class 11 notes

Class 11 Physics Notes Chapter 11 (Chapter 11) – Examplar Problems (English) Book

Philoid

Philoid

— 10 min read

Examplar Problems (English)
Alright class, let's begin our detailed study of Chapter 11: Thermal Properties of Matter. This chapter is fundamental for understanding how matter behaves when subjected to heat and temperature changes, and it's a frequent source of questions in various government exams. Pay close attention to the concepts, definitions, and formulas.

Chapter 11: Thermal Properties of Matter - Detailed Notes

1. Introduction: Heat and Temperature

  • Temperature: A measure of the degree of hotness or coldness of a body. It determines the direction of heat flow between two bodies in thermal contact (heat flows from higher temperature to lower temperature).
    • It's related to the average kinetic energy of the molecules of the substance.
    • SI Unit: Kelvin (K)
    • Common Units: Degree Celsius (°C), Degree Fahrenheit (°F)
  • Heat: The form of energy transferred between systems or a system and its surroundings by virtue of a temperature difference.
    • It is energy in transit. A body does not contain heat; it contains internal energy.
    • SI Unit: Joule (J)
    • Common Unit: calorie (cal) (1 cal = 4.186 J)

2. Measurement of Temperature

  • Thermometer: A device used to measure temperature. Its working is based on some physical property of a substance (thermometric property) that changes continuously with temperature (e.g., length of a liquid column, pressure of a gas at constant volume, volume of a gas at constant pressure, electrical resistance).
  • Temperature Scales:
    • Celsius (°C): Ice point (freezing point of water) = 0°C; Steam point (boiling point of water) = 100°C.
    • Fahrenheit (°F): Ice point = 32°F; Steam point = 212°F.
    • Kelvin (K): Absolute scale. Ice point = 273.15 K; Steam point = 373.15 K. Absolute zero (0 K = -273.15 °C) is the theoretically lowest possible temperature.
  • Conversion Formulas:
    • T(K) = T(°C) + 273.15
    • T(°F) = (9/5) * T(°C) + 32
    • (T(°C) - 0) / 100 = (T(°F) - 32) / 180 = (T(K) - 273.15) / 100
    • Key Point: Temperature differences have the same magnitude on Celsius and Kelvin scales (ΔT(K) = ΔT(°C)). A difference of 1°C is equal to a difference of 1.8°F.

3. Thermal Expansion

  • The increase in the dimensions (length, area, volume) of a body due to an increase in its temperature.
  • Linear Expansion: Increase in length.
    • ΔL = α L₀ ΔT
    • L = L₀ (1 + α ΔT)
    • Where: ΔL = change in length, L₀ = original length, ΔT = change in temperature, α = coefficient of linear expansion (Unit: K⁻¹ or °C⁻¹). α depends on the nature of the material.
  • Area (Superficial) Expansion: Increase in area.
    • ΔA = β A₀ ΔT
    • A = A₀ (1 + β ΔT)
    • Where: ΔA = change in area, A₀ = original area, β = coefficient of area expansion (Unit: K⁻¹ or °C⁻¹).
  • Volume (Cubical) Expansion: Increase in volume.
    • ΔV = γ V₀ ΔT
    • V = V₀ (1 + γ ΔT)
    • Where: ΔV = change in volume, V₀ = original volume, γ = coefficient of volume expansion (Unit: K⁻¹ or °C⁻¹).
  • Relation between α, β, and γ (for isotropic solids):
    • β ≈ 2α
    • γ ≈ 3α
    • α : β : γ ≈ 1 : 2 : 3
  • Thermal Stress: If a rod is prevented from expanding or contracting, stress is developed due to temperature change. Thermal Stress = Y α ΔT (where Y is Young's Modulus).
  • Anomalous Expansion of Water: Water contracts on heating from 0°C to 4°C. It has its maximum density (and minimum volume) at 4°C. Above 4°C, it expands on heating, like most other substances. This property is crucial for the survival of aquatic life in cold climates.

4. Specific Heat Capacity (s or c)

  • The amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin).
    • Q = m s ΔT
    • Where: Q = heat supplied/removed, m = mass, s = specific heat capacity, ΔT = change in temperature.
    • SI Unit: J kg⁻¹ K⁻¹
    • Depends on the nature of the substance and its phase (e.g., specific heat of water is different from ice or steam). Water has a very high specific heat capacity (≈ 4186 J kg⁻¹ K⁻¹), making it useful as a coolant.
  • Molar Specific Heat Capacity (C): Heat energy required to raise the temperature of one mole of a substance by one degree.
    • Q = n C ΔT (where n is the number of moles)
    • C = M s (where M is the molar mass)
    • SI Unit: J mol⁻¹ K⁻¹
    • For gases, we define specific heat at constant volume (Cv) and constant pressure (Cp). Generally, Cp > Cv.

5. Calorimetry

  • The measurement of heat.
  • Principle of Calorimetry (Law of Mixtures): When two bodies at different temperatures are brought into thermal contact, heat flows from the body at higher temperature to the body at lower temperature until they reach thermal equilibrium (same temperature), provided no heat is lost to the surroundings.
    • Heat Lost by hot body = Heat Gained by cold body
    • m₁ s₁ (T₁ - T_final) = m₂ s₂ (T_final - T₂) (assuming T₁ > T₂)
  • Water Equivalent: The mass of water that would require the same amount of heat as the calorimeter (or another object) for the same temperature rise. Water Equivalent (W) = m_object * s_object / s_water.

6. Change of State

  • Matter exists in different phases (solid, liquid, gas). A transition from one phase to another is called a change of state.
  • These changes occur at constant temperature and pressure.
  • Melting: Solid to Liquid (at melting point)
  • Freezing/Fusion: Liquid to Solid (at freezing point)
  • Boiling/Vaporization: Liquid to Gas (at boiling point)
  • Condensation: Gas to Liquid (at condensation point)
  • Sublimation: Solid directly to Gas (e.g., camphor, dry ice)
  • Deposition: Gas directly to Solid
  • Latent Heat (L): The amount of heat absorbed or released during a change of state per unit mass, without any change in temperature.
    • Q = m L
    • SI Unit: J kg⁻¹
    • Latent Heat of Fusion (Lf): Heat required to change unit mass from solid to liquid at the melting point. (For ice: ≈ 3.33 × 10⁵ J kg⁻¹)
    • Latent Heat of Vaporization (Lv): Heat required to change unit mass from liquid to gas at the boiling point. (For water: ≈ 22.6 × 10⁵ J kg⁻¹)
    • Note: Lv is generally much higher than Lf.

7. Heat Transfer

  • The process by which heat energy flows from a region of higher temperature to a region of lower temperature.

  • Modes of Heat Transfer: Conduction, Convection, Radiation.

    • a) Conduction:

      • Heat transfer through a material medium without the actual movement of the particles of the medium from their mean positions. Primarily occurs in solids.
      • Mechanism: Vibration of particles and collision between adjacent particles (in insulators); drift of free electrons (in metals - hence metals are good conductors).
      • Steady State: The state where the temperature at every point of the conductor becomes constant (does not change with time).
      • Fourier's Law of Heat Conduction: The rate of heat flow (dQ/dt or H) through a conductor in steady state is directly proportional to the area of cross-section (A) and the temperature gradient (dT/dx).
        • H = dQ/dt = - K A (dT/dx)
        • K = Coefficient of Thermal Conductivity (Unit: W m⁻¹ K⁻¹ or J s⁻¹ m⁻¹ K⁻¹). It measures the ability of a material to conduct heat. High K for conductors, low K for insulators.
        • The negative sign indicates that heat flows in the direction of decreasing temperature.
      • For a uniform rod of length L with ends at temperatures T₁ and T₂ (T₁ > T₂):
        • H = K A (T₁ - T₂) / L
      • Thermal Resistance (Rth): Analogy to electrical resistance.
        • Rth = L / (K A)
        • H = (T₁ - T₂) / Rth
        • Series combination: R_eq = R₁ + R₂ + ...
        • Parallel combination: 1/R_eq = 1/R₁ + 1/R₂ + ...

    • b) Convection:

      • Heat transfer by the actual movement of the particles of the medium. Occurs in fluids (liquids and gases).
      • Mechanism: Heated fluid becomes less dense and rises, while cooler, denser fluid sinks, setting up convection currents.
      • Natural Convection: Due to density differences caused by heating (e.g., land and sea breezes, ventilation, heating water in a pot from below).
      • Forced Convection: Fluid is forced to move by external agents like pumps or fans (e.g., car radiator fan, air conditioners, convection ovens).

    • c) Radiation:

      • Heat transfer through electromagnetic waves. Does not require a material medium. Can occur through vacuum.
      • All bodies above absolute zero temperature radiate energy.
      • Blackbody: An ideal body that absorbs all incident radiation and emits the maximum possible radiation at a given temperature (perfect absorber and perfect emitter).
      • Absorptivity (a), Reflectivity (r), Transmissivity (t): a + r + t = 1. For a perfect blackbody, a = 1, r = 0, t = 0.
      • Emissivity (e): Ratio of emissive power of a body to the emissive power of a blackbody at the same temperature (0 ≤ e ≤ 1). e = 1 for a blackbody, e = 0 for a perfect reflector.
      • Kirchhoff's Law: At a given temperature, the ratio of emissive power to absorptive power is constant for all bodies and is equal to the emissive power of a perfect blackbody at that temperature. (Good absorbers are good emitters).
      • Stefan-Boltzmann Law: The total energy radiated per unit time per unit area by a perfect blackbody is directly proportional to the fourth power of its absolute temperature (T).
        • E = σ T⁴ (Energy per unit area per unit time)
        • Total Energy radiated by a blackbody of surface area A: P = σ A T⁴
        • For any other body (emissivity e): P = e σ A T⁴
        • σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
      • Net Rate of Heat Loss by Radiation: If a body at temperature T is placed in surroundings at temperature T₀ (T > T₀):
        • P_net = e σ A (T⁴ - T₀⁴)
      • Wien's Displacement Law: The wavelength (λm) corresponding to the maximum intensity of radiation emitted by a blackbody is inversely proportional to its absolute temperature (T).
        • λm T = b (Wien's constant)
        • b ≈ 2.898 × 10⁻³ m K
        • As temperature increases, the peak of the radiation curve shifts towards shorter wavelengths (e.g., a heated iron bar first glows red, then orange, yellow, and finally white/bluish-white).
      • Newton's Law of Cooling: For small temperature differences between a body and its surroundings, the rate of cooling (rate of loss of heat) is directly proportional to the temperature difference.
        • dQ/dt = - k (T - T₀) (where k is a constant depending on the body and surroundings)
        • Also expressed as: dT/dt = - K' (T - T_avg) where T_avg is the average temperature during cooling.
        • This is an approximation of the Stefan-Boltzmann law for (T - T₀) << T₀.

Multiple Choice Questions (MCQs)

  1. A bimetallic strip is made of aluminum and steel (α_Al > α_Steel). On heating, the strip will:
    a) Remain straight
    b) Bend with aluminum on the convex side (outer side)
    c) Bend with steel on the convex side (outer side)
    d) Get twisted

  2. Water has its maximum density at:
    a) 0 °C
    b) 100 °C
    c) 4 °C
    d) -4 °C

  3. Two rods, one of aluminum and the other made of steel, having the same initial length, are heated equally. Which of the following statements is true? (α_Al > α_Steel)
    a) Steel rod will expand more
    b) Aluminum rod will expand more
    c) Both rods will expand equally
    d) Both rods will contract

  4. The amount of heat required to change 1 kg of ice at 0°C completely into water at 0°C is called:
    a) Specific heat capacity of ice
    b) Specific heat capacity of water
    c) Latent heat of fusion of ice
    d) Latent heat of vaporization of water

  5. Two spheres made of the same material have radii in the ratio 1:2. Both are heated to the same temperature and allowed to cool in the same surroundings. The ratio of their initial rates of cooling will be:
    a) 1:2
    b) 1:4
    c) 1:1
    d) 2:1

  6. According to Wien's displacement law, if the temperature of a blackbody doubles, the wavelength corresponding to maximum emission (λm) becomes:
    a) Double
    b) Half
    c) Four times
    d) One-fourth

  7. In which mode of heat transfer is the medium NOT required?
    a) Conduction
    b) Convection
    c) Radiation
    d) Both Conduction and Convection

  8. A good absorber of heat is also a:
    a) Poor emitter
    b) Good emitter
    c) Good reflector
    d) Poor conductor

  9. Two identical rods are connected in series. Their thermal conductivities are K₁ and K₂. The thermal conductivity of the combination is:
    a) (K₁ + K₂) / 2
    b) K₁ + K₂
    c) 2 K₁ K₂ / (K₁ + K₂)
    d) K₁ K₂ / (K₁ + K₂)

  10. The SI unit of coefficient of thermal conductivity (K) is:
    a) W m K⁻¹
    b) W m⁻¹ K⁻¹
    c) J kg⁻¹ K⁻¹
    d) J mol⁻¹ K⁻¹

Answers to MCQs:

  1. b) Aluminum expands more, forcing the strip to bend with aluminum on the longer, outer (convex) side.
  2. c) Due to anomalous expansion.
  3. b) Since ΔL = α L₀ ΔT and L₀, ΔT are the same, the rod with higher α (Aluminum) expands more.
  4. c) Definition of Latent Heat of Fusion.
  5. c) Rate of cooling (dT/dt) depends on (Area/Mass) for bodies of same material cooled by radiation/convection. Rate of heat loss ∝ Area (A = 4πr²). Mass ∝ Volume (V = 4/3 πr³). So, Rate ∝ A/m ∝ r²/r³ = 1/r. Rate of cooling ∝ (Rate of heat loss) / (m * s) ∝ (eσAT⁴) / (ms). For radiation, Rate ∝ A/m ∝ 1/r. Thus, ratio of rates of cooling = r₂/r₁ = 2/1. Correction: Let's re-evaluate using Newton's Law or Stefan's Law. Rate of heat loss dQ/dt ∝ Area (r²). Rate of cooling dT/dt = (dQ/dt) / (ms). Mass m ∝ Volume (r³). Specific heat s is same. So, dT/dt ∝ (r²) / (r³) = 1/r. Ratio of initial rates of cooling = (1/r₁) / (1/r₂) = r₂/r₁ = 2/1. Let me recheck the common interpretation. Rate of heat loss P = eσA(T⁴ - T₀⁴). Rate of cooling dT/dt = P/(ms) = eσA(T⁴ - T₀⁴) / (ρVs). Since e, σ, T, T₀, ρ, s are same, dT/dt ∝ A/V ∝ r²/r³ = 1/r. So the ratio of initial rates of cooling (Body 1 / Body 2) = (1/r₁) / (1/r₂) = r₂/r₁ = 2/1. Let's assume the question means ratio of rates of heat loss. Rate of heat loss ∝ Area ∝ r². Ratio = r₁²/r₂² = (1/2)² = 1:4. If it means rate of cooling (dT/dt), it's 2:1. Often, "rate of cooling" implies dT/dt. Let me assume rate of heat loss was intended, as 1:4 is a common option structure. Reconsidering standard problems: Rate of cooling dT/dt = [eσA(T⁴-T₀⁴)] / [mc]. dT/dt ∝ A/m ∝ r²/r³ = 1/r. Ratio of rates of cooling = (dT/dt)₁ / (dT/dt)₂ = r₂/r₁ = 2/1. Okay, the ratio of rates of cooling is 2:1. Let me check if I made a mistake in the options or the standard interpretation. Let's re-read the question - "ratio of their initial rates of cooling". This is dT/dt. It is proportional to 1/r. So the ratio is r₂/r₁ = 2/1. Option (d). Self-correction: Let me re-verify the dependency. dT/dt = -(eAσ/ms)(T⁴-T₀⁴). dT/dt ∝ A/m = (4πr²)/(ρ * 4/3 πr³) = 3/(ρr). Since ρ is same, dT/dt ∝ 1/r. Ratio = (dT/dt)₁ / (dT/dt)₂ = r₂/r₁ = 2/1. Yes, the ratio is 2:1.
  6. b) From Wien's Law, λm T = constant. If T doubles, λm must become half.
  7. c) Radiation uses electromagnetic waves.
  8. b) According to Kirchhoff's Law.
  9. c) For series combination, R_eq = R₁ + R₂ => (L+L)/(K_eq * A) = L/(K₁A) + L/(K₂A). => 2/K_eq = 1/K₁ + 1/K₂ = (K₁ + K₂)/(K₁ K₂). => K_eq = 2 K₁ K₂ / (K₁ + K₂).
  10. b) From H = K A (ΔT/L), K = H * L / (A * ΔT). Units = (J/s) * m / (m² * K) = J s⁻¹ m⁻¹ K⁻¹ = W m⁻¹ K⁻¹.

Correction for Q5: The ratio of initial rates of cooling (dT/dt) is proportional to 1/r. Therefore, the ratio is r₂/r₁ = 2/1. The correct option is (d).

Corrected Answers:

  1. b
  2. c
  3. b
  4. c
  5. d
  6. b
  7. c
  8. b
  9. c
  10. b

Make sure you understand the derivation of formulas, especially for thermal resistance combinations and the application of Stefan-Boltzmann and Wien's laws. Good luck with your preparation!

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