Thermal Properties of Matter

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Home Concept Boosters CBSE CBSE Class 11 Physics Thermal Properties of Matter

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How to Use This Page

Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Thermal Properties of Matter for CBSE Class 11 Physics.

Key Concepts

Class 11 · Physics · Thermal Properties

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Heat, Temperature, and Energy Transfer

Class 11 · Physics

Temperature Scales & Conversion Formula

Temperature is the degree of hotness or coldness. The relationship between Celsius ($T_C$), Fahrenheit ($T_F$), and Kelvin ($T_K$) is linear.

$$\frac{T_C}{100} = \frac{T_F – 32}{180} = \frac{T_K – 273.15}{100}$$

Thermal Expansion Formula

Most substances expand on heating. The fractional change in dimension is proportional to the change in temperature ($\Delta T$). Coefficients of linear ($\alpha$), area ($\beta$), and volume ($\gamma$) expansion are related.

$$\Delta L = \alpha L \Delta T \quad | \quad \Delta A = \beta A \Delta T \quad | \quad \Delta V = \gamma V \Delta T$$ $$\alpha : \beta : \gamma = 1 : 2 : 3$$

Specific Heat Capacity ($s$ or $c$) Formula

The amount of heat required to raise the temperature of unit mass of a substance by $1^\circ\text{C}$ (or $1\text{ K}$). It dictates how “stubborn” a material is to temperature change.

$$Q = ms\Delta T \quad \implies \quad s = \frac{Q}{m\Delta T}$$

Latent Heat ($L$) Formula

The heat energy required to change the state of unit mass of a substance without any change in temperature (e.g., solid to liquid is Latent Heat of Fusion, $L_f$).

$$Q = mL$$

Principle of Calorimetry Formula

Based on the law of conservation of energy. In an isolated system, the heat lost by the hotter body is exactly equal to the heat gained by the colder body until thermal equilibrium is reached.

$$\text{Heat Lost} = \text{Heat Gained}$$

Thermal Conductivity ($K$) Formula

Heat transfer through a solid by conduction. The rate of heat flow ($H = dQ/dt$) is proportional to the cross-sectional area ($A$) and the temperature gradient ($\Delta T/L$).

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

Newton’s Law of Cooling Formula

The rate of loss of heat from a body is directly proportional to the temperature difference between the body and its surroundings, provided the difference is small.

$$-\frac{dT}{dt} = k(T – T_s)$$

Wien’s Displacement Law Formula

The wavelength corresponding to maximum intensity of emission of black body radiation ($\lambda_{\text{max}}$) is inversely proportional to the absolute temperature ($T$) of the body.

$$\lambda_{\text{max}} T = b \quad (b = 2.89 \times 10^{-3} \text{ m}\cdot\text{K})$$

Stefan-Boltzmann Law Formula

The total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature. For a non-perfect black body, emissivity ($e$) is included.

$$E = \sigma T^4 \quad | \quad P = e\sigma A T^4$$

Concept Deep Dive

Anomalous Expansion of Water

Why lakes don’t freeze from the bottom up

Core Concept

Almost all liquids expand when heated and contract when cooled. Water is the bizarre exception. When water is cooled from room temperature, it contracts normally until it reaches $4^\circ\text{C}$. At $4^\circ\text{C}$, water reaches its maximum density.

If you cool it further from $4^\circ\text{C}$ down to $0^\circ\text{C}$, it actually expands! Because ice at $0^\circ\text{C}$ is less dense than water at $4^\circ\text{C}$, ice floats. In cold climates, the surface of a lake freezes, forming an insulating layer of ice, while the denser $4^\circ\text{C}$ water sinks to the bottom, allowing fish and aquatic life to survive the harsh winter.

Newton’s Law of Cooling

The exponential decay of a hot cup of tea

Calculus Application

If you leave a boiling cup of tea ($100^\circ\text{C}$) in a room ($20^\circ\text{C}$), it cools very rapidly at first because the temperature difference is huge ($80^\circ\text{C}$). As the tea cools down to say, $30^\circ\text{C}$, the rate of cooling drastically slows down because the difference is now only $10^\circ\text{C}$.

Mathematically, integrating $-\frac{dT}{dt} = k(T – T_s)$ gives $T(t) = T_s + (T_0 – T_s)e^{-kt}$. This proves that the cooling curve is an exponential decay curve. It theoretically takes an infinite amount of time for the tea to reach the exact temperature of the room!

Compare & Contrast

✗ Specific Heat ($s$)

Heat required to change the temperature.
Phase of the substance remains constant (e.g., warming liquid water).
Formula: $Q = ms\Delta T$.
Unit: $\text{J/kg}\cdot\text{K}$
Depends heavily on the material (Water has a very high $s$).

✓ Latent Heat ($L$)

Heat required to change the phase/state.
Temperature remains perfectly constant (e.g., melting ice at $0^\circ\text{C}$).
Formula: $Q = mL$.
Unit: $\text{J/kg}$
Hidden heat used solely to break intermolecular bonds.

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Remember

A steam burn at $100^\circ\text{C}$ is far more severe than a boiling water burn at $100^\circ\text{C}$. Why? Because steam carries an enormous amount of extra hidden energy (Latent Heat of Vaporization, $L_v \approx 2.26 \times 10^6 \text{ J/kg}$) that it releases into your skin as it condenses back into water!

Common Mistakes to Avoid

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Mistake 1

Using Celsius in Radiation Laws: Newton’s Law of Cooling works fine with Celsius because it relies on temperature differences ($\Delta T$). However, Stefan-Boltzmann Law ($E \propto T^4$) and Wien’s Law ($\lambda_{\text{max}} T = b$) strictly require Absolute Temperature in Kelvin. Plugging $27^\circ\text{C}$ directly into $T^4$ is a guaranteed zero mark.

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Mistake 2

Ignoring the Latent Heat Step in Calorimetry: If you drop $-10^\circ\text{C}$ ice into $50^\circ\text{C}$ water, you cannot just write $m_1 s_{\text{ice}} (T – (-10)) = m_2 s_{\text{water}} (50 – T)$. Ice must first warm to $0^\circ\text{C}$, then melt into water, and then that new water warms to final $T$. The correct left side is: $m_1 s_{\text{ice}} (10) + m_1 L_f + m_1 s_{\text{water}} (T – 0)$.

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Mistake 3

Heat Capacity vs. Specific Heat Capacity: Heat capacity ($C$ or $S$) is the heat required for the whole body ($Q = C\Delta T$). Specific heat capacity ($s$ or $c$) is for a unit mass ($Q = ms\Delta T$). Watch out for the word “specific”!

Exam Tips

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Tip 1

Average Form of Newton’s Law of Cooling: For numericals where a body cools from $T_1$ to $T_2$ in time $t$, use the approximation: $\frac{T_1 – T_2}{t} = k\left( \frac{T_1 + T_2}{2} – T_s \right)$. This avoids complex logarithmic calculations and is perfectly acceptable in board exams!

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Tip 2

Star Colors and Wien’s Law: Red stars are cool, blue stars are hot. Because $\lambda_{\text{max}} \propto 1/T$, a higher temperature shifts the peak emission to a shorter wavelength (blue/violet end of the spectrum).

Expected Exam Questions

Board Pattern Questions

Class 11 · Thermal Properties · CBSE Exam

Class 11 · Physics

Why is a small gap left between consecutive iron rails of a railway track? [1 mark]

Answer To allow for linear thermal expansion during summer. 📝

Explanation

During the hot summer months, the iron rails expand due to thermal expansion ($\Delta L = \alpha L \Delta T$). If no gap is left between the rails, the huge thermal stress generated will cause the tracks to bend and buckle, which can derail trains.

A pan filled with hot food cools from $94^\circ\text{C}$ to $86^\circ\text{C}$ in 2 minutes when the room temperature is $20^\circ\text{C}$. How long will it take to cool from $71^\circ\text{C}$ to $69^\circ\text{C}$? [3 marks]

Answer $t = 0.7$ minutes (or $42$ seconds) 📝

Explanation

Using the average form of Newton’s Law of Cooling: $\frac{T_1 – T_2}{t} = k\left( \frac{T_1 + T_2}{2} – T_s \right)$.
Case 1: $T_1=94$, $T_2=86$, $t=2 \text{ min}$, $T_s=20$.
$\frac{94 – 86}{2} = k\left( \frac{94 + 86}{2} – 20 \right) \implies \frac{8}{2} = k(90 – 20) \implies 4 = k(70) \implies k = \frac{4}{70}$.
Case 2: $T_1=71$, $T_2=69$, $t=?$
$\frac{71 – 69}{t} = k\left( \frac{71 + 69}{2} – 20 \right) \implies \frac{2}{t} = k(70 – 20) \implies \frac{2}{t} = k(50)$.
Substitute $k = 4/70$: $\frac{2}{t} = \left(\frac{4}{70}\right) \times 50 \implies \frac{2}{t} = \frac{200}{70} = \frac{20}{7}$.
$20t = 14 \implies t = \frac{14}{20} = 0.7 \text{ minutes}$ (or $42$ seconds).

If the temperature of a black body is increased from $27^\circ\text{C}$ to $327^\circ\text{C}$, by what factor does the energy emitted per second increase? [2 marks]

Answer Increases by a factor of 16. 📝

Explanation

By Stefan-Boltzmann Law, the energy emitted per second is $E \propto T^4$, where $T$ MUST be in Kelvin.
Initial Temp $T_1 = 27 + 273 = 300 \text{ K}$.
Final Temp $T_2 = 327 + 273 = 600 \text{ K}$.
Ratio $\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{600}{300}\right)^4 = (2)^4 = 16$.
The energy emitted increases by 16 times.

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