Thermal Properties of Matter: Class 11 Physics NCERT Guide

Fundamental Concepts: Heat, Temperature, and Internal Energy

Temperature

A measure of the degree of hotness or coldness of a body. On a microscopic level, it is proportional to the average kinetic energy of the atoms or molecules of the substance.

Heat

The form of energy that is transferred between two systems (or a system and its surroundings) by virtue of a temperature difference. Heat is energy in transit, not energy stored in a body. Its SI unit is the Joule (J).

Internal Energy

The sum of the kinetic and potential energies of all the molecules within a substance. When you supply heat to a substance, you increase its internal energy.

Thermal Equilibrium

A state in which two objects in thermal contact cease to have any net exchange of heat energy. This occurs when they reach the same temperature.

Understanding Thermal Expansion

Most substances expand when heated and contract when cooled. But why does this happen? When you supply heat energy, the atoms and molecules of the substance absorb this energy and vibrate more vigorously about their mean positions. This increased vibration increases the average distance between adjacent molecules, causing the material to expand in all directions. This phenomenon is called thermal expansion.

In solids, we study this expansion in three ways:

1. Linear Expansion (α): Increase in length. The change in length (ΔL) is proportional to the original length (L) and the change in temperature (ΔT). Formula: ΔL = αLΔT
2. Area Expansion (β): Increase in surface area. The change in area (ΔA) is proportional to the original area (A) and ΔT. Formula: ΔA = βAΔT
3. Volume Expansion (γ): Increase in volume. The change in volume (ΔV) is proportional to the original volume (V) and ΔT. Formula: ΔV = γVΔT

For isotropic solids (which expand uniformly in all directions), the coefficients are related: β ≈ 2α and γ ≈ 3α. This relationship is crucial for solving many numerical problems. An interesting exception is the anomalous expansion of water, which contracts when heated from 0°C to 4°C, reaching its maximum density at 4°C.

Worked Examples on Calorimetry

• Problem 1: 200 g of water at 80°C is mixed with 300 g of water at 20°C. Assuming no heat is lost to the surroundings, find the final temperature of the mixture. (Specific heat of water, c = 4.186 J/g°C) Solution: Step 1: Principle of Calorimetry. The core principle here is that the heat lost by the hot water will be equal to the heat gained by the cold water. Heat Lost = Heat Gained Step 2: Identify bodies and variables. Hot water: mass (m₁) = 200 g, initial temperature (T₁) = 80°C. Cold water: mass (m₂) = 300 g, initial temperature (T₂) = 20°C. Let the final equilibrium temperature be T_f. Step 3: Write expressions for heat lost and gained. Heat Lost by hot water: Q₁ = m₁c(T₁ - T_f) = 200 c (80 - T_f) Heat Gained by cold water: Q₂ = m₂c(T_f - T₂) = 300 c (T_f - 20) Step 4: Equate and solve for T_f. 200 c (80 - T_f) = 300 c (T_f - 20) The specific heat 'c' cancels out on both sides. 2 (80 - T_f) = 3 (T_f - 20) 160 - 2T_f = 3T_f - 60 160 + 60 = 3T_f + 2T_f 220 = 5T_f T_f = 220 / 5 = 44°C * Final Answer: The final temperature of the mixture is 44°C.
• Problem 2: A 50 g block of metal at 100°C is dropped into a calorimeter containing 100 g of water at 20°C. The final temperature is 25°C. Find the specific heat capacity of the metal. (Specific heat of water = 4.2 J/g°C) Solution: Step 1: Apply the Principle of Calorimetry. Heat Lost by metal = Heat Gained by water. (We assume the calorimeter has negligible heat capacity). Step 2: List the knowns. Metal: m_m = 50 g, T_initial_m = 100°C, c_m = ? Water: m_w = 100 g, T_initial_w = 20°C, c_w = 4.2 J/g°C Final Temperature: T_f = 25°C Step 3: Formulate the equation. m_m c_m (T_initial_m - T_f) = m_w c_w (T_f - T_initial_w) Step 4: Substitute values and solve for c_m. 50 c_m (100 - 25) = 100 4.2 (25 - 20) 50 c_m 75 = 100 4.2 5 3750 c_m = 2100 c_m = 2100 / 3750 c_m = 0.56 J/g°C Final Answer: The specific heat capacity of the metal is 0.56 J/g°C.

The Three Modes of Heat Transfer

1. Conduction — This is the transfer of heat through a substance without any movement of the substance itself. It occurs primarily in solids. When one end of a metal rod is heated, the atoms at that end vibrate more vigorously and collide with their neighbors, transferring kinetic energy. This chain reaction continues down the rod. The rate of heat flow is given by Fourier's Law: dQ/dt = -kA(dT/dx), where k is the thermal conductivity of the material.
2. Convection — This mode involves heat transfer by the actual movement of matter. It occurs in fluids (liquids and gases). When a fluid is heated, it expands, becomes less dense, and rises. The cooler, denser fluid from above sinks to take its place, gets heated, and rises. This creates a continuous circulation called a convection current. Examples include boiling water and sea breezes.
3. Radiation — This is the transfer of heat in the form of electromagnetic waves, mainly in the infrared region. Unlike conduction and convection, radiation does not require a medium and can travel through a vacuum. This is how we receive heat from the Sun. Every object above absolute zero (0 K) radiates energy. The rate of radiation is given by the Stefan-Boltzmann Law: P = σεAT⁴.

Exam Traps and Key Formulas to Remember

Pay close attention to these points to avoid losing marks in your exams:

• Temperature vs. Heat: Never use these terms interchangeably. Temperature is a measure of average kinetic energy, while heat is the energy transferred due to a temperature difference.
• Units: Always be consistent with units. If specific heat is in J/kg°C, ensure mass is in kg. A common trap is mixing grams and kilograms.
• Sign Convention in Calorimetry: A simple way is to always use Heat Lost = Heat Gained and set up the temperature differences as positive values: (T_hot - T_final) and (T_final - T_cold). This avoids confusion with negative signs.
• Change of State: When a substance changes state (e.g., ice melting), its temperature remains constant. Use the latent heat formula Q = mL, not Q = mcΔT. Many problems involve both, so calculate heat for temperature change and state change separately.

Must-Know Formulas:

1. Linear Expansion: ΔL = αLΔT
2. Specific Heat Capacity: Q = mcΔT
3. Latent Heat: Q = mL (L_f for fusion, L_v for vaporization)
4. Conduction Rate: H = dQ/dt = kA(T₁-T₂)/L
5. Stefan-Boltzmann Law: P = σεAT⁴

Practice Questions with Solutions

• Q: A steel ruler is 1 m long at 20°C. What is its length on a hot day when the temperature is 45°C? The coefficient of linear expansion for steel is 1.2 x 10⁻⁵ /°C. A: Step 1: Identify the given values. Original length, L = 1 m Initial temperature, T₁ = 20°C Final temperature, T₂ = 45°C Coefficient of linear expansion, α = 1.2 x 10⁻⁵ /°C Step 2: Calculate the change in temperature, ΔT. ΔT = T₂ - T₁ = 45°C - 20°C = 25°C Step 3: Use the formula for linear expansion to find the change in length, ΔL. ΔL = αLΔT ΔL = (1.2 x 10⁻⁵ /°C) (1 m) (25°C) ΔL = 30 x 10⁻⁵ m = 0.0003 m Step 4: Calculate the final length. Final Length = Original Length + ΔL Final Length = 1 m + 0.0003 m = 1.0003 m Final answer: The length of the ruler on the hot day will be 1.0003 m.
• Q: How much heat is required to convert 10 g of ice at 0°C to steam at 100°C? (Given: Latent heat of fusion of ice, L_f = 334 J/g; Specific heat of water, c = 4.2 J/g°C; Latent heat of vaporization of water, L_v = 2260 J/g) A: Step 1: Break the problem into three stages. Stage 1 (Q₁): Melting 10 g of ice at 0°C to water at 0°C. Stage 2 (Q₂): Heating 10 g of water from 0°C to 100°C. Stage 3 (Q₃): Vaporizing 10 g of water at 100°C to steam at 100°C. Step 2: Calculate the heat for each stage. Q₁ = mL_f = 10 g 334 J/g = 3340 J Q₂ = mcΔT = 10 g 4.2 J/g°C (100 - 0)°C = 10 4.2 100 = 4200 J Q₃ = mL_v = 10 g 2260 J/g = 22600 J Step 3: Calculate the total heat required. Total Heat, Q = Q₁ + Q₂ + Q₃ Q = 3340 J + 4200 J + 22600 J = 30140 J Final answer: The total heat required is 30140 J or 30.14 kJ.
• Q: Two rods, A and B, of different materials are welded together. Their thermal conductivities are k₁ and k₂. The rods have the same length and area of cross-section. If the free end of A is at 100°C and the free end of B is at 0°C, what is the temperature at the junction? A: Step 1: Understand the steady-state condition. In steady state, the rate of heat flow (H) through rod A is equal to the rate of heat flow through rod B. H₁ = H₂ Step 2: Use the formula for heat conduction. Rate of heat flow, H = kA(ΔT)/L. Let T be the temperature at the junction. For rod A: H₁ = k₁A(100 - T)/L For rod B: H₂ = k₂A(T - 0)/L Step 3: Equate the two expressions and solve for T. k₁A(100 - T)/L = k₂A(T - 0)/L The terms 'A' and 'L' cancel out. k₁(100 - T) = k₂T 100k₁ - k₁T = k₂T 100k₁ = k₁T + k₂T 100k₁ = T(k₁ + k₂) T = 100k₁ / (k₁ + k₂) Final answer: The temperature at the junction is T = 100k₁ / (k₁ + k₂).
• Q: A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C. A: Step 1: State Newton's Law of Cooling. The rate of cooling is proportional to the temperature difference between the body and the surroundings. Using the approximate form: (T₁ - T₂)/t = K [((T₁ + T₂)/2) - T₀], where T₀ is the surrounding temperature. Step 2: Apply the law for the first case. T₁ = 80°C, T₂ = 50°C, t = 5 min, T₀ = 20°C (80 - 50)/5 = K [((80 + 50)/2) - 20] 30/5 = K [(130/2) - 20] 6 = K [65 - 20] 6 = K 45 => K = 6/45 = 2/15 Step 3: Apply the law for the second case. T₁ = 60°C, T₂ = 30°C, T₀ = 20°C, t = ? (60 - 30)/t = K [((60 + 30)/2) - 20] 30/t = (2/15) [(90/2) - 20] 30/t = (2/15) [45 - 20] 30/t = (2/15) 25 30/t = 50/15 = 10/3 t = (30 3) / 10 = 9 minutes. Final answer: It will take 9 minutes to cool from 60°C to 30°C.

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