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Thermal Physics
Thermal Expansion of Solids
Thermal expansion of solids is the tendency of solid matter to change in volume, area, and length in response to a change in temperature. This phenomenon arises due to the asymmetry of the interatomic potential energy well and the increased anharmonicity of atomic vibrations at elevated temperatures.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Linear, superficial, and cubical expansion coefficients (alpha, beta, gamma), their interrelations, determination of expansion effects in simple structures, and pendulum clock time-loss problems.
PGT extension
Temperature dependence of expansion coefficients, anisotropic expansion in non-cubic crystals, stress and strain generated by thermal constraints, and thermal bending of bimetallic strips.
Prerequisites
Concept of Temperature
Kinetic theory of gases
Conductors
Microscopic Origin of Thermal Expansion
Solids consist of atoms vibrating about their equilibrium positions inside an asymmetric potential well.
As temperature rises, the total vibrational energy increases, raising the average interatomic distance due to the steeper repulsive wall compared to the attractive wall.
Harmonic approximation yields zero thermal expansion; anharmonicity of lattice vibrations is strictly required for thermal expansion.
Types of Thermal Expansion
Linear expansion refers to the fractional change in length per unit change in temperature, quantified by the coefficient of linear expansion (α).
Superficial (area) expansion refers to the fractional change in surface area per unit change in temperature, quantified by β.
Cubical (volume) expansion refers to the fractional change in volume per unit change in temperature, quantified by γ.
Interrelation Between Expansion Coefficients
For isotropic solids, the coefficients are related by the standard geometric ratios: $\beta = 2\alpha$ and $\gamma = 3\alpha$.
These relations are derived by neglecting higher-order terms of the expansion coefficient multiplied by the temperature change (ΔT).
For anisotropic crystals, expansion coefficients vary with crystallographic axes, leading to tensor representations of thermal expansion.
Thermal Stress and Bimetallic Strips
If a solid is rigidly clamped and prevented from expanding or contracting, thermal stress develops inside the material, proportional to Young's modulus and the temperature change.
A bimetallic strip consists of two metals with different expansion coefficients riveted together, causing the strip to bend toward the metal with the lower coefficient upon heating.
Derivations
Relation between Superficial and Linear Expansion Coefficients
Conditions: Isotropic solid sheet undergoing a small uniform temperature change ΔT
Consider a rectangular sheet of initial dimensions L0 and breadth B0 at temperature T0.
Upon heating by ΔT, the new dimensions become $L = L_{0}(1 + \alpha\DeltaT)$ and $B = B_{0}(1 + \alpha\DeltaT)$.
The new area $A = L \times B = L_{0}B_{0}(1 + \alpha\DeltaT)^{2} = A_{0}(1 + 2\alpha\DeltaT + \alpha^{2}(\DeltaT)^{2})$.
Neglecting the small second-order term α2(ΔT)2, we get $A = A_{0}(1 + 2\alpha\DeltaT)$.
By definition of superficial expansion, $A = A_{0}(1 + \beta\DeltaT)$, therefore equating the coefficients gives $\beta = 2\alpha$.
Result: \beta = 2\alpha
Thermal Stress in a Constrained Rod
Conditions: A uniform metallic rod of length L, cross-sectional area A, and Young's modulus Y fixed rigidly between two immovable walls
Let the temperature of the rod be increased by ΔT. If free, the expansion would be ΔLfree = L α ΔT.
Since the walls prevent this expansion, a compressive strain is induced in the rod given by Strai$n = \DeltaL / L = \alpha \DeltaT$.
Using Hooke's law, Stres$s = Y \times Strain = Y \alpha \DeltaT$.
The total compressive force (F) exerted by the rod on the walls is given by $F = Stress \times A = Y A \alpha \DeltaT$.
Result: F = YA\alpha\Delta T
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Linear Thermal Expansion
$$L = L_0 (1 + \alpha \Delta T)$$
L = Final length, L0 = Initial length, $\alpha = Coefficient of linear expansion$, ΔT = Change in temperature m (length), K-1 (α)
αΔT << 1, isotropic material
Area Thermal Expansion
$$A = A_0 (1 + \beta \Delta T)$$
A = Final area, A0 = Initial area, $\beta = Coefficient of area expansion$ m2 (area), K-1 (β)
β ≈ 2α
Volume Thermal Expansion
$$V = V_0 (1 + \gamma \Delta T)$$
V = Final volume, V0 = Initial volume, $\gamma = Coefficient of volume expansion$ m3 (volume), K-1 (γ)
γapp = Apparent expansion coefficient, γreal = Real expansion coefficient of liquid, γvessel = Volume expansion coefficient of container K-1
Container expands along with the liquid
Graphs and experiments
Interatomic Potential Energy vs Interatomic Distance
Draw: An asymmetric potential well curve U(r) versus interatomic separation r, showing horizontal energy levels representing vibrational states.
Interpret: The asymmetry of the well causes the mean separation (average of turning points) to shift to larger values as vibrational energy increases with temperature.
Exam use: Theoretical explanation of why solids expand when heated.
Length vs Temperature Graph
Draw: A straight line plot of length L versus temperature T with a positive slope equal to L0α.
Interpret: The slope of the curve represents the absolute linear expansion rate for a given initial length.
Exam use: Determining the coefficient of linear expansion from experimental data points.
Searches' Apparatus / Pullinger's Apparatus
Draw: Schematic setup of a metallic rod enclosed in a steam jacket with a micrometer screw gauge measuring linear expansion.
Interpret: Direct measurement of expansion ΔL corresponding to steam temperature change enables precise evaluation of α.
Exam use: Experimental determination and viva-voce questions on laboratory measurement of expansion coefficients.
Solved numericals
TGT core
A metal rod of length 2 m at 20 °C is heated to 100 °C. If the coefficient of linear expansion of the metal is 1.2 x 10-5 K-1, calculate the final length of the rod.
Identify given parameters: L0 = 2 m, T1 = 20 °C, T2 = 100 °C, $\alpha = 1.2 x 10^{-5} K^{-1}$.
Calculate temperature change: Δ$T = T_{2} - T_{1} = 100 - 20 = 80 °C (or K)$.
Apply linear expansion formula: $L = L_{0}(1 + \alpha\DeltaT)$.
Substitute values: $L = 2 [1 + (1.2 x 10^{-5}) x 80] = 2 [1 + 9.6 x 10^{-4}] = 2 [1 + 0.00096] = 2 x 1.00096 = 2.00192 m$.
Answer: 2.00192 m
PGT extension
A brass rod of cross-sectional area 4 cm2 is clamped firmly at both ends so that it cannot expand. If the temperature is raised by 50 °C, calculate the thermal force developed in the rod. (Given for brass: $Y = 1.0 x 10^{11} N/m^{2}$, $\alpha = 2.0 x 10^{-5} K^{-1})$
Identify given parameters: $A = 4 cm^{2} = 4 x 10^{-4} m^{2}$, ΔT = 50 K, $Y = 1.0 x 10^{11} N/m^{2}$, $\alpha = 2.0 x 10^{-5} K^{-1}$.
Recall thermal force formula: $F = Y A \alpha \DeltaT$.
Substitute the numerical values: $F = (1.0 x 10^{11}) x (4 x 10^{-4}) x (2.0 x 10^{-5}) x 50$.
Simplify the exponents: 1011 x 10-4 x 10-5 = 102 = 100.
Multiply coefficients: 1.0 x 4 x 2.0 x 50 = 400.
Calculate final force: F = 400 x 100 = 40,000 $N = 4 x 10^{4} N$.
Answer: 4 x 10^4 N
PGT extension
A pendulum clock keeps correct time at 20 °C. The pendulum consists of a brass rod of linear expansion coefficient $\alpha = 1.9 x 10^{-5} K^{-1}$. How much time will the clock lose or gain in one day if the room temperature rises to 40 °C?
Identify given parameters: $\alpha = 1.9 x 10^{-5} K^{-1}$, Δ$T = 40 - 20 = 20 K$, Time in one day (t) = 24 x 3600 s.
Use fractional time loss relation: Δt / $t = (1/2) \alpha \DeltaT$.
Substitute values: Δ$t = (1/2) x (1.9 x 10^{-5}) x 20 x (86400)$.
Calculate intermediate product: (1/2) x $20 = 10; \Deltat = 1.9 x 10^{-5} x 10 x 86400 = 1.9 x 10^{-4} x 86400$.
Compute final value: Δt = 16.416 seconds.
Answer: Loss of 16.42 seconds per day
Assumptions and limitations
The coefficients of expansion (α, β, γ) are assumed to be constant over the moderate temperature range considered, whereas in reality, they depend slightly on temperature.
Higher-order terms in temperature change (such as α2(ΔT)2) are neglected under the assumption of small temperature variations (ΔT << 1 / α).
Isotropy is assumed for simple geometric relations between linear, superficial, and cubical expansion; anisotropic crystals require tensorial treatment.
Common misconceptions
Common error: A cavity or hole inside a solid object contracts when the solid object is heated.
Correct: The hole expands in the exact same proportion as the surrounding solid material.
Thermal expansion applies uniformly to all dimensions of the body, meaning empty space within the body expands as if it were filled with the solid.
Common error: Water exhibits regular thermal expansion behavior across all temperature ranges.
Correct: Water exhibits anomalous expansion between 0 °C and 4 °C, contracting upon heating in this specific interval.
Due to hydrogen bonding and open cage-like molecular clustering, water has maximum density at 4 °C.
Common error: The coefficient of volume expansion is strictly three times the linear expansion coefficient for all materials.
Correct: The relation $\gamma = 3\alpha holds strictly$ for isotropic solids; for anisotropic or single crystals, $\gamma = \alpha_{x} + \alpha_{y} + \alpha_{z}$.
Anisotropic media possess different expansion properties along orthogonal crystallographic axes.
Quick revision
The fundamental relations for isotropic solids are $\beta = 2\alpha$ and $\gamma = 3\alpha$.
Thermal stress developed in a rigidly fixed rod is given by $\sigma = Y\alpha\DeltaT$.
Density of a solid decreases with temperature according to ρ ≈ ρ0(1 - γΔT).
A pendulum clock runs slow (loses time) at higher temperatures because the pendulum rod expands, increasing its effective length and time period.
Must remember
Thermal expansion of solids originates fundamentally from the anharmonicity of lattice vibrations within asymmetric interatomic potential wells.
For isotropic solids, the coefficients of linear ($\alpha$), superficial ($\beta$), and cubical ($\gamma$) expansion are related by $\beta = 2alpha$ and $\gamma = 3alpha$.
Cavities or holes inside a solid object expand when heated in the exact same proportion as the surrounding solid material.
Thermal stress developed in a rigidly clamped solid is given by $\sigma = Y \alpha \delta T$, directly proportional to Young's modulus and temperature change.
A bimetallic strip bends upon heating toward the metal with the lower coefficient of linear expansion.
Glossary
Coefficient of Linear Expansion ($alpha$)Fractional change in length per unit change in temperature, measured in $\text{K}^{-1}$.
Coefficient of Superficial Expansion ($\beta$)Fractional change in surface area per unit change in temperature, approximately equal to $2alpha$ for isotropic solids.
Coefficient of Cubical Expansion ($gamma$)Fractional change in volume per unit change in temperature, approximately equal to $3alpha$ for isotropic solids.
Thermal Stress ($sigma$)Stress developed within a solid when thermal expansion or contraction is prevented by rigid boundary constraints.
AnharmonicityDeparture of lattice atomic vibrations from a simple harmonic oscillator model, which is strictly required for thermal expansion.
1. Which of the following is strictly required for thermal expansion of solids to occur?
Explanation: Harmonic approximation yields zero thermal expansion; anharmonicity of lattice vibrations is strictly required because the repulsive wall is steeper than the attractive wall, raising the average interatomic distance upon heating.
2. For an isotropic solid, what is the geometric interrelation between the coefficient of superficial expansion ($\beta$) and the coefficient of linear expansion ($\alpha$)?
Explanation: For isotropic solids, superficial expansion considers two dimensions, yielding $\beta \approx 2alpha$ by neglecting higher-order terms.
3. If a solid metal plate with a circular hole is heated uniformly, what happens to the size of the hole?
Explanation: Thermal expansion applies uniformly to all dimensions of a body; empty space or cavities expand as if they were filled with the solid material.
4. What is the expression for thermal stress ($\sigma$) developed in a solid rigidly clamped at both ends when its temperature changes by $\delta T$?
Explanation: Thermal stress is given by $\sigma = Y \alpha \delta T$, where $Y$ is Young's modulus, $\alpha$ is the linear expansion coefficient, and $\delta T$ is the temperature change.
5. A bimetallic strip is formed by riveting together two metals, Metal A and Metal B, where $\alpha_A > \alpha_B$. When the strip is heated, which way does it bend?
Explanation: Upon heating, the metal with the higher expansion coefficient ($\alpha_A$) expands more than Metal B, causing the strip to bend toward the metal with the lower coefficient ($\alpha_B$).
6. How does the density ($\rho$) of a solid vary with a temperature increase $\delta T$, given a volume expansion coefficient $\gamma$?
Explanation: Since mass remains constant and volume increases as $V = V_0(1 + \gamma \delta T)$, the density satisfies $\rho = \frac{\rho_0}{1 + \gamma \delta T} \approx \rho_0(1 - \gamma \delta T)$.
7. What is the fractional time loss per second of a pendulum clock whose pendulum rod has a linear expansion coefficient $\alpha$ when subjected to a temperature rise $\delta T$?
Explanation: The fractional time loss is given by $\frac{\delta t}{t} = \frac{1}{2} \alpha \delta T$, derived from the period of a simple pendulum $T = 2pisqrt{L/g}$.
8. For anisotropic single crystals, how is the cubical expansion coefficient $\gamma$ related to the principal linear expansion coefficients $\alpha_x$, $\alpha_y$, and $\alpha_z$ along orthogonal axes?
Explanation: For anisotropic media, tensor representations show that the volume expansion coefficient is the sum of the linear expansion coefficients along the principal orthogonal crystallographic axes: $\gamma = \alpha_x + \alpha_y + \alpha_z$.
9. What is the SI unit of the coefficient of linear expansion ($\alpha$)?
Explanation: Since $\alpha = \frac{\delta L}{L_0 \delta T}$, its unit is fractional length per Kelvin, which simplifies to $\text{K}^{-1}$ (or ${}^circ\text{C}^{-1}$).
10. What force $F$ is generated in a uniform solid rod of cross-sectional area $A$ and Young's modulus $Y$ when its thermal expansion is entirely prevented and its temperature is raised by $\delta T$?
Explanation: Thermal force is obtained by multiplying thermal stress ($\sigma = Y \alpha \delta T$) by the cross-sectional area $A$, yielding $F = Y A \alpha \delta T$.
11. Which of the following conditions is required for the validity of the relation $L = L_0 (1 + \alpha \delta T)$?
Explanation: The linear expansion formula is derived by binomial expansion, keeping only first-order terms, which requires $\alpha \delta T ll 1$.
12. Why does water exhibit anomalous expansion between $0^circ\text{C}$ and $4^circ\text{C}$?
Explanation: Water exhibits anomalous expansion between $0^circ\text{C}$ and $4^circ\text{C}$ due to hydrogen bonding and open cage-like molecular arrangements that collapse upon heating, causing contraction in this specific temperature range.
13. In the relation for apparent expansion of a liquid in a vessel, $\gamma_{app} = \gamma_{real} - \gamma_{vessel}$, what does $\gamma_{vessel}$ represent?
Explanation: $\gamma_{vessel}$ represents the volume (cubical) expansion coefficient of the container holding the liquid.
14. What happens to the average interatomic distance in a solid as its temperature increases?
Explanation: Due to the asymmetry of the interatomic potential well (steeper repulsive wall than attractive wall), higher vibrational energy raises the average interatomic distance.
15. If a metal rod of length $L_0$ is heated such that its temperature increases by $\delta T$, what is the first-order change in its length ($\delta L$)?
Explanation: From $L = L_0(1 + \alpha \delta T)$, the change in length is $\delta L = L - L_0 = L_0 \alpha \delta T$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A uniform metal rod of length $L$ and linear expansion coefficient $\alpha$ is heated through $\delta T^circ\text{C}$. What is the percentage change in its length?
Explanation: Percentage change in length is given by $\frac{\delta L}{L_0} \times 100% = \alpha \delta T \times 100%$.
2. A pendulum clock loses $10\text{ s}$ per day when the temperature rises by $20^circ\text{C}$. What is the coefficient of linear expansion ($\alpha$) of the pendulum rod?
Explanation: Fractional time loss per second is $\frac{\delta t}{t} = \frac{10}{24 \times 3600} = \frac{1}{2}\alpha \delta T$. Substituting $\delta T = 20^circ\text{C}$ gives $\alpha = \frac{20}{24 \times 3600 \times 10} \approx 2.31 \times 10^{-5} \text{ K}^{-1}$.
3. Two rods of length $L_1$ and $L_2$ have linear expansion coefficients $\alpha_1$ and $\alpha_2$ respectively. If the difference in their lengths is independent of temperature, what is the relation between them?
Explanation: The lengths at temperature $T$ are $L_1(1 + \alpha_1 \delta T)$ and $L_2(1 + \alpha_2 \delta T)$. The difference is $(L_1 - L_2) + (L_1alpha_1 - L_2alpha_2)\delta T$. For this to be independent of $\delta T$, $L_1alpha_1 - L_2alpha_2 = 0$, so $L_1alpha_1 = L_2alpha_2$.
4. A solid cube is heated, increasing its temperature by $\delta T$. What is the approximate percentage increase in its volume if its linear expansion coefficient is $\alpha$?
Explanation: Volume expansion is governed by $\gamma = 3alpha$. Thus, the fractional change in volume is $\frac{\delta V}{V_0} = \gamma \delta T = 3alpha \delta T$, which gives a percentage increase of $3alpha \delta T \times 100%$.
5. A metal wire of Young's modulus $Y$ and cross-sectional area $A$ is rigidly fixed between two walls at $20^circ\text{C}$. If the temperature drops to $0^circ\text{C}$, what is the tensile force developed in the wire?
Explanation: The temperature change is $\delta T = 20^circ\text{C} - 0^circ\text{C} = 20^circ\text{C}$. The tensile force is $F = Y A \alpha \delta T = Y A \alpha (20^circ\text{C})$.
6. The density of a substance at $0^circ\text{C}$ is $\rho_0$ and its volume expansion coefficient is $\gamma$. What is its density at $T^circ\text{C}$?
Explanation: Using the variation of density formula with $\delta T = T$, $\rho = \frac{\rho_0}{1 + \gamma T}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which atomic potential well characteristic is responsible for thermal expansion in solids?
Explanation: An asymmetric potential well where the repulsive wall is steeper than the attractive wall causes the average interatomic distance to increase as vibrational energy rises.
2. If the coefficient of linear expansion of a material is $\alpha = 2 \times 10^{-5} \text{ K}^{-1}$, what is its coefficient of cubical expansion ($\gamma$)?
3. A circular metallic disc has a circular hole at its center. When the disc is heated, what happens to the diameter of the hole and the thickness of the disc?
Explanation: Thermal expansion is uniform across all dimensions of a solid body, meaning linear dimensions (diameter) and thickness both increase upon heating.
4. What is the thermal stress developed in a brass rod of Young's modulus $10^{11} \text{ Pa}$ and linear expansion coefficient $2 \times 10^{-5} \text{ K}^{-1}$ when heated by $50^circ\text{C}$ with fixed ends?
5. In an anisotropic crystal, if the linear expansion coefficients along the $x$, $y$, and $z$ axes are $\alpha_x$, $\alpha_y$, and $\alpha_z$ respectively, the cubical expansion coefficient $\gamma$ is given by:
Explanation: For anisotropic single crystals, the volume expansion coefficient is the trace of the thermal expansion tensor: $\gamma = \alpha_x + \alpha_y + \alpha_z$.
6. Why does a bimetallic strip bend when heated?
Explanation: A bimetallic strip bends because the two component metals possess different coefficients of linear expansion, leading to unequal fractional changes in length upon heating.
7. A pendulum clock keeps correct time at $20^circ\text{C}$. If the temperature drops to $0^circ\text{C}$, how will the clock behave?
Explanation: When the temperature drops, the pendulum rod contracts (shortens), decreasing its time period $T = 2pisqrt{L/g}$, which causes the clock to run faster (gain time).
8. Which of the following approximations is used in deriving the area expansion relation $A = A_0(1 + \beta \delta T)$ from linear expansion?
Explanation: Starting from $A = L_0(1 + \alpha \delta T) \times W_0(1 + \alpha \delta T) = A_0(1 + 2alpha \delta T + \alpha^2 (\delta T)^2)$, the term $\alpha^2 (\delta T)^2$ is neglected because $\alpha \delta T ll 1$.
9. If mass density $\rho_0$ at $0^circ\text{C}$ changes to $\rho$ at $\delta T$, which binomial approximation yields $\rho \approx \rho_0(1 - \gamma \delta T)$?
Explanation: From $\rho = \frac{\rho_0}{1 + \gamma \delta T} = \rho_0(1 + \gamma \delta T)^{-1}$, applying the binomial approximation $(1 + x)^{-1} \approx 1 - x$ gives $\rho \approx \rho_0(1 - \gamma \delta T)$ since $\gamma \delta T ll 1$.
10. What happens to the apparent expansion coefficient of a liquid ($\gamma_{app}$) if the container vessel has a higher cubical expansion coefficient than the liquid itself?
Explanation: Since $\gamma_{app} = \gamma_{real} - \gamma_{vessel}$, if $\gamma_{vessel} > \gamma_{real}$, then $\gamma_{app} < 0$, causing the liquid level in the container to drop initially when heated.