Thermal Properties of Matter Class 11 Physics NCERT Solutions & Concepts

Temperature, Heat, and Thermal Expansion Mechanisms

In class 11 physics thermal properties matter, we distinguish between heat and temperature. Heat is thermal energy in transit from a body at a higher temperature to a body at a lower temperature. Temperature, measured in Kelvin (K) or Celsius (°C), indicates the average kinetic energy of the molecules in a substance. When a substance absorbs heat, its molecules vibrate more vigorously, leading to an increase in inter-atomic separation. This macroscopically manifests as thermal expansion. Solids experience linear expansion (coefficient $\alpha$), area expansion (coefficient $\beta$), and volume expansion (coefficient $\gamma$). For isotropic solids, these expansion coefficients are related by the ratio $\alpha : \beta : \gamma = 1 : 2 : 3$. A critical exception to standard thermal expansion is the anomalous expansion of water: water contracts when heated from 0°C to 4°C, reaching its maximum density at 4°C. This unique property keeps aquatic life safe, as ice forms on the surface of lakes while dense liquid water remains underneath at 4°C.

Important Terms & Formulas

Specific Heat Capacity (c)

The amount of heat energy required to raise the temperature of a unit mass of a substance by 1°C or 1 K. Formula: c = Q / (m * ΔT), measured in J/kg·K or cal/g·°C.

Latent Heat (L)

The heat absorbed or released by a substance during a change in its physical state (phase change) that occurs without a change in temperature. Formula: Q = m * L.

Calorimetry

The quantitative measurement of heat exchange between bodies. According to the Principle of Calorimetry, when two bodies at different temperatures are mixed in an isolated system, Heat Lost by Hotter Body = Heat Gained by Colder Body.

Thermal Conductivity (K)

The property of a material to conduct heat, defined by the rate of heat flow: H = K A (T1 - T2) / d, where A is the cross-sectional area, d is the length, and T1 - T2 is the temperature gradient.

How to Solve Calorimetry Problems

1. Identify Phases and Initial Temperatures — List all substances in the mix, their respective masses (m), initial temperatures (T), specific heat capacities (c), and latent heats (L).
2. Set Up Heat Exchange Equations — Determine which bodies will lose heat (hotter bodies cooling down) and which will gain heat (colder bodies warming up or changing state). Write down equations: Q = m c ΔT for temperature change and Q = m * L for phase transitions.
3. Apply Conservation of Energy — Formulate the equilibrium equation: Total Heat Gained = Total Heat Lost. Keep all terms on both sides of the equation in the same unit system (either SI or CGS).
4. Calculate the Final Equilibrium Temperature — Solve the algebraic equation for the final temperature (T_f) or unknown mass. Always ensure T_f lies between the lowest and highest initial temperatures.

Step-by-Step Solved Numericals

• Example 1: A steel rail has a length of 20 m at 20°C. Find its length on a hot summer day when the temperature rises to 45°C. (Coefficient of linear expansion of steel = 1.2 x 10^-5 /°C). Step 1: Write down given values. Initial length (L0) = 20 m Initial temperature (T1) = 20°C Final temperature (T2) = 45°C Temperature difference (ΔT) = 45 - 20 = 25°C Step 2: Apply the formula for linear expansion. ΔL = α L0 ΔT ΔL = (1.2 x 10^-5) 20 25 ΔL = 0.006 m (or 6 mm) Step 3: Calculate the final length. L = L0 + ΔL = 20 + 0.006 = 20.006 m. Final Answer: The length of the rail at 45°C is 20.006 m.
• Example 2: Calculate the heat required to convert 10 g of ice at 0°C into water at 50°C. (Given: Latent heat of fusion of ice L_f = 80 cal/g, specific heat of water c_w = 1 cal/g·°C). Step 1: Identify the heat processes. Phase Change (Ice to Water at 0°C): Q1 = m L_f Temperature Change (Water at 0°C to 50°C): Q2 = m c_w ΔT Step 2: Calculate Q1. Q1 = 10 g 80 cal/g = 800 cal Step 3: Calculate Q2. Q2 = 10 g 1 cal/g·°C (50 - 0)°C = 500 cal Step 4: Find total heat energy. Q_total = Q1 + Q2 = 800 + 500 = 1300 calories. Final Answer: The total heat required is 1300 calories.

Exam Traps & Conversion Tips

1. Unit Compatibility Trap: Never mix SI and CGS units in calorimetry calculations. Specific heat is often given in cal/g·°C, whereas heat capacity might be given in J/kg·K. Convert using: $1 \text{ cal} \approx 4.186 \text{ J}$.

1. Phase Changes: Remember that temperature stays completely constant during a phase change. Do not apply the formula $Q = mc\Delta T$ during melting or vaporization. Only use $Q = mL$.

1. Expansion Coefficients: Make sure you read the question carefully to see if $\alpha$ (linear), $\beta$ (areal), or $\gamma$ (volume) is given. If the question asks for volume expansion but gives linear expansion, multiply by 3 ($\gamma = 3\alpha$).

Practice Questions with Solutions

• Q: A copper rod of length 1.5 m has its temperature increased by 80°C. Calculate the fractional change in its length if the linear coefficient of expansion is 1.7 x 10^-5 /°C. A: Step 1: State the formula for fractional change in length. Fractional change = ΔL / L0 = α ΔT Step 2: Substitute the given values. α = 1.7 x 10^-5 /°C ΔT = 80°C Step 3: Perform the calculation. ΔL / L0 = (1.7 x 10^-5) 80 = 1.36 x 10^-3 Final answer: The fractional change in the rod's length is 1.36 x 10^-3.
• Q: Why are gaps left between successive rails on railway tracks? A: Step 1: Explain the physical process involved. During summers, high temperatures cause the steel rails to undergo linear thermal expansion. Step 2: Discuss what happens if no gap is left. If no gaps are provided, the expansion would be restricted, creating immense thermal stress. This would buckle, bend, or warp the rails, leading to catastrophic train derailments. Final answer: Gaps are left to allow safe thermal expansion of the rails during seasonal temperature rises without causing structural damage.
• Q: How much heat is lost when 100 g of water at 80°C cools down to 20°C? (Take specific heat capacity of water as 4.2 J/g·°C). A: Step 1: State the heat transfer equation. Q = m c ΔT Step 2: Substitute the known values. m = 100 g c = 4.2 J/g·°C ΔT = T_final - T_initial = 20°C - 80°C = -60°C Step 3: Solve for Q. Q = 100 4.2 (-60) = -25200 J (The negative sign denotes heat release/loss) Final answer: The heat lost by the water is 25,200 Joules (or 25.2 kJ).
• Q: Explain the basic difference between conduction, convection, and radiation. A: Step 1: Define Conduction. Conduction is the process of heat transfer where energy is transferred from particle to particle by molecular collisions without any actual bulk movement of the matter itself. It is dominant in solids. Step 2: Define Convection. Convection involves the actual bulk movement of heated fluid molecules from regions of high temperature to low temperature. It occurs only in fluids (liquids and gases). Step 3: Define Radiation. Radiation is the process of heat transfer via electromagnetic waves. Unlike conduction and convection, it does not require a material medium and travels at the speed of light. Final answer: Conduction transfers heat through atomic contact without bulk motion; convection moves heat through actual fluid displacement; radiation transfers heat via electromagnetic waves requiring no physical medium.

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