Quick Revision Notes & Formula Sheet

Class 11 Physics — Chapter: Thermal Properties of Matter (तापीय गुण)

1. Temperature and Heat (तापमान एवं ऊष्मा)

Heat ($Q$): A form of energy that flows from a body at a higher temperature to a body at a lower temperature.
- SI Unit: Joule ($\text{J}$)
- Practical Unit: Calorie ($\text{cal}$)
- Relation: $1\text{ cal} = 4.186\text{ J} \approx 4.2\text{ J}$
Temperature ($T$): The degree of hotness or coldness of a body.
- SI Unit: Kelvin ($\text{K}$)

Temperature Scale Conversions

$$\frac{C - 0}{100} = \frac{F - 32}{180} = \frac{K - 273.15}{100}$$

Simplified Formula: $$\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273.15}{5}$$

2. Thermal Expansion (तापीय प्रसार)

Increase in dimensions of a body due to an increase in temperature.

A. Linear Expansion (रेखीये प्रसार)

Expansion in length.

Formula: $\Delta L = L_0 \cdot \alpha \cdot \Delta T$
Final Length: $L = L_0 (1 + \alpha \cdot \Delta T)$
Coefficient of Linear Expansion ($\alpha$): $$\alpha = \frac{\Delta L}{L_0 \cdot \Delta T} \quad [\text{Unit: } \text{K}^{-1} \text{ or } {^\circ}\text{C}^{-1}]$$

B. Superficial / Area Expansion (क्षेत्रीय प्रसार)

Expansion in surface area.

Formula: $\Delta A = A_0 \cdot \beta \cdot \Delta T$
Final Area: $A = A_0 (1 + \beta \cdot \Delta T)$
Coefficient of Area Expansion ($\beta$): $$\beta = \frac{\Delta A}{A_0 \cdot \Delta T} \quad [\text{Unit: } \text{K}^{-1}]$$

C. Volume / Cubical Expansion (आयतन प्रसार)

Expansion in volume.

Formula: $\Delta V = V_0 \cdot \gamma \cdot \Delta T$
Final Volume: $V = V_0 (1 + \gamma \cdot \Delta T)$
Coefficient of Volume Expansion ($\gamma$): $$\gamma = \frac{\Delta V}{V_0 \cdot \Delta T} \quad [\text{Unit: } \text{K}^{-1}]$$

Relationship between Coefficients

$$\alpha : \beta : \gamma = 1 : 2 : 3$$ $$\beta = 2\alpha, \quad \gamma = 3\alpha$$

Thermal Stress (तापीय प्रतिबल)

When a rod fixed between two rigid supports is heated: $$\text{Thermal Strain} = \frac{\Delta L}{L} = \alpha \cdot \Delta T$$ $$\text{Thermal Stress} = Y \cdot \alpha \cdot \Delta T \quad (Y = \text{Young's Modulus})$$ $$\text{Force developed } (F) = Y \cdot A \cdot \alpha \cdot \Delta T$$

Anomalous Expansion of Water (जल का असंगत प्रसार)

Water contracts when heated from $0^\circ\text{C}$ to $4^\circ\text{C}$.
Water has maximum density and minimum volume at $4^\circ\text{C}$ ($\rho = 1000\text{ kg/m}^3$).

3. Specific Heat Capacity & Calorimetry (विशिष्ट ऊष्मा धारिता एवं कैलोरीमिति)

A. Heat Capacity / Thermal Capacity ($S$)

Amount of heat required to raise the temperature of a whole body by $1^\circ\text{C}$ or $1\text{ K}$. $$S = \frac{\Delta Q}{\Delta T} \quad [\text{Unit: } \text{J/K}]$$

B. Specific Heat Capacity ($c$)

Amount of heat required to raise the temperature of unit mass of a substance by $1^\circ\text{C}$ or $1\text{ K}$. $$c = \frac{\Delta Q}{m \cdot \Delta T} \quad \implies \quad \Delta Q = m \cdot c \cdot \Delta T$$

SI Unit: $\text{J kg}^{-1}\text{K}^{-1}$
For Water: $c_{\text{water}} = 1\text{ cal/g}^\circ\text{C} = 4186\text{ J kg}^{-1}\text{K}^{-1}$

C. Molar Specific Heat Capacity ($C$)

Heat required to raise the temperature of $1\text{ mole}$ of a substance by $1\text{ K}$. $$C = \frac{\Delta Q}{n \cdot \Delta T}$$

At Constant Volume ($C_v$): Molar specific heat at constant volume.
At Constant Pressure ($C_p$): Molar specific heat at constant pressure.
Mayer's Relation: $$C_p - C_v = R \quad (R = \text{Universal Gas Constant})$$

D. Principle of Calorimetry

Based on the Law of Conservation of Energy: $$\text{Heat Lost by Hot Body} = \text{Heat Gained by Cold Body}$$ (Provided there is no heat exchange with the surroundings)

4. Change of State & Latent Heat (अवस्था परिवर्तन एवं गुप्त ऊष्मा)

Latent Heat ($L$)

The heat energy required to change the state of unit mass of a substance at a constant temperature. $$Q = m \cdot L \quad \implies \quad L = \frac{Q}{m}$$

SI Unit: $\text{J/kg}$

Latent Heat of Fusion ($L_f$): Solid $\rightarrow$ Liquid transition.
- For Ice: $L_f \approx 80\text{ cal/g} = 3.33 \times 10^5\text{ J/kg}$
Latent Heat of Vaporization ($L_v$): Liquid $\rightarrow$ Gas transition.
- For Water: $L_v \approx 540\text{ cal/g} = 2.26 \times 10^6\text{ J/kg}$

5. Modes of Heat Transfer (ऊष्मा स्थानांतरण की विधियाँ)

Mode	Mechanism	Medium Required?
Conduction (चालन)	Heat transfer by vibration of particles without actual particle movement.	Yes (Solids)
Convection (संवहन)	Heat transfer by actual physical movement of fluid particles.	Yes (Fluids)
Radiation (विकिरण)	Heat transfer in the form of electromagnetic waves.	No (Vacuum)

6. Thermal Conduction Formulas

Rate of Heat Flow / Thermal Current ($H$)

$$H = \frac{dQ}{dt} = \frac{K \cdot A \cdot (T_1 - T_2)}{L}$$

Where:

$K$ = Thermal Conductivity of the material (ऊष्मीय चालकता) [Unit: $\text{W m}^{-1}\text{K}^{-1}$]
$A$ = Cross-sectional area
$L$ = Length / Thickness of the rod
$T_1 - T_2$ = Temperature difference

Thermal Gradient

$$\text{Temperature Gradient} = \frac{T_1 - T_2}{L} = \frac{dT}{dx}$$

Thermal Resistance ($R_{th}$)

$$R_{th} = \frac{L}{K \cdot A}$$

Heat flow analogy to Ohm's Law: $H = \frac{\Delta T}{R_{th}}$

7. Radiation Laws (विकिरण नियम)

A. Stefan-Boltzmann Law

Energy radiated per unit area per second by a perfectly black body is directly proportional to the fourth power of its absolute temperature. $$E = \sigma \cdot T^4$$

For a general body of emissivity $e$ ($0 < e < 1$) and area $A$: $$P = e \cdot \sigma \cdot A \cdot T^4$$
Net Power Loss in Surroundings at $T_0$: $$P_{\text{net}} = e \cdot \sigma \cdot A \cdot (T^4 - T_0^4)$$
Stefan's Constant ($\sigma$): $\sigma \approx 5.67 \times 10^{-8}\text{ W m}^{-2}\text{K}^{-4}$

B. Wien's Displacement Law

The wavelength $\lambda_m$ corresponding to maximum energy emission is inversely proportional to the absolute temperature $T$. $$\lambda_m \cdot T = b$$

Wien's Constant ($b$): $b \approx 2.898 \times 10^{-3}\text{ m}\cdot\text{K}$

C. Newton's Law of Cooling (न्यूटन का शीतलन नियम)

The rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings (for small temperature differences).

Differential Form: $$\frac{dQ}{dt} = -k(T - T_0)$$
Temperature Change Form: $$\frac{T_1 - T_2}{t} = K \left[ \frac{T_1 + T_2}{2} - T_0 \right]$$