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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. Q H. B. W. D. C. Q L. The most efficient process: the Carnot cycle.
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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
QH B W D C QL 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. P A Adiabatic Isotherm Animation V
Furthermore, replace each 3D oscillator (i.e. each atom) with three independent 1D oscillators (x, y, z). 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……. Consider each atom in solid as moving independently of its neighbours.
E3 E2 E1 E0 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: U(x) x The Einstein model of a solid revisited Energies of the atomic simple harmonic oscillators comprising the solid are quantised.
According to the equipartition of energy theorem what is the average thermal energy per degree of freedom? • kT • ½ kT • 3kT/2 • Don’t know
Which means that the average thermal energy of an ideal gas molecule is……? • 3kT/2 • kT • 3 kT • 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
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
C (JK-1) 3R ? Why do we measure a value of 3R only at high temperatures? How does this compare to the experimental results? T (K) Specific heats and equipartition of energy Specific heat capacity, C = = 3R for a solid according to the equipartition of energy theorem. Only at high temperatures is a value of 3R for C observed.
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
k is the coefficient of thermal conductivity and A is the cross-sectional area. dT/dz is the temperature gradient. Rate of heat flow (heat current) Thermal conductivity in solids NB Note that the following equation, which we defined previously for gases also holds for solids:
U(r) r Thermal expansion The final topic we’ll cover (briefly) in the module brings us back to Lecture 1 when we considered potential energy curves….. 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:
10 cm, kbrick 15 cm, kair 10 cm, kbrick T = 0°C T = 25°C 1 2 3 NB Note that heat current must be conserved as we pass from brick to air to brick. 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, kbrick = 0.8 Wm-1K-1 and kair = 0.023 Wm-1K-1]
T2 T1 ? 10 cm, kbrick 15 cm, kair 10 cm, kbrick How do we determine T1 and T2? T = 0°C T = 25°C 1 2 3 Solve to get T1 and T2 and then use any expression for dQ/dt to get dQ/dt = 3.7 Wm-2
T2 T1 10 cm, kbrick 15 cm, kair 10 cm, kbrick T = 0°C T = 25°C 1 2 3 dQ/dt R1 R2 R3 Alternatively….. • Consider: • I equivalent to dQ/dt, • V equivalent to temperature difference • R equivalent to thermal resistance
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!
Thermal & Kinetic paper, ’06/’07: Q5 – Worked solution • Thermal resistance, R = Dx/(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 = DT/REq {1} Heat current 332 W {+}
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) Ans: 47°C DT = ? So, Dl = ? Ans:6.5 mm
Is it possible for the temperature of an ideal gas to rise without heat flowing into the gas? • Yes • No • Don’t know
Must the temperature of an ideal gas necessarily change as a result of hear flow into or out of it? • Yes • No • Don’t know
…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: • an adiabatic compression until the pressure is 3.0 x 104 Pa; • an isobaric expansion to a volume of 0.4 m3; • an isothermal expansion until the pressure reaches 1.0 x 104 Pa; • 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].
