like for a real substance that undergoes phase changes? Critical point P Isotherm at temperature T3> T2 Saturated vapour – gas starts to liquefy Liquid All molecules now in liquid phase Gas Isotherm at temperature T2 V

From ideal to real gases Behaviour of real gases only approximates that of ideal gases at low pressures where the average intermolecular separation is large. Need to modify the ideal gas equation to take account of: nonzero volume of molecules intermolecular forces U(r) r E0 r0 The van der Waals equation of state takes into account these two factors: Depends on intermolecular force, related to E0 Due to non-zero volume of molecules and intermolecular repulsion at short distances

From ideal to real gases: isotherms The van der Waals equation of state can be written as: Inflection of curve at critical point: From this we can determine PC, VC and Tc P V

The Einstein model of a solid revisited (from Lecture 10) Back in Lecture 1, the model to the right was introduced. In Lecture 10, we simplified this model further and considered the Einstein model……. Consider each atom in solid as moving independently of its neighbours. Furthermore, replace each 3D oscillator (i.e. each atom) with three independent 1D oscillators (x, y, z).

The Einstein model of a solid revisited (from Lecture 10) U(x) x E0 E2 E1 E3 Energies of the atomic simple harmonic oscillators comprising the solid are quantised. 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:

Specific heats and equipartition of energy: revision According to the equipartition of energy theorem what is the average thermal energy per degree of freedom? ? Ans: ½ kT Which means that the average thermal energy of an ideal gas molecule is……? ? Ans: (3/2) kT Which in turn means that the average thermal energy of 1 mole of an ideal gas is……? ? Ans: (3/2) NA kT = (3/2) RT 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? Ans:3 RT

? Specific heats and equipartition of energy Specific heat capacity, CV = = 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?

Specific heats, equipartition, and quantisation Remember our discussions of specific heats (for gases) and blackbody radiation…….? Energy 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. E3 E2 E1 E0

Heat transfer in solids: normal modes of vibration However, are the quantised vibrational energy levels associated with the vibrations of individual atoms? Energy of vibration of individual atom is very large because restoring forces are high if the rest of the atoms remain fixed in place. …but the atoms can vibrate collectively. All atoms vibrate at the same frequency: normal modes of vibration. (See your Vibrations & Waves notes). Just as for a guitar string fundamental frequency determined by length of ‘string’ Each vibrational mode can only have certain discrete energy values, En:

Vibrational energy,Boltzmann factors and phonons As the temperature is lowered, the probability of exciting a vibrational mode of frequency n is reduced by a factor exp (-hn/kT) ….but if these waves extend the entire length of the solid then why don’t all solids have ‘perfect’ thermal conductivity ? Not quite correct to think of atom waves extending unimpeded across entire length of solid. Have fluctuations/ variations in thermal energy in solid – this sets up waves of slightly different frequency – superposition of different waves: wavepacket (again, see your Vibrations & Waves notes) A vibrational wavepacket in a solid is a phonon.

Thermal conductivity in solids Phonons can be thought of as particles – quantised packets of vibration that travel through the lattice. Crystals are not perfect – phonons get scattered as they travel through the lattice. Phonons have a mean free path in a solid which is much smaller than the size of the sample. Movement of phonons from hot to cold regions: analogous to motion of molecules causing heat transfer in gases

Thermal conductivity in solids Thermal conductivity in nonmetals – all heat transfer is by phonons. Thermal conductivity in metals – much higher due to free electrons – electronic contribution to thermal conductivity vastly outweighs phonon contribution (speed of electrons >> speed of sound). NB Note that the following equation, which we defined in Lecture 9 for gases also holds for solids: k is the coefficient of thermal conductivity and A is the cross-sectional area. Rate of heat flow (heat current) dT/dz is the temperature gradient.

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, a:

Some worked examples 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, kbrick = 0.8 Wm-1K-1 and kair = 0.023 Wm-1K-1] [Exam 2000-2001 Section A Question 3] T = 25°C T = 0°C 10 cm, kbrick 15 cm, kair 1 2 3 NB Note that heat current must be conserved as we pass from brick to air to brick.

? How do we determine T1 and T2? T = 25°C T = 0°C 10 cm, kbrick 15 cm, kair 1 2 3 T1 T2 How do we determine T1 and T2? ? 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 T = 25°C T = 0°C 10 cm, kbrick 15 cm, kair 1 2 3 T1 T2 R1 R2 R3 dQ/dt

Thermal ‘circuits’ RTOTAL =R1 + R2 + R3 In analogy with V = IR: dQ/dt = DT/RTOTAL Make sure that you can get the same answer as before!

Some worked examples on thermal conductivity… A compound slab of material consists of a 20 cm layer of wood and a 20 cm layer of polystyrene. The temperature of the outer surface of the wood is 5°C and that of the polystyrene is 30°C. Determine the temperature at the interface between the two materials and calculate the heat flow per unit area through the compund slab. [For this temperature range, kbrick = 0.8 Wm-1K-1 and kair = 0.023 Wm-1K-1] Exactly the same ideas as before. You should try this one before the next lecture. The temperature at the interface is: 7.8°C and the heat flow is 1.1 Wm-2. I hear, and I forget I see, and I remember I do, and I understand. Chinese Proverb

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) DT = ? Ans: 47°C So, Dl = ? Ans: 6.5 mm