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अध्याय 10: द्रव्य के तापीय गुण (Thermal Properties of Matter)

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झारखण्ड बोर्ड (JAC) • कक्षा XI • Physics • अध्याय 10

अनुमानित समय: 45 Mins

प्रगति: अध्ययनरत

द्रव्य के तापीय गुण (Thermal Properties of Matter)

In Class 11 Physics, "Thermal Properties of Matter" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

🌡️ Have You Ever Wondered?

Why do lakes freeze from the top down in freezing winters allowing fish to swim happily below thick ice, or why do railway lines have tiny deliberate gaps between steel tracks? Thermal expansion and anomalous water expansion sustain Earth's biosphere.

यह अध्याय क्यों महत्वपूर्ण है

In Class 11 Physics, "Thermal Properties of Matter" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Heat and temperature from Class 7.
States of matter and phase changes.
Thermal conduction.

इस अध्याय के लक्ष्य

Distinguish between Temperature (thermal state) and Heat (transferred energy).
Analyze Thermal Expansion: Linear ($\alpha$), Superficial ($\beta = 2\alpha$), and Cubical ($\gamma = 3\alpha$).
Explain the Anomalous Expansion of Water (maximum density at $4^\circ\text{C}$) and its ecological significance.
Define Specific Heat Capacity ($c = \frac{\Delta Q}{m\Delta T}$), Molar heat capacity, and Latent Heat ($Q = mL$).
State Newton's Law of Cooling: $\frac{dT}{dt} = -k(T - T_0)$.

अध्याय रूपरेखा एवं प्रगति

1 1. Thermal Expansion & Anomalous Wa...

2 2. Calorimetry & Latent Heat

3 3. Newton's Law of Cooling

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Thermal Expansion & Anomalous Water Expansion

Most materials expand when heated:
• Linear: $\Delta L = \alpha L \Delta T$.
• Area: $\Delta A = \beta A \Delta T = 2\alpha A \Delta T$.
• Volume: $\Delta V = \gamma V \Delta T = 3\alpha V \Delta T$.
Anomalous Expansion of Water: Between $0^\circ\text{C}$ and $4^\circ\text{C}$, water contracts on heating! It reaches maximum density at $4^\circ\text{C}$ ($1000\text{ kg/m}^3$). Thus, cold $0^\circ\text{C}$ ice floats at the surface, insulating the liquid $4^\circ\text{C}$ bottom where aquatic marine life survives!

2. Calorimetry & Latent Heat

Principle of Calorimetry: Heat lost by hot body = Heat gained by cold body.
• Specific Heat ($c$): $\Delta Q = m c \Delta T$. (Water has high $c = 4186\text{ J/kg}\cdot\text{K}$, ideal as car engine coolant).
• Latent Heat ($L$): Heat absorbed during phase change at constant temperature: $\mathbf{Q = mL}$ (Latent heat of fusion $L_f$, vaporization $L_v$).

3. Newton's Law of Cooling

The rate of loss of heat of a hot body is directly proportional to the temperature difference between the body and its surroundings: $$\mathbf{\frac{dT}{dt} = -k(T - T_0)}$$

Thermal Properties of Matter - Key Conceptual & Analytical Model

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Thermal Expansion Coefficients: $\alpha : \beta : \gamma = 1 : 2 : 3$.

मुख्य बिंदु 2

Anomalous Water Expansion: Water density peaks at $4^\circ\text{C}$, preserving aquatic winter habitats.

मुख्य बिंदु 3

Specific Heat of Water: Highest among common liquids ($4186\text{ J/kg}\cdot\text{K}$), buffering climate.

मुख्य बिंदु 4

Latent Heat: Hidden thermodynamic heat driving isothermal phase transformations ($Q = mL$).

मुख्य बिंदु 5

Newton's Law of Cooling: Exponential thermal decay scaling with ambient temperature differential.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

Explain why water pipes burst in severe cold winters.

उत्तर एवं व्याख्या देखें

उत्तर: When water cools below $4^\circ\text{C}$ and freezes into ice at $0^\circ\text{C}$, it expands anomalously by about 9% in volume. Enclosed inside rigid metallic pipes, this thermal expansion exerts colossal hydrostatic pressure, rupturing the pipes.
Anomalous expansion below 4°C expands ice volume.

Why does a burn from steam at $100^\circ\text{C}$ cause far more severe tissue damage than boiling water at $100^\circ\text{C}$?

उत्तर एवं व्याख्या देखें

उत्तर: Because steam at $100^\circ\text{C}$ contains an extra $2.26 \times 10^6\text{ J/kg}$ of latent heat of vaporization compared to liquid water at the same temperature, releasing far more thermal energy upon condensing on the skin.
Steam contains latent heat of vaporization (2.26 × 10^6 J/kg).

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of same length and diameter. What is the change in length of the combined rod at $250^\circ\text{C}$ if original length is at $40^\circ\text{C}$? ($\alpha_{\text{brass}} = 2.0 \times 10^{-5}\text{ K}^{-1}, \alpha_{\text{steel}} = 1.2 \times 10^{-5}\text{ K}^{-1}$).

उत्तर एवं व्याख्या देखें

उत्तर: $\Delta T = 250 - 40 = 210^\circ\text{C}$. $\Delta L = L_1 \alpha_1 \Delta T + L_2 \alpha_2 \Delta T = (0.50)(210)[2.0 \times 10^{-5} + 1.2 \times 10^{-5}] = 105 [3.2 \times 10^{-5}] = 3.36 \times 10^{-3}\text{ m} = 3.36\text{ mm}$.
Elongation is 3.36 mm.

State the Principle of Calorimetry. Under what conditions is it valid?

उत्तर एवं व्याख्या देखें

उत्तर: In an isolated system with no heat lost to surroundings or absorbed by chemical reactions: Heat lost by hot bodies = Heat gained by cold bodies.
Heat lost = Heat gained in insulated system.

State Newton's Law of Cooling and write its mathematical equation.

उत्तर एवं व्याख्या देखें

उत्तर: The rate of loss of heat by radiation from a body is directly proportional to the excess temperature of the body over that of its surroundings: $\frac{dQ}{dt} = -k(T - T_0)$ or $\frac{dT}{dt} = -K(T - T_0)$ where $T - T_0$ is small.
dT/dt = -k(T - T0).

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