Thermal Expansion Problems with Constraints

Hello! Welcome to the first lesson of our second module. We've successfully navigated the principles of calorimetry, and now we shift our focus from how heat affects temperature and phase to how it alters the physical dimensions of matter.

Today's lesson addresses the learning outcome: Solve problems involving linear, areal, and volumetric thermal expansion with constraints. We'll start with the fundamental formulas for thermal expansion and then, crucially for your JEE Advanced preparation, focus on how to solve problems where this expansion is restricted. These "constrained" systems are where simple physics principles intersect to create challenging and interesting problems.

1. The Fundamentals of Thermal Expansion

When most materials are heated, the increased kinetic energy of their atoms or molecules causes them to vibrate with greater amplitude, pushing them further apart on average. This microscopic change leads to a macroscopic expansion of the object.

We quantify this expansion in one, two, or three dimensions. For a concise overview of the basic definitions and formulas, please study the following material.

[PDF] Elasticity, Thermal Expansion, Calorimetry & Heat Transfer

These notes from 'Elasticity, Thermal Expansion, Calorimetry & Heat Transfer' by MyRankers cover the essential formulas for linear, areal (superficial), and volume expansion.

Read the section titled 'THERMAL EXPANSION' up to the discussion of 'Contraction on heating'. Focus on the definitions and formulas for linear, superficial, and volume expansion.

From the reading, the key equations are:

Linear Expansion: $Δ L = L_{0} α Δ T$ or $L = L_{0} (1 + α Δ T)$
Areal Expansion: $Δ A = A_{0} β Δ T$ or $A = A_{0} (1 + β Δ T)$
Volumetric Expansion: $Δ V = V_{0} γ Δ T$ or $V = V_{0} (1 + γ Δ T)$

Here, $α$ , $β$ , and $γ$ are the coefficients of linear, areal, and volumetric expansion, respectively.

The Relationship Between $α$ , $β$ , and $γ$

For an isotropic material (which expands uniformly in all directions), these coefficients are related. Given your math background, you'll see this is a direct result of the binomial approximation.

Consider a square plate of initial side length $L_{0}$ . Its initial area is $A_{0} = L_{0}^{2}$ . When heated by $Δ T$ , the new side length is $L = L_{0} (1 + α Δ T)$ . The new area is:
$A = L^{2} = [L_{0} (1 + α Δ T)]^{2} = L_{0}^{2} (1 + 2 α Δ T + (α Δ T)^{2})$
Since $α$ is very small (typically $\sim 1 0^{- 5} K^{- 1}$ ), the $(α Δ T)^{2}$ term is negligible. So,
$A \approx A_{0} (1 + 2 α Δ T)$
Comparing this to $A = A_{0} (1 + β Δ T)$ , we find $β \approx 2 α$ .

A similar argument for a cube of volume $V_{0} = L_{0}^{3}$ yields $γ \approx 3 α$ . Thus, we have the crucial relationship:
$α : β : γ \approx 1 : 2 : 3$

A note on cavities: A common exam question involves the expansion of an object with a hole in it (e.g., a metal washer). The hole expands exactly as if it were filled with the same material as the object. Think of it as a photographic enlargement.

2. Thermal Expansion with Constraints

This is where the problems become interesting for JEE Advanced. A constraint is anything that prevents an object from expanding or contracting freely. Solving these problems involves combining the idea of thermal expansion with principles from mechanics (forces, stress) or hydrostatics (buoyancy).

The general strategy is:

Calculate the free expansion $Δ L$ or $Δ V$ that would occur without the constraint.
Identify the constraint and how it counteracts this expansion.
Formulate an equilibrium equation based on the forces or pressures involved.

Type 1: Mechanical Constraints (Rigid and Elastic)

Rigid Constraint: Imagine a rod of length $L$ placed between two immovable walls. If heated, it tries to expand by $Δ L = Lα Δ T$ . Since the walls prevent this, the rod experiences a compressive strain of $ϵ = Δ L / L = α Δ T$ . This strain induces a thermal stress in the rod:
$σ = Y ϵ = Y α Δ T$
where $Y$ is Young's Modulus. This stress creates a force $F = A σ = A Y α Δ T$ pushing against the walls. We will explore thermal stress in more detail in our next lesson.

Elastic Constraint: What if the constraint isn't perfectly rigid? Consider a rod attached to a spring.

Thermal Stress in a Rod-Spring System — This JEE Advanced problem shows a rod heated while attached to a spring. The thermal expansion of the rod is partially accommodated by the compression of the spring and partially by the compression of the rod itself.

Let's analyze this setup:

Free Expansion: If the spring weren't there, the rod would expand to the right by $Δ L_{t h er ma l} = Lα Δ T$ .
Constraint: The spring pushes back. In equilibrium, the rod settles at a final extension $x$ (where $x < Δ L_{t h er ma l}$ ).
Equilibrium Equation: The spring is compressed by $x$ , so it exerts a force $F = k x$ on the rod. This same force causes an elastic compression in the rod itself, amounting to $Δ L_{e l a s t i c} = \frac{F L}{A Y} = \frac{k xL}{A Y}$ .
The final expansion $x$ is the free thermal expansion minus the elastic compression caused by the spring force:
$x = Δ L_{t h er ma l} - Δ L_{e l a s t i c}$
$x = Lα Δ T - \frac{k xL}{A Y}$
You can now solve this equation for $x$ , the force $F = k x$ , or the stress $σ = F / A$ .

Type 2: Composite Rods

When two rods of different materials are rigidly joined (e.g., riveted), they are constrained to have the same final length. Since their free expansions would be different, one rod gets stretched (tensile stress) and the other gets compressed (compressive stress).

The following video provides an excellent, detailed analysis of this exact scenario.

3. Advance Illustration | Thermal Expansion & Calorimetry | Elongation of Riveted Rods |Ashish Arora

In this 'Advance Illustration' from Physics Galaxy, Ashish Arora breaks down the problem of finding the final length of a composite system of riveted rods. The approach is a perfect model for solving constrained expansion problems.

Watch the entire video (00:00 - 07:28). Pay close attention to how the analysis is structured: Calculating the individual 'free' expansions. Setting up the relationship between the final common expansion and the elastic deformations (one stretch, one compression). Applying force equilibrium (Stress_Y * Area_Y = 2 * Stress_X * Area_X, assuming equal areas). Solving for the final expansion.

The core principle is the balance of internal forces. The total tensile force on the material(s) with lower $α$ must equal the total compressive force on the material(s) with higher $α$ .

Type 3: Buoyancy as a Constraint

When an object is immersed in a fluid, its apparent weight depends on the buoyant force, which in turn depends on the fluid's density and the object's submerged volume. Heating the system changes both of these quantities, leading to a new equilibrium.

This video demonstrates how to solve such a problem.

4. Advance Illustration | Thermal Expansion & Calorimetry | Weighing of a Metal in a Liquid

This video, also from Physics Galaxy, analyzes how the apparent weight of a metal piece changes when it and the surrounding liquid are heated. This is a practical example of volumetric expansion affecting buoyant force.

Watch the full video (00:00 - 04:20). Focus on how the volume of the metal at two different temperatures is determined using the buoyant force (Weight in air - Apparent weight) and the liquid's density at each temperature. The final step connects these two volumes using the thermal expansion formula.

Test your understanding! (JEE Advanced 2004)

Problem: A cubical block of coefficient of linear expansion $α_{s}$ is submerged partially in a liquid with a coefficient of volume expansion $γ_{l}$ . On increasing the temperature of the system by $Δ T$ , the height of the cube inside the liquid remains unchanged. Find the relation between $α_{s}$ and $γ_{l}$ .

Show answer

Let the initial side length of the cube be $L_{0}$ , its density be $ρ_{s}$ , and the liquid density be $ρ_{l, 0}$ . Let the submerged height be $h$ .

The constraint is that the block is in floating equilibrium: Weight = Buoyant Force.
$Weight = Volume_{c u b e} \times ρ_{s} \times g = L_{0}^{3} ρ_{s} g$
$Buoyant Force = Volume_{s u bm er g e d} \times ρ_{l, 0} \times g = (L_{0}^{2} h) ρ_{l, 0} g$
So, $L_{0}^{3} ρ_{s} g = L_{0}^{2} h ρ_{l, 0} g ⟹ L_{0} ρ_{s} = h ρ_{l, 0}$ --- (1)

Now, we increase the temperature by $Δ T$ . The new quantities are denoted by a prime (').

New side length of cube: $L^{'} = L_{0} (1 + α_{s} Δ T)$
New area of cube base: $A^{'} = L_{0}^{2} (1 + β_{s} Δ T) = L_{0}^{2} (1 + 2 α_{s} Δ T)$
New liquid density: $ρ_{l}^{'} = \frac{ρ _{l, 0}}{1 + γ _{l} Δ T} \approx ρ_{l, 0} (1 - γ_{l} Δ T)$
The mass and thus weight of the cube do not change.
The submerged height $h$ remains the same (this is the key piece of information).

The new buoyant force must still equal the weight.
$Weight = Buoyant Force^{'}$
$L_{0}^{3} ρ_{s} g = A^{'} h ρ_{l}^{'} g$
$L_{0}^{3} ρ_{s} = [L_{0}^{2} (1 + 2 α_{s} Δ T)] h [ρ_{l, 0} (1 - γ_{l} Δ T)]$

Substitute $L_{0} ρ_{s} = h ρ_{l, 0}$ from equation (1):
$L_{0}^{3} ρ_{s} = (h ρ_{l, 0}) L_{0} (1 + 2 α_{s} Δ T) (1 - γ_{l} Δ T)$
$1 = (1 + 2 α_{s} Δ T) (1 - γ_{l} Δ T)$
$1 = 1 + 2 α_{s} Δ T - γ_{l} Δ T - (2 α_{s} γ_{l} Δ T^{2})$
Ignoring the very small $Δ T^{2}$ term:
$0 = (2 α_{s} - γ_{l}) Δ T$
This gives the relation: $γ_{l} = 2 α_{s}$ .

This problem perfectly illustrates the method: set up the initial equilibrium, apply the expansion formulas to the new state, and enforce the equilibrium condition again.

Conclusion

Today we laid the groundwork for analyzing the effects of temperature change on the dimensions of objects. While the basic formulas are simple, the real challenge and depth, especially for JEE Advanced, lie in applying them to constrained systems.

Key Takeaways:

Thermal expansion can be linear, areal, or volumetric, with coefficients related by $α : β : γ \approx 1 : 2 : 3$ .
Solving problems with constraints requires a combination of thermal expansion formulas and principles from other areas of physics like mechanics or fluid statics.
The core strategy is to compare the state of free expansion with the final state under constraint and use equilibrium conditions (force balance, pressure balance) to find the solution.

Preview of the Next Lesson:
We got a brief introduction to thermal stress today. In our next lesson, we will dive deep into this concept. We will formally analyze thermal stress problems in constrained systems and composite rods, building directly on the foundation we've established.

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