Thermal Physics & Calorimetry: Master Notes

Chapter Overview

Thermal physics encompasses heat capacity, phase transitions, and thermal transport mechanisms. In IIT-JEE (Advanced), core problems focus on thermal resistance networks (composite rods & concentric spherical shells), Stefan-Boltzmann radiative exchange, Newton's Law of Cooling approximations, and calorimetry phase equilibrium.

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1. Thermal Expansion & Thermal Stress

• Linear Expansion: $\mathrm{\Delta }L=L_0\alpha \mathrm{\Delta }T$
• Area Expansion: $\mathrm{\Delta }A=A_0\beta \mathrm{\Delta }T\approx A_0(2\alpha )\mathrm{\Delta }T$
• Volume Expansion: $\mathrm{\Delta }V=V_0\gamma \mathrm{\Delta }T\approx V_0(3\alpha )\mathrm{\Delta }T$
• Thermal Stress on a Clamped Rod:$$\boxed{{\displaystyle \sigma =\frac{F}{A}=Y\alpha \mathrm{\Delta }T}}$$

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2. Heat Conduction & Thermal Resistance Networks

$$\text{Fourier’s Law: }\frac{dQ}{dt}=-kA\frac{dT}{dx}⟹\boxed{{\displaystyle H=\frac{dQ}{dt}=\frac{\mathrm{\Delta }T}{R_{\text{th}}}}}$$

Where the Thermal Resistance is defined as:

$$\boxed{{\displaystyle R_{\text{th}}=\frac{L}{kA}}}$$

Multi-Mode DiagramThermal Physics: Conduction & Composite Slabs

Option 1: Publication-Grade Scientific Vector SVG

Heat current H = dQ/dt through series composite thermal rods with junction temperature Tj.

Heat Current: $H = \frac{K A \Delta T}{L} = \frac{\Delta T}{R_{\text{th}}}$

Option 2: Interactive 60 FPS Canvas Physics Engine

Simulate steady-state temperature gradient across thermal junctions.

Conductivity K₁: 60 W/mK

Junction Temp ($T_j$):68.0 °C

Option 3: Interactive 3D Spatial Viewport

3D thermal block model.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Thermal Resistance:$H = \frac{T_1 - T_2}{R_{\text{th}}}$

Option 5: Reactive Parametric Physics with Live Calculus

Junction temperature calculator.

Conductivity K₁: 40

: 45

T_junction ($T_j$):$50.0 °C$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

Geometry	Thermal Resistance Formula $R_{\text{th}}$
Uniform Rod of Length $L$, Area $A$	$R_{\text{th}}=\frac{L}{kA}$
Concentric Spherical Shell (Radii $r_1,r_2$)	$R_{\text{th}}=\frac{1}{4\pi k}(\frac{1}{r_1}-\frac{1}{r_2})$
Coaxial Cylindrical Tube (Radii $r_1,r_2$, Length $L$)	$R_{\text{th}}=\frac{\ln (r_2/r_1)}{2\pi kL}$

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3. Radiation Laws & Newton's Law of Cooling

• Stefan-Boltzmann Law: Total radiated power by a body of emissivity $e$ and area $A$ in surroundings $T_0$:

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

  (where $\sigma =5.67\times {10}^{-8}{\text{ W/m}}^2{\text{K}}^4$ is Stefan's constant).

• Wien's Displacement Law: Peak emission wavelength ${\lambda }_{max}$:

  $$\boxed{{\displaystyle {\lambda }_{max}T=b\approx 2.898\times {10}^{-3}\text{ m}\cdot \text{K}}}$$

• Newton's Law of Cooling (Small temperature excess $\mathrm{\Delta }T\ll T_0$):

  $$\frac{dT}{dt}=-K(T-T_0)⟹\boxed{{\displaystyle \frac{T_1-T_2}{t}=K(\frac{T_1+T_2}{2}-T_0)}}$$

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2021 (Paper 1) — Concentric Shell Heat Conduction

Question:
A spherical shell of inner radius $R$ and outer radius $2R$ has thermal conductivity $k$. The inner surface is maintained at temperature $T_1={100}^{\circ }\text{C}$ and the outer surface at $T_2={20}^{\circ }\text{C}$. Find:

1. The steady-state rate of heat flow $H$.
2. The temperature $T(r)$ at radial distance $r=1.5R$.

Step-by-Step Solution:

• 1. Thermal Resistance of Spherical Shell:

  $$R_{\text{th}}=\frac{1}{4\pi k}(\frac{1}{R}-\frac{1}{2R})=\frac{1}{4\pi k}\left(\frac{1}{2R}\right)=\frac{1}{8\pi kR}$$$$H=\frac{T_1-T_2}{R_{\text{th}}}=\frac{100-20}{\frac{1}{8\pi kR}}=\boxed{{\displaystyle 640\pi kR}}$$

• 2. Temperature at $r=1.5R=\frac{3}{2}R$: Thermal resistance from $R$ to $r$:

  $$R_{\text{th}}(r)=\frac{1}{4\pi k}(\frac{1}{R}-\frac{1}{1.5R})=\frac{1}{4\pi k}(\frac{1}{R}-\frac{2}{3R})=\frac{1}{12\pi kR}$$$$T_1-T(r)=H\cdot R_{\text{th}}(r)=(640\pi kR)\left(\frac{1}{12\pi kR}\right)=\frac{640}{12}=\frac{160}{3}\approx {53.33}^{\circ }\text{C}$$$$T(r)=100-53.33=\boxed{{\displaystyle {46.67}^{\circ }\text{C}}}$$

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PYQ 2: JEE Advanced 2020 — Stefan's Radiation & Spherical Blackbody

Question:
A black spherical body of radius $r_1=10\text{ cm}$ at temperature $T_1=2000\text{ K}$ radiates into space. A second black spherical body of radius $r_2$ at temperature $T_2=1000\text{ K}$ radiates the same total power as the first body. Find the radius $r_2$.

Step-by-Step Solution:

Total radiated power by a blackbody ($e=1$):

$$P=\sigma A{T}^4=\sigma (4\pi r^2)T^4$$

Given $P_1=P_2$:

$$4\pi \sigma r_1^2{T}_1^4=4\pi \sigma r_2^2{T}_2^4⟹r_1^2{T}_1^4=r_2^2{T}_2^4$$$$r_2=r_1{\left(\frac{T_1}{T_2}\right)}^2=10\text{ cm}\times {\left(\frac{2000}{1000}\right)}^2=10\times (2{)}^2=\boxed{{\displaystyle 40\text{ cm}}}$$

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5. High-Yield Formula Sheet

Entity	Formula	Notes
Thermal Resistance	$R_{\text{th}}=\frac{L}{kA}$	Series: $R_1+R_2$
Thermal Stress	$\sigma =Y\alpha \mathrm{\Delta }T$	Clamped expansion
Stefan's Law	$P=e\sigma A{T}^4$	Total radiant power
Wien's Law	${\lambda }_{max}T=b$	$b=2.898\times {10}^{-3}\text{ m}\cdot \text{K}$
Newton's Cooling	$\frac{T_1-T_2}{t}=K(\frac{T_1+T_2}{2}-T_0)$	Approximation