Thermal Properties of Matter

Temperature is the physical quantity that determines the direction of net heat flow between two bodies in thermal contact — heat flows from a body at higher temperature to one at lower temperature until thermal equilibrium is reached. The zeroth law of thermodynamics states that if two systems A and B are separately in thermal equilibrium with a third system C, then A and B are also in thermal equilibrium with each other. This law provides the logical basis for using a thermometer (system C) to compare the temperatures of other bodies.

A thermometer measures temperature through some thermometric property that changes uniformly with hotness — for example, the length of a mercury column, the pressure of a gas at constant volume, or the resistance of a wire. The most reliable device is the constant-volume gas thermometer, because different gases give nearly identical temperature values (unlike liquid-in-glass thermometers, whose readings vary slightly with the liquid used).

In the SI system, the size of one kelvin is fixed by choosing the triple point of water (the unique temperature and pressure at which ice, water and water vapour coexist in equilibrium) as exactly 273.16 K. For a constant-volume gas thermometer, the temperature of a body is then defined as:

T = 273.16 K × (P / P_tr)

where P is the gas pressure at the unknown temperature and P_tr is the gas pressure at the triple point of water, both measured at constant volume. As the amount of gas used is made smaller and smaller (so the gas behaves more ideally), readings from different gases converge to the same value — this limiting value defines the ideal-gas temperature scale, which is independent of the properties of any particular substance.

The everyday Celsius scale is related to the Kelvin scale by T_C = T − 273.15, and to the Fahrenheit scale by T_F = (9/5)T_C + 32.

Worked Example: A constant-volume gas thermometer reads a pressure of 1.500 × 10⁵ Pa at the triple point of water. When inserted into a hot bath, it reads 2.050 × 10⁵ Pa. Find the temperature of the bath.

Given: P_tr = 1.500 × 10⁵ Pa, P = 2.050 × 10⁵ Pa

Formula: T = 273.16 K × (P / P_tr)

Substitution: T = 273.16 × (2.050 × 10⁵ / 1.500 × 10⁵) = 273.16 × 1.3667

Result: T ≈ 373.3 K (≈ 100.2 °C, close to the steam point)

Most substances expand on heating because the average separation between their constituent atoms/molecules increases as thermal energy (vibrational kinetic energy) increases. For a solid rod of original length L₀ heated through ΔT, the new length is:

L = L₀(1 + α ΔT)

where α is the coefficient of linear expansion (unit: K⁻¹ or °C⁻¹), assumed constant over the temperature range considered. Similarly, for area A₀ and volume V₀:

A = A₀(1 + β ΔT), V = V₀(1 + γ ΔT)

where β is the coefficient of superficial (area) expansion and γ is the coefficient of cubical (volume) expansion. For an isotropic solid it can be shown that:

β ≈ 2α and γ ≈ 3α, so that α : β : γ = 1 : 2 : 3

Liquids have only a volume expansion coefficient γ (they have no fixed shape). Gases expand far more than solids or liquids for the same temperature change; for an ideal gas at constant pressure, γ = 1/T.

Anomalous expansion of water: Unlike most liquids, water contracts on heating from 0 °C to 4 °C and expands only above 4 °C — its density is therefore maximum at 4 °C, not at 0 °C. This is why ice forms first at the top of a lake and floats, insulating the water below and allowing aquatic life to survive winter.

Worked Example: A steel rod is exactly 1.000 m long at 0 °C. If the coefficient of linear expansion of steel is α = 1.2 × 10⁻⁵ °C⁻¹, find its length at 100 °C.

Given: L₀ = 1.000 m, α = 1.2 × 10⁻⁵ °C⁻¹, ΔT = 100 °C

Formula: L = L₀(1 + α ΔT)

Substitution: ΔL = L₀ α ΔT = 1.000 × 1.2 × 10⁻⁵ × 100 = 1.2 × 10⁻³ m

Result: ΔL = 1.2 mm, so L = 1.0012 m at 100 °C

When heat ΔQ is supplied to a body and its temperature rises by ΔT, the heat capacity of the body is S = ΔQ/ΔT. The heat capacity per unit mass is the specific heat capacity, s = ΔQ/(mΔT), so that:

ΔQ = m s ΔT

The SI unit of specific heat capacity is J kg⁻¹ K⁻¹. Water has an unusually high specific heat capacity (about 4186 J kg⁻¹ K⁻¹), which is why large water bodies moderate the climate of nearby land and why water is used as a coolant.

For a gas, heat can be supplied at constant volume or at constant pressure, giving two distinct molar specific heats, C_V and C_P. Since a gas does work while expanding at constant pressure, C_P is always greater than C_V; for an ideal gas they are related by Mayer's relation, C_P − C_V = R, where R is the universal gas constant.

Calorimetry is the technique of measuring heat exchanges. If a hot body is mixed with a cold body inside an insulated container (calorimeter) so that no heat is lost to the surroundings, then by conservation of energy:

Heat lost by the hot body(ies) = Heat gained by the cold body(ies) and the calorimeter

Worked Example: A 0.20 kg block of metal at 150 °C is dropped into 0.50 kg of water at 25 °C in a calorimeter of negligible heat capacity. The mixture settles at a final temperature of 30 °C. Find the specific heat capacity of the metal (s_water = 4186 J kg⁻¹ K⁻¹).

Given: m_metal = 0.20 kg, ΔT_metal = 150 − 30 = 120 °C; m_water = 0.50 kg, ΔT_water = 30 − 25 = 5 °C

Formula: m_metal s_metal ΔT_metal = m_water s_water ΔT_water

Substitution: 0.20 × s_metal × 120 = 0.50 × 4186 × 5 = 10465 J, so 24 s_metal = 10465

Result: s_metal ≈ 436 J kg⁻¹ K⁻¹ (close to the known specific heat of iron)

When a solid is heated, its temperature rises until it starts melting; while melting continues, the temperature remains constant at the melting point even though heat is still being absorbed — this heat breaks the intermolecular bonds of the solid lattice rather than raising kinetic energy/temperature. Similarly, a liquid boils and turns into vapour at a constant boiling point. The heat absorbed or released per unit mass during such a change of state, at constant temperature, is called the latent heat, L:

Q = m L

The specific latent heat of fusion of ice is L_f = 3.34 × 10⁵ J kg⁻¹, and the specific latent heat of vaporisation of water is L_v = 2.256 × 10⁶ J kg⁻¹. A substance may also pass directly from solid to vapour without becoming liquid — this is called sublimation (e.g. dry ice, camphor, iodine).

Both melting point and boiling point depend on pressure. Increasing pressure generally lowers the melting point of ice (this allows regelation — ice melts under high local pressure and refreezes when the pressure is removed) but raises the boiling point of water (used in pressure cookers to cook food faster). At a particular pressure and temperature, called the triple point, the solid, liquid and vapour phases of a substance coexist in equilibrium; for water this is 273.16 K at about 611.73 Pa.

Worked Example: Find the total heat needed to convert 1.0 kg of ice at −10 °C completely to steam at 100 °C (s_ice = 2100 J kg⁻¹K⁻¹, s_water = 4186 J kg⁻¹K⁻¹, L_f = 3.34 × 10⁵ J kg⁻¹, L_v = 2.256 × 10⁶ J kg⁻¹).

Given: m = 1.0 kg; four stages: heating ice, melting, heating water, vaporising

Formula and substitution:

• Heat ice from −10 °C to 0 °C: Q₁ = m s_ice ΔT = 1.0 × 2100 × 10 = 21000 J
• Melt ice at 0 °C: Q₂ = m L_f = 1.0 × 3.34 × 10⁵ = 334000 J
• Heat water from 0 °C to 100 °C: Q₃ = m s_water ΔT = 1.0 × 4186 × 100 = 418600 J
• Vaporise water at 100 °C: Q₄ = m L_v = 1.0 × 2.256 × 10⁶ = 2256000 J

Result: Total Q = Q₁ + Q₂ + Q₃ + Q₄ = 21000 + 334000 + 418600 + 2256000 = 3029600 J ≈ 3.03 × 10⁶ J

Heat can travel from a hotter region to a colder one by three mechanisms: conduction, convection and radiation. In conduction, heat is transferred through a material by successive collisions between vibrating atoms/molecules (and free electrons in metals), without any net movement of the material itself. For a rod or slab of cross-sectional area A and length L, with its two faces maintained at temperatures T₁ and T₂ (T₁ > T₂), the steady-state rate of heat flow (heat current) is:

H = ΔQ/Δt = K A (T₁ − T₂) / L

where K is the thermal conductivity of the material (SI unit: W m⁻¹ K⁻¹). Metals have high K (good conductors), while air, wood and most porous materials have low K (good insulators; trapped air is an especially poor conductor, which is why woollen clothes, thermos flasks and double-glazed windows work).

The quantity L/(KA) behaves like an electrical resistance to heat flow and is called thermal resistance, R = L/(KA), so that H = ΔT/R. For several slabs of a composite wall placed in series (same area A, same heat current through each), the equivalent thermal resistance simply adds: R_total = R₁ + R₂ + …

In convection, heat is carried by the actual bulk movement of a heated fluid (liquid or gas) from one place to another. Natural convection arises because a heated fluid expands, becomes less dense, and rises, being replaced by cooler, denser fluid (e.g. land and sea breezes, boiling water, atmospheric circulation). Forced convection uses a pump or fan to move the fluid (e.g. car radiators, forced-air heating).

Worked Example: A composite wall has two slabs in series: slab 1 has thickness L₁ = 2 cm and K₁ = 0.2 W m⁻¹K⁻¹; slab 2 has thickness L₂ = 3 cm and K₂ = 0.4 W m⁻¹K⁻¹. Area A = 1 m², temperature difference across the composite = 50 °C. Find the steady-state heat current.

Given: L₁ = 0.02 m, K₁ = 0.2 W m⁻¹K⁻¹; L₂ = 0.03 m, K₂ = 0.4 W m⁻¹K⁻¹; A = 1 m²; ΔT = 50 °C

Formula: R = L/(KA); R_total = R₁ + R₂; H = ΔT/R_total

Substitution: R₁ = 0.02/(0.2×1) = 0.1 K/W; R₂ = 0.03/(0.4×1) = 0.075 K/W; R_total = 0.175 K/W

Result: H = 50/0.175 ≈ 285.7 W

Unlike conduction and convection, radiation does not need a material medium — energy is transferred as electromagnetic waves and can travel through vacuum (this is how the sun heats the earth). All bodies above 0 K continuously emit thermal radiation, and also absorb radiation falling on them. A black body is an idealised object that absorbs all radiation incident on it and, at a given temperature, emits the maximum possible amount of radiation for that temperature.

The Stefan–Boltzmann law states that the power radiated per unit area (radiant emittance) of a black body is proportional to the fourth power of its absolute temperature:

E = σ T⁴

where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ is the Stefan–Boltzmann constant. A real (non-black) body with surface area A and emissivity e (0 < e ≤ 1, e = 1 for a perfect black body) radiates at a rate P = e σ A T⁴. If the body is surrounded by an enclosure at temperature T₀, it also absorbs radiation from the surroundings, so the net rate of loss of thermal energy is:

P_net = e σ A (T⁴ − T₀⁴)

The wavelength at which a black body radiates most strongly, λ_m, shifts to shorter wavelengths as temperature increases — this is Wien's displacement law:

λ_m T = b

where b = 2.9 × 10⁻³ m K is Wien's constant. This is why an object glows red when moderately hot and bluish-white when very hot, and is used in pyrometry/thermal imaging to estimate the temperature of distant hot bodies such as stars.

Newton's law of cooling is an approximate form of the radiation loss law, valid only when the temperature difference between a body and its surroundings is small. It states that the rate of loss of heat (or temperature) of a body is directly proportional to the temperature difference between the body and its surroundings:

dT/dt = −k (T − T₀)

For a finite time interval in which temperature falls from T₁ to T₂ in time t, the average form commonly used in numericals is:

(T₁ − T₂) / t = k [ (T₁ + T₂)/2 − T₀ ]

Worked Example: A body cools from 80 °C to 70 °C in 5 minutes when the surrounding temperature is 30 °C. Using Newton's law of cooling, find the time taken to cool from 70 °C to 60 °C.

Given: T₁ = 80 °C, T₂ = 70 °C, t₁ = 5 min, T₀ = 30 °C

Formula: (T₁ − T₂)/t = k[(T₁+T₂)/2 − T₀]

Substitution (first interval): (80−70)/5 = k[(80+70)/2 − 30] ⟹ 2 = k(75−30) = 45k ⟹ k = 2/45 ≈ 0.0444 min⁻¹

Substitution (second interval, 70 °C to 60 °C): (70−60)/t₂ = 0.0444 × [(70+60)/2 − 30] = 0.0444 × 35 = 1.556

Result: t₂ = 10/1.556 ≈ 6.43 minutes