Chapter 11 — Thermal Properties of Matter

Heat is energy in transit; temperature is the measure that decides which way it flows. This chapter sets up temperature scales, the ideal-gas law as a primary thermometer, thermal expansion, calorimetry, change-of-state, and the three modes of heat transfer that culminate in the Stefan–Boltzmann law and Newton's law of cooling.

Concept Map

• 11.1 Temperature & heat; thermometers; scales
• 11.2 Ideal gas law; absolute scale
• 11.3 Thermal expansion — linear, area, volume; anomalous water
• 11.4 Heat capacity, specific & molar
• 11.5 Calorimetry — principle of mixtures
• 11.6 Change of state — latent heat; phase diagram; triple point
• 11.7 Heat transfer — conduction, convection, radiation
• 11.8 Blackbody radiation; emissivity

11.1 Temperature & Heat; Thermometers; Scales

Definition

Heat $Q$: energy transferred between two systems (or system & surroundings) due to a temperature difference. SI unit: joule (J); 1 cal = 4.186 J.

Temperature $T$: the property that determines the direction of heat flow. Two bodies at the same temperature exchange no net heat.

A thermometer is any device with a measurable property (volume of liquid, resistance, pressure of a fixed-volume gas, EMF of a thermocouple, colour of radiation) that varies monotonically with temperature.

Scales — conversions

Three standard scales:

Linear conversions:

$\boxed{\ \frac{t_C}{100} = \frac{t_F - 32}{180} = \frac{T - 273.15}{100}\ }$

Useful relations: $t_F = \tfrac{9}{5} t_C + 32,\qquad T = t_C + 273.15.$

Worked Example

Convert $40^\circ$C to Fahrenheit and Kelvin.

$t_F = \tfrac{9}{5}(40) + 32 = 104^\circ\text{F},\quad T = 40 + 273.15 = 313.15\ \text{K}.$

Pitfalls

• Kelvin uses no "$^\circ$" symbol — "$300$ K", not "$300^\circ$K".
• The Kelvin step is the same size as the Celsius step — so $\Delta T_K = \Delta T_C$ but $T_K = T_C + 273.15$.
• $-40^\circ$C $= -40^\circ$F (the famous crossover).

11.2 Ideal Gas Law; Absolute Scale

The Ideal Gas Law

For a fixed mass of an ideal gas:

$\boxed{\ PV = nRT\ }$

with $R = 8.314$ J/(mol·K), $n$ = moles. Equivalently $PV = N k_B T$ with $k_B = 1.38\times 10^{-23}$ J/K.

Special cases:

• Boyle ($T$ const): $PV =$ const.
• Charles ($P$ const): $V/T =$ const.
• Gay-Lussac ($V$ const): $P/T =$ const.
• Avogadro ($P,T$ same): equal volumes contain equal molecules.

Absolute (Kelvin) Scale

Extrapolating Charles's law plot of $V$ vs $t_C$ for any gas at low pressure, $V \to 0$ at $t_C = -273.15^\circ$C — defining absolute zero. The Kelvin scale takes this as its origin:

$T = t_C + 273.15.$

Constant-Volume Gas Thermometer

Pressure of a low-density gas at constant volume defines temperature via

$T = 273.16\ \text{K} \times \frac{P}{P_{tp}},$

where $P_{tp}$ is the pressure at the triple point of water ($T = 273.16$ K). This is the SI primary thermometer.

Worked Example

A balloon at $27^\circ$C contains 2 L of air at 1 atm. Heated to $127^\circ$C at constant pressure. New volume?

$\frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow V_2 = 2 \cdot \frac{400}{300} = 2.67\ \text{L}.$

Pitfalls

• $T$ must be in kelvin for $PV = nRT$, not Celsius.
• "Standard temperature" can mean $0^\circ$C (STP) or $25^\circ$C (chemistry). Check the convention.

11.3 Thermal Expansion

Definitions

For small temperature change $\Delta T$:

• Linear expansion (1-D): $\Delta L = \alpha L\, \Delta T$, coefficient $\alpha$ in K⁻¹.
• Area expansion (2-D): $\Delta A = \beta A\, \Delta T$.
• Volume expansion (3-D): $\Delta V = \gamma V\,\Delta T$.

Derivation — relation $\alpha : \beta : \gamma = 1 : 2 : 3$

Take an isotropic cube of side $L$. Each side becomes $L(1+\alpha\Delta T)$.

Area: $A' = L^2(1+\alpha\Delta T)^2 \approx L^2(1 + 2\alpha\Delta T) \Rightarrow \Delta A/A = 2\alpha\Delta T \Rightarrow \beta = 2\alpha.$

Volume: $V' = L^3(1+\alpha\Delta T)^3 \approx L^3(1 + 3\alpha\Delta T) \Rightarrow \gamma = 3\alpha.$

$\boxed{\ \alpha : \beta : \gamma = 1 : 2 : 3\ }$

Valid in the small-strain limit ($\alpha\Delta T \ll 1$, almost always true).

Anomalous Expansion of Water

Between $0^\circ$C and $4^\circ$C, water contracts on heating; above $4^\circ$C it expands normally. At $4^\circ$C water has maximum density ($\approx 1000$ kg/m³). Reason: residual hydrogen-bonded "ice-like" clusters collapse, reducing volume, dominating over thermal kinetic expansion in this narrow band.

Consequence: a freezing pond freezes from the top. The $4^\circ$C-densest water sits at the bottom; ice (lower density) floats; aquatic life survives below.

Worked Example

A steel ruler is calibrated at $25^\circ$C. Length read on a hot day at $40^\circ$C is $1.000$ m. True length? ($\alpha_\text{steel} = 1.2\times 10^{-5}$ K⁻¹.)

The ruler itself has expanded: each mark is further apart. So the actual length is

$L_\text{true} = (1.000)(1 + 1.2\times 10^{-5}(15)) = 1.00018\ \text{m}.$

Pitfalls

• A hole in a metal plate expands on heating — the cavity behaves like a virtual piece of the same metal.
• Volume coefficient for an ideal gas at constant pressure is $\gamma = 1/T$ (not constant in temperature) — much larger than for solids.

11.4 Heat Capacity, Specific & Molar

Definitions

Heat capacity $C$: heat needed to raise temperature by 1 K — extensive, depends on amount.

$Q = C\,\Delta T.$

Specific heat capacity $c$ (per unit mass):

$\boxed{\ Q = m\,c\,\Delta T\ }$

SI unit J/(kg·K).

Molar specific heat $C_m$:

$Q = n\,C_m\,\Delta T,\quad C_m = M c\quad\text{(M = molar mass)}.$

For gases two values exist:

• $C_V$ (constant volume): heat goes entirely to internal energy.
• $C_P$ (constant pressure): heat raises internal energy and does $PdV$ work.

Mayer's relation (Ch 12): $C_P - C_V = R$.

Water's exceptionally high specific heat moderates Earth's climate and makes it ideal as a coolant.

Worked Example

$200$ g of water heated from $20^\circ$C to $80^\circ$C. Heat required?

$Q = (0.2)(4186)(60) = 50.2\times 10^3\ \text{J} = 50.2\ \text{kJ}.$

Pitfalls

• Specific heat is temperature dependent (Dulong-Petit, then $T^3$ at low $T$ — not in NCERT).
• For gases always specify $C_V$ or $C_P$ — they differ by $R$.

11.5 Calorimetry

Principle of Mixtures

In an isolated calorimeter (no exchange with surroundings):

$\boxed{\ \text{Heat lost by hotter body} = \text{Heat gained by colder body}\ }$

Provided no phase change occurs:

$m_1 c_1 (T_1 - T) = m_2 c_2 (T - T_2).$

Worked Example — temperature of mixing

$100$ g of water at $80^\circ$C is mixed with $300$ g of water at $20^\circ$C. Final temperature?

$100(80 - T) = 300(T - 20) \Rightarrow 8000 - 100T = 300T - 6000 \Rightarrow T = 35^\circ\text{C}.$

Worked Example — with a calorimeter

A copper calorimeter of mass $100$ g contains $200$ g of water at $25^\circ$C. A lead piece of $50$ g at $100^\circ$C is dropped in. Find final temperature. ($c_\text{Cu} = 390, c_\text{water} = 4186, c_\text{Pb} = 130$.)

Heat lost by lead = heat gained by water + calorimeter: $50(130)(100 - T) = 200(4186)(T - 25) + 100(390)(T - 25),$ $6500(100 - T) = (8.372\times 10^5 + 3.9\times 10^4)(T - 25),$ $6500(100 - T) = 8.76\times 10^5 (T - 25).$

$T \approx 25.55^\circ$C — almost no change because the lead's heat capacity is tiny.

Pitfalls

• Always convert masses to kg if using SI $c$.
• If the colder body undergoes melting / boiling during mixing, latent heat must be added separately — see next section.
• Water equivalent of a calorimeter: $W = m\,c/c_\text{water}$ — heat absorbed by the vessel expressed as an equivalent mass of water.

11.6 Change of State; Latent Heat

Definition

During a phase change (solid↔liquid, liquid↔gas) temperature stays constant; the heat supplied is used to rearrange molecular bonds.

$\boxed{\ Q = m\,L\ }$

$L$ = latent heat (J/kg).

• Latent heat of fusion $L_f$: ice → water at $0^\circ$C: $L_f = 3.34\times 10^5$ J/kg.
• Latent heat of vaporisation $L_v$: water → steam at $100^\circ$C: $L_v = 2.26\times 10^6$ J/kg.

That's why steam burns are far worse than boiling-water burns: every kg of steam condensing on skin releases $L_v$ before further cooling.

Phase Diagram (intro)

     P
     ^         | (Solid)
     |  Solid  |
     |---------+--- critical pt
     |         |  Liquid       /
     | triple  +--------------+ critical
     |  pt   __|              :
     |     /   |   Gas        : Supercritical
     |____|____|______________:______
                                T

• Triple point: unique $(P,T)$ at which all three phases coexist (water: $611.657$ Pa, $273.16$ K = $0.01^\circ$C).
• Critical point: above $T_c$ no liquid–gas distinction exists.
• Sublimation: solid → gas directly (dry ice CO₂, naphthalene).

Effect of pressure

• Melting point of ice decreases with pressure ($-7.4\times 10^{-3}$ °C/atm) — water is anomalous.
• For most substances melting point rises with pressure.
• Boiling point always rises with pressure (pressure cooker).

Worked Example

How much heat to convert $50$ g of ice at $-10^\circ$C into steam at $100^\circ$C?

Five segments:

1. Warm ice $-10\to 0$: $Q_1 = (0.05)(2100)(10) = 1050$ J.
2. Melt ice: $Q_2 = (0.05)(3.34\times 10^5) = 16700$ J.
3. Warm water $0 \to 100$: $Q_3 = (0.05)(4186)(100) = 20930$ J.
4. Vaporise: $Q_4 = (0.05)(2.26\times 10^6) = 113000$ J.

Total $Q = 151.7$ kJ.

Pitfalls

• During phase change temperature is constant; do not add $mc\Delta T$ over the transition.
• Heating curve of water: 5 segments (ice warm, melt, water warm, boil, steam warm) — each with its own slope or plateau.

11.7 Heat Transfer

Conduction — Fourier's Law

In a steady-state slab of thickness $L$, cross-section $A$, with hot face at $T_1$, cold at $T_2$:

$\boxed{\ \frac{dQ}{dt} = -kA\,\frac{dT}{dx} = kA\,\frac{T_1 - T_2}{L}\ }$

$k$ = thermal conductivity, W/(m·K).

Thermal resistance: $R_\text{th} = L/(kA)$; in series $R_\text{tot} = \sum R_i$; in parallel $1/R_\text{tot} = \sum 1/R_i$.

Convection

Bulk movement of a heated fluid (no formula required, just qualitative). Two types:

• Natural convection: density change drives flow (rising hot air, sea breezes, monsoons).
• Forced convection: fan, pump, blood circulation.

Radiation — Stefan-Boltzmann

Every body at temperature $T > 0$ K emits electromagnetic radiation. For a perfect blackbody the total emitted power per unit area is

$\boxed{\ \frac{P}{A} = \sigma T^4\ }$

$\sigma = 5.67\times 10^{-8}$ W/(m²·K⁴).

For a real body of emissivity $e$ ($0\le e\le 1$):

$\frac{P}{A} = e\sigma T^4.$

Net radiation between body (T) and surroundings ($T_0$):

$P_\text{net} = e\sigma A (T^4 - T_0^4).$

Wien's Displacement Law

The wavelength of peak emission shifts with temperature:

$\boxed{\ \lambda_\text{max}\,T = b\ },\qquad b = 2.898\times 10^{-3}\ \text{m·K}.$

Examples: Sun ($T\approx 5800$ K) peaks at $\lambda_\text{max} \approx 500$ nm (green-yellow, eye-matched). A blacksmith's iron at $1000$ K peaks at $2.9\ \mu$m (infrared) — glow we see is the tail of the distribution.

Newton's Law of Cooling

When a body is only slightly hotter than its surroundings, $T - T_0 \ll T_0$:

$T^4 - T_0^4 = (T_0 + \Delta T)^4 - T_0^4 \approx 4 T_0^3 \Delta T.$

So the rate of heat loss is proportional to $\Delta T$:

$-\frac{dT}{dt} = k(T - T_0).$

Solving:

$\boxed{\ T(t) - T_0 = (T_i - T_0)\,e^{-kt}\ }$

— exponential decay to ambient.

Worked Example — Conduction

A copper rod 50 cm long, cross-section $4\ \text{cm}^2$, with ends at $100^\circ$C and $0^\circ$C ($k = 385$).

$\frac{dQ}{dt} = \frac{(385)(4\times 10^{-4})(100)}{0.5} = 30.8\ \text{W}.$

Worked Example — Radiation

A blackbody sphere of radius $5$ cm at $500$ K. Radiated power?

$A = 4\pi(0.05)^2 = 0.0314\ \text{m}^2,$ $P = \sigma A T^4 = (5.67\times 10^{-8})(0.0314)(500)^4 = 111\ \text{W}.$

Worked Example — Newton's law

A cup of tea cools from $80^\circ$C to $60^\circ$C in $5$ min, surroundings $25^\circ$C. Time to cool from $60$ to $40$?

Using mean-temperature form:

$\frac{80 - 60}{5} = k(70 - 25) \Rightarrow k = \frac{4}{45} = 0.0889\ \text{min}^{-1}.$ $\frac{60 - 40}{t} = k(50 - 25) = 0.0889(25) = 2.22 \Rightarrow t = 9.0\ \text{min}.$

Pitfalls

• Stefan-Boltzmann uses kelvin, raised to the fourth power — sensitivity to $T$ is huge.
• Wien's law: $\lambda T =$ const, not $\lambda/T =$ const.
• Newton's law is linear — strictly valid only for small $\Delta T$.

11.8 Blackbody & Emissivity

Blackbody

An idealised object that absorbs all incident radiation (none reflected, none transmitted) and re-emits a universal spectrum depending only on $T$. Realised approximately by a small hole in a heated cavity (Hohlraum).

• Total emitted power: $\sigma T^4$ per unit area.
• Spectral distribution: Planck's law (not in NCERT formal scope, but qualitatively a curve peaking at $\lambda_\text{max}$).

Emissivity

For a real body, emissivity $e$ is the ratio of its emission to that of a blackbody at the same temperature, at every wavelength (grey body approximation: $e$ independent of $\lambda$).

• Polished silver: $e \approx 0.02$.
• Soot, lampblack: $e \approx 0.95$.

Kirchhoff's law: good emitters are good absorbers — $a = e$.

Worked Example

A car radiator at $90^\circ$C ($363$ K) in surroundings of $300$ K, surface area $0.5\ \text{m}^2$, emissivity $0.9$. Net radiative loss?

$P = (0.9)(5.67\times 10^{-8})(0.5)(363^4 - 300^4) = (0.9)(5.67\times 10^{-8})(0.5)(1.737\times 10^{10} - 8.1\times 10^{9}),$ $= (0.9)(5.67\times 10^{-8})(0.5)(9.27\times 10^{9}) = 237\ \text{W}.$

Pitfalls

• Blackbody is a mathematical idealisation — the Sun is "approximately blackbody" with $e\sim 1$.
• Emissivity may differ from absorptivity only across different wavelengths; at the same wavelength they are equal (Kirchhoff).

Solved Problems

1. A glass beaker of $100$ ml is full at $20^\circ$C. Heated to $80^\circ$C. Volume of water that overflows? ($\gamma_\text{glass} = 27\times 10^{-6}$, $\gamma_\text{water} = 2.1\times 10^{-4}$ K⁻¹.)

$\Delta V = V(\gamma_\text{water} - \gamma_\text{glass})\Delta T = 100(2.1\times 10^{-4} - 2.7\times 10^{-5})(60) = 1.10\ \text{ml}.$

2. A pendulum clock is correct at $20^\circ$C. How much does it lose per day at $35^\circ$C if the pendulum is brass ($\alpha = 1.9\times 10^{-5}$ K⁻¹)?

$\frac{\Delta T}{T} = \tfrac{1}{2}\alpha\Delta\theta = \tfrac{1}{2}(1.9\times 10^{-5})(15) = 1.43\times 10^{-4},$

so $\Delta T = 1.43\times 10^{-4} \times 86400 \approx 12.3$ s/day (loss).

3. $30$ g of ice at $0^\circ$C is added to $200$ g of water at $30^\circ$C. Final temperature?

Heat from water → melt ice + warm cold water:

$200(4186)(30 - T) = 30(3.34\times 10^{5}) + 30(4186)T.$ $8.372\times 10^5 (30 - T) = 10020 \times 10^3 + 125580 T...$

Working in cal (easier): $200(1)(30-T) = 30(80) + 30(1)T$, $6000 - 200T = 2400 + 30T$, $T = 15.65^\circ$C.

4. Two rods of equal length, copper and steel, joined end to end. Free end of copper at $100^\circ$C, free end of steel at $0^\circ$C. Temperature at the junction? ($k_\text{Cu} = 385, k_\text{St} = 50$.)

Steady state, same flow: $\frac{k_\text{Cu}(100 - T_j)}{L} = \frac{k_\text{St}(T_j - 0)}{L},$ $385(100 - T_j) = 50 T_j \Rightarrow T_j = \frac{38500}{435} = 88.5^\circ\text{C}.$

5. A blackbody radiates $100$ W at $300$ K. Power at $600$ K?

$\frac{P_2}{P_1} = (600/300)^4 = 16 \Rightarrow P_2 = 1600\ \text{W}.$

6. Wien's law for the cosmic microwave background, $T = 2.73$ K. Peak wavelength?

$\lambda_\text{max} = \frac{2.898\times 10^{-3}}{2.73} = 1.06\times 10^{-3}\ \text{m} \approx 1.06\ \text{mm}\quad\text{(microwave)}.$

7. A cup of coffee at $90^\circ$C in a room at $25^\circ$C reaches $80^\circ$C in $1$ min. Estimated time to reach $50^\circ$C?

Approximate Newton's law: $\ln[(T-T_0)/(T_i-T_0)] = -kt$.

$\ln(55/65) = -k(1) \Rightarrow k = 0.167$/min. To reach $50$: $\ln(25/65)=-kt \Rightarrow t = 5.7$ min.

JEE/NEET Edge Cases

• Bimetallic strip: two metals of different $\alpha$ riveted together; bends on heating because one expands more — basis of thermostats.
• Pendulum clock keeps slower time in summer (longer pendulum); compensated by gridiron or invar pendulum.
• Hole in a metal plate expands (the cavity expands, contrary to first guess).
• Density change under heating: $\rho' = \rho/(1+\gamma\Delta T)$.
• Thermal stress = $Y\alpha\Delta T$ (revisit Ch 9).
• Combined cooling: Newton's law applies only for small $\Delta T$; for large differences use Stefan-Boltzmann directly.
• Ratio $\alpha_\text{gas} \gg \alpha_\text{solid}$: at constant $P$, $\gamma_\text{gas} = 1/T \approx 1/300 \approx 3\times 10^{-3}$ K⁻¹, vs $\sim 10^{-5}$ for solids.
• Heating curve plateau must be calculated separately from latent heat, never $mc\Delta T$.

Quick Recap

• Kelvin = Celsius $+ 273.15$; $\Delta T_K = \Delta T_C$.
• Ideal gas: $PV = nRT$.
• Linear/area/volume expansion in ratio $1:2:3$ for isotropic solid.
• Anomalous water: densest at $4^\circ$C; ice floats.
• $Q = mc\Delta T$ for warming; $Q = mL$ for phase change.
• Principle of mixtures balances heat lost = heat gained.
• Fourier: $dQ/dt = kA\Delta T/L$.
• Stefan-Boltzmann: $P = e\sigma A T^4$; net $\propto T^4 - T_0^4$.
• Wien: $\lambda_\text{max} T = b$.
• Newton's law of cooling: exponential approach to ambient for small $\Delta T$.

Formula Sheet