Thermal Expansion and Calorimetry

Solids, liquids, and gases all expand on heating. Calorimetry combines specific heat and latent heat to track energy exchange when bodies at different temperatures are mixed.

Concept

For a solid rod, linear expansion gives $\Delta L = L_0 \alpha \Delta T$. Area expansion uses $\beta = 2\alpha$ and volume expansion uses $\gamma = 3\alpha$ (for isotropic solids).

For liquids in containers, the apparent expansion equals $\gamma_{liq} - \gamma_{container}$.

The calorimetry principle: heat lost by hot bodies equals heat gained by cold bodies (in an isolated system): $\sum m_i c_i \Delta T_i + \sum m_j L_j = 0$

Specific heat: $Q = mc\Delta T$. Latent heat (phase change at constant $T$): $Q = mL$.

Derivation

Consider a bimetallic strip of metals with coefficients $\alpha_1 > \alpha_2$, each of length $L_0$ and thickness $d$. On heating by $\Delta T$:

• New lengths: $L_1 = L_0(1+\alpha_1\Delta T)$, $L_2 = L_0(1+\alpha_2\Delta T)$.
• The strip bends into an arc of radius $R$ with the longer metal on the outside.

For thin strips, arc-length matching gives: $R = \frac{d}{(\alpha_1 - \alpha_2)\Delta T}$

This is the basis of thermostats.

JEE Worked Example

Problem: 50 g of ice at $-10^\circ$C is added to 200 g of water at $40^\circ$C in a copper calorimeter of mass 100 g. Find the final temperature. ($c_{ice}=0.5$, $c_w=1.0$, $c_{Cu}=0.1$ cal/g°C, $L_{fusion}=80$ cal/g.)

Solution: Heat needed to warm ice to 0°C: $50 \times 0.5 \times 10 = 250$ cal. Heat to melt ice: $50 \times 80 = 4000$ cal. Total: 4250 cal.

Heat available from water+calorimeter cooling from 40°C to 0°C: $(200 \times 1 + 100 \times 0.1)\times 40 = 8400$ cal.

Excess heat: $8400 - 4250 = 4150$ cal warms the now-melted water plus calorimeter: $4150 = (250 + 10)T \Rightarrow T \approx 15.96^\circ\text{C}$

Traps

• Volume coefficient $\gamma = 3\alpha$ only for isotropic solids; anisotropic crystals have different $\alpha$ along axes.
• Water has anomalous expansion between 0°C and 4°C; density is maximum at 4°C.
• For a hole inside a metal plate, the hole expands (treat as if filled with the same material).
• Latent heat absorbs/releases energy at constant temperature — don't add $mc\Delta T$ during a phase change.
• Always check whether ice fully melts before assuming final $T > 0$°C.

Key Takeaways

• $\Delta L = L_0\alpha\Delta T$; $\gamma = 3\alpha$ for isotropic solids.
• Apparent expansion of liquid = real $-$ container's.
• Energy balance: heat lost = heat gained; include phase changes via $mL$.
• Always verify the final state (solid/liquid/mixed) before solving.