Thermal Properties of Matter

1. Temperature

• Temperature is a physical quantity that expresses quantitatively the hotness or coldness of a body. It is measured with a thermometer.
• Thermometers are calibrated on different temperature scales, each defined by two fixed points (ice point & steam point of water).

2. Temperature Scales

Upper line : water boils  ·  lower line : water freezes  ·  red line : same temperature on all scales.

Denominators = number of divisions between the two fixed points. (Exactly, K = C + 273.15 ; 273 is the usual approximation.)

Graph of C vs F :

Straight line : slope +ve (5/9), C-intercept −ve (−160/9 ≈ −17.8 °C).

• A temperature difference has the same value in °C and K : 20 °C → 50 °C is a rise of 30 °C = 30 K (293 K → 323 K).

3. Thermal Expansion

• Tendency of matter to change its length, area, volume and density in response to a change in temperature (particles move farther apart on heating).

Photographic (isotropic) expansion :

• The object expands equally in every direction → no change in shape (like an enlarged photograph).
• Every linear dimension changes by the same percentage; every area changes by the same percentage; angles do not change.

4. Linear Expansion

• dl = α l dT ⇒ exactly l = l₀ e^αΔT (α constant) ; if α varies, l = l₀ e^{∫α dT}.
• Since αΔT ≪ 1 : l = l₀(1 + αΔT + (αΔT)²/2! + …) ≈ l₀(1 + αΔT)

• Valid for every linear dimension : radius, diameter, diagonal, sides.
• α = coefficient of linear expansion ; unit /°C or /K.

5. Superficial (Areal) Expansion

A = ab = a₀(1 + αΔT) · b₀(1 + αΔT) = a₀b₀(1 + 2αΔT + (αΔT)²) ; (αΔT)² ≈ 0

• β = coefficient of areal expansion. % change in area = βΔT × 100 = 2 × % change in length.

6. Volume Expansion

V = abc = a₀b₀c₀(1 + αΔT)³ ≈ V₀(1 + 3αΔT)

• γ = coefficient of volume expansion. % change in volume = γΔT × 100 = 3 × % change in length.

7. Variation of Density

• On heating, volume changes but mass stays the same ⇒ density changes.

ρ = mV = mV₀(1 + γΔT)

Useful approximations (αΔT ≪ 1) :

• (αΔT)² , (αΔT)³ … ≈ 0
• (1 + α₁ΔT)(1 + α₂ΔT) ≈ 1 + (α₁ + α₂)ΔT
• 1 + α₁ΔT1 + α₂ΔT = (1 + α₁ΔT)(1 + α₂ΔT)⁻¹ ≈ 1 + (α₁ − α₂)ΔT

8. Non-isotropic Expansion

• Different α along x, y, z :

(β of a face = sum of the two linear α's lying in that face.)

9. Summary Table

10. Pendulum Clock

• Here ΔT = change in time period. With a temperature change Δθ (g constant) :

• Period increases (ΔT +ve) → clock becomes slow (loses time) — e.g. in summer.
• Period decreases (ΔT −ve) → clock becomes fast (gains time) — e.g. in winter.
• Time lost / gained per day = ½ α Δθ × 86400 s.
• Clock loses 12 s a day ⇒ ΔTT = + 1224 × 3600 ; gains 12 s a day ⇒ ΔTT = − 1224 × 3600

11. Equivalent α (rods end to end)

l₁(1 + α₁ΔT) + l₂(1 + α₂ΔT) = (l₁ + l₂)[1 + l₁α₁ + l₂α₂l₁ + l₂ΔT]

12. Length Measured by an Expanding Scale

• True length of object at T = l₀(1 + α_oΔT) ; each scale division also grows by (1 + α_sΔT).
• Reading = l₀(1 + α_oΔT)1 + α_sΔT

l₀ = reading at T₀ (scale correct) ; α_o = α of object ; α_s = α of scale.

• Wooden object (α_o ≈ 0) on steel scale : l = l₀(1 − α_sΔT) → reading decreases on heating.
• Steel object on wooden scale (α_s ≈ 0) : l = l₀(1 + α_oΔT) → reading increases on heating.

13. Liquid in a Vessel

• Vessel : A = A₀(1 + 2α_sΔT), H = H₀(1 + α_sΔT), capacity V = V₀(1 + 3α_sΔT).
• Liquid : volume = A₀h₀(1 + γ_LΔT).

h = new volume of liquidnew base area = A₀h₀(1 + γ_LΔT)A₀(1 + 2α_sΔT)

• γ_L > 2α_s → level rises ; γ_L < 2α_s → level falls ; γ_L = 2α_s → level unchanged.

γ_L − 3α_s compares liquid volume with vessel capacity; γ_L − 2α_s gives the liquid height. Both are consistent.

e.g. 500 cm³ liquid in a vessel of base 100 cm² ⇒ h = 500/100 = 5 cm.

14. Bimetallic Strip

• Two thin strips of different metals (same length l₀, each of thickness d) are bonded face to face along their length. On heating, the strip bends into an arc.

• R = radius of the interface ; outer strip mid-line R + d/2, inner R − d/2 :

(R + d/2)θ = l₀(1 + α₂ΔT) ,  (R − d/2)θ = l₀(1 + α₁ΔT)

R + d/2R − d/2 = 1 + α₂ΔT1 + α₁ΔT ≈ 1 + (α₂ − α₁)ΔT ⇒ d ≈ R(α₂ − α₁)ΔT

• The metal with larger α is on the convex (outer) side ; e.g. brass outside, steel inside. On cooling it bends the other way.

15. Elasticity (recap)

16. Thermal Strain & Thermal Stress

(a) Rod free to expand :

• l = l₀(1 + αΔT). The rod simply takes its new natural length ⇒ thermal stress = 0, thermal force = 0 (no elastic strain).

(b) Rod between fixed walls, heated :

• Expansion that should have happened : Δl = l₀αΔT ; the walls prevent it.

• Heated rod between walls → compressive force on the walls.
• Wire fixed between walls and cooled by ΔT → same magnitudes, but tension in the wire.
• F is independent of the length of the rod.

17. Heat

• Heat is energy in transit, transferred due to a temperature difference, flowing on its own from a hotter body to a colder one.
• Bodies contain thermal (internal) energy, not "heat".

18. Specific Heat Capacity (s)

• Energy needed to raise the temperature of unit mass of a substance by 1 °C (or 1 K). Depends only on the material (and slightly on temperature).

SI unit : J kg⁻¹ K⁻¹ (J/kg °C). Use the integral when s varies with T.

19. Calorie & Values for Water

• 1 calorie = heat needed to raise the temperature of 1 g of water by 1 °C (14.5 °C → 15.5 °C).

(J = mechanical equivalent of heat)

s of water is not exactly constant : ≈ 1.004 cal/g °C near 0 °C, a shallow minimum ≈ 1 at intermediate T, rising again towards 100 °C.

20. Heat Capacity (H)

• Energy needed to raise the temperature of the whole object by 1 °C (1 K).

Unit of H : J/°C or J/K.

21. Water Equivalent (W)

• Mass of water that needs the same heat as the object for the same temperature rise.

H_object = m_ws_w = (m_w g)(1 cal/g °C) ⇒ W = mss_w

e.g. vessel of water equivalent 150 g ⇒ H = 150 cal/°C (≈ 630 J/°C).

22. Latent (Hidden) Heat (L)

• During a change of state the temperature stays constant (at the melting / boiling point) ; heat absorbed or released :

(80 × 4.2 = 336 ; 540 × 4.2 = 2268. Accepted values ≈ 334 kJ/kg and 2256 kJ/kg.)

23. Temperature vs Heat Supplied

• Same phase : Q = msΔT ⇒ slope of T–Q graph :

• Larger heat capacity → smaller slope.
• Flat parts = change of state at constant T ; their length = mL.

Example : 1 g ice at −80 °C heated to steam

Drawn to scale : ice and steam (s = ½) are twice as steep as water (s = 1) ; boiling plateau = 540/80 = 6.75 × melting plateau.

24. Law of Mixtures (Calorimetry)

• When two bodies at different temperatures are mixed, heat flows from hot to cold until both reach the same final temperature.
• Assuming no heat loss to the surroundings : Heat lost = Heat gained (conservation of energy).
• Final temperature is the same for both ; final phase may or may not be the same.

(cold body T₁, hot body T₂, final temperature T)

Equivalently (sum of heat changes = 0) :

• If a phase change occurs, add the mL terms on the appropriate side (and check whether there is enough heat for the full change — the final state may be a mixture at 0 °C or 100 °C).

25. Modes of Heat Transfer

26. Conduction & Steady State

• Steady state : temperature of every element stays constant with time (dT/dt = 0) ⇒ for every element, heat in = heat out.
• Syllabus : rod is insulated (no heat loss to surroundings). (With side losses, T would not fall linearly.)
• At steady state the rate of heat flow ∝ area A and ∝ temperature gradient dT/dx.

Fourier's law :

• P = thermal power (heat current), unit J/s = W ; k = thermal conductivity (W m⁻¹ K⁻¹).
• Minus sign : heat flows towards decreasing temperature.
• At steady state P is the same through every cross-section.

27. Uniform Rod (k, A constant)

• P, k, A uniform ⇒ dT/dx uniform ⇒ T falls linearly along the rod.

28. Thermal Resistance (Ohm's law analogy)

T₁ − T₂ = P · lkA

29. Non-uniform Cross-section

• P same everywhere, k uniform ⇒

• Where area is large the slope is small; where area is small the slope is steep ⇒ T–x curve is concave down for a rod narrowing from T₁ to T₂.

30. Series Combination

l₁ + l₂k_eqA = l₁k₁A + l₂k₂A

Junction temperature : T = T₁/R₁ + T₂/R₂1/R₁ + 1/R₂

31. Parallel Combination

Same ΔT across both : P₁ = T₁ − T₂R₁, P₂ = T₁ − T₂R₂ ; P = P₁ + P₂

k_eq(A₁ + A₂)l = k₁A₁l + k₂A₂l

32. Junction Rule & Wheatstone Bridge

• If R₁R₃ = R₂R₄ (i.e. R₁/R₂ = R₄/R₃) then T_C = T_D ⇒ no heat flows through R₅ — simply remove it.

33. Hollow Sphere (radial flow)

Thin shell element of radius r, thickness dr : dR = drk(4πr²)

R = ∫_a^b dr4πkr² = 14πk[−1r]_a^b

Heat current P = (T₁ − T₂)/R.

34. Hollow Cylinder (radial flow)

Element : thin cylindrical shell, radius r, thickness dr, area 2πrL : dR = drk(2πrL)

35. Frustum (axial flow)

r = a + b − aLx ⇒ dr = b − aLdx ;  dR = dxkπr²

R = L(b − a)kπ∫_a^bdrr² = L(b − a)kπ(1a − 1b)

Like a cylinder R = L/(kπr²) with r² replaced by ab (geometric mean of the end radii). P = (T₁ − T₂)/R.

36. Anomalous Expansion of Water

• From 0 °C to 4 °C water contracts (density increases) — the open crystal structure of ice breaks down and molecules pack closer.
• Above 4 °C normal expansion : volume increases, density decreases.
• Water has maximum density at 4 °C.

37. Freezing of a Lake

• Lakes freeze from the top downwards (water at 4 °C, being densest, stays at the bottom).

Top of ice at −T₀, bottom at 0 °C ⇒ ΔT = T₀. Heat conducted up = latent heat released by a new layer dx :

kAT₀x = Ldmdt = ρ(A dx)Ldt ⇒ ∫_h₀^h x dx = kT₀ρL∫₀^t dt

k = conductivity of ice, ρ = density of ice, L = latent heat of fusion. Starting from h₀ = 0, times to freeze successive equal thicknesses (0→x, x→2x, 2x→3x) are in the ratio 1 : 3 : 5.

38. Radiation

• Heat transfer through electromagnetic waves (like light) — needs no medium (e.g. Sun → Earth through space).

39. Prevost's Theory of Heat Exchange

• Every body at any temperature above 0 K continuously emits radiation and also absorbs radiation from the surroundings.
• Rate of emission depends on surface area, absolute temperature and nature of the surface.
• Ice cube : P_emitted < P_absorbed → warms up. Hot coffee : P_emitted > P_absorbed → cools down (even in a vacuum chamber).

40. Nature of Surface

P = P_r + P_a + P_t ⇒ P_rP + P_aP + P_tP = 1

r = reflectivity, a = absorptivity, t = transmissivity (e.g. r = 0.2 ⇒ 20 % reflected).

41. Emissivity (e)

Both at the same temperature, same shape & size, under the same conditions. e is unitless ; 0 ≤ e ≤ 1 ; e = 1 for a black body.

42. Kirchhoff's Law

• At the same temperature, the emissivity of a surface equals its absorptivity.

• Good absorbers are good emitters (black surfaces) ; poor absorbers are poor emitters (white / shiny surfaces).

43. Stefan–Boltzmann Law

• Radiant energy emitted per second per unit area by a black body ∝ T⁴ (T in kelvin).

σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ ≈ 173 × 10⁻⁸ ; A = surface area ; E = emissive power.

44. Body in Surroundings

• Emitted : P = σAeT⁴ ; Absorbed : P_a = σAaT₀⁴ = σAeT₀⁴ (a = e, Kirchhoff).

45. Steady State : Power in = Power out

• Case 1 – body heated by an electric heater of power P :

• Case 2 – body heated by a light beam of intensity I : same equation with P = I × (projected area) ; for a sphere of radius r, P = I · πr².

46. Wien's Displacement Law

• Emissive power E : radiation emitted per second per unit area (E = σeT⁴).
• Spectral emissive power E_λ : radiation emitted per second per unit area per unit wavelength range at wavelength λ.
• λ_m = wavelength at which E_λ is maximum.

• As T increases, λ_m decreases (peak shifts to shorter wavelength) and the peak rises.
• Area under the E_λ–λ curve = E = σeT⁴.

47. Sun–Earth System (Solar Constant)

• Solar constant : average solar energy received per second per unit area (normal to rays) at the top of Earth's atmosphere ≈ 1361 W/m².

R₀, T₀ = radius & temperature of Sun (e = 1) ; R = radius of Earth ; r = Sun–Earth distance.

⇒ I = σT₀⁴ R₀²r² (gives T₀ of Sun if I, R₀, r are known)

πR² = projected area of Earth facing the rays.

48. Newton's Law of Cooling

• Rate of cooling ∝ temperature difference between body and surroundings, provided the difference is small compared with T₀ (in kelvin).

i.e. P = HC(T − T₀), H = ms = heat capacity.

Derivation from Stefan's law :

Let T = T₀ + ΔT, ΔT ≪ T₀ :

− dTdt = σAems[(T₀ + ΔT)⁴ − T₀⁴] = σAeT₀⁴ms[(1 + ΔTT₀)⁴ − 1]

≈ σAeT₀⁴ms · 4ΔTT₀ = 4σAeT₀³ms(T − T₀)

C is constant for a given body in given surroundings.

• Stefan's law (∝ T⁴ − T₀⁴) is exact ; Newton's law (∝ T − T₀) is an approximation. e.g. 30 °C body in 20 °C room : 10 K ≪ 293 K ✓. (Exact solution : T − T₀ = (T_i − T₀)e^−Ct.)

Average-temperature approximation :