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Physics (SSC, Railway, Police & All State exam)Chapter Unit

Heat

Introduction to Heat

  1. Definition:

    • Heat is a form of energy transferred between two bodies due to a temperature difference.
    • SI Unit: Joule (JJJ).
    • Common Unit: Calorie (calcalcal), where 1 cal=4.186 J1 \, cal = 4.186 \, J1cal=4.186J.
  2. Temperature:

    • A measure of the average kinetic energy of particles in a substance.
    • SI Unit: Kelvin (KKK).
    • Other Units: Celsius (∘C^\circ C∘C), Fahrenheit (∘F^\circ F∘F).
    • Conversion: T(K)=T(∘C)+273.15T(K) = T(^{\circ}C) + 273.15T(K)=T(∘C)+273.15 T(∘F)=95T(∘C)+32T(^{\circ}F) = \frac{9}{5} T(^{\circ}C) + 32T(∘F)=59​T(∘C)+32

Modes of Heat Transfer

  1. Conduction:

    • Transfer of heat through a substance without the movement of particles.
    • Governing Law: Q=kAΔTd⋅tQ = \frac{kA \Delta T}{d} \cdot tQ=dkAΔT​⋅t
      • QQQ: Heat transferred.
      • kkk: Thermal conductivity.
      • AAA: Cross-sectional area.
      • ΔT\Delta TΔT: Temperature difference.
      • ddd: Thickness.
      • ttt: Time.
  2. Convection:

    • Transfer of heat through the movement of fluids (liquids or gases).
    • Types:
      • Natural Convection: Due to density differences.
      • Forced Convection: Using external forces like fans or pumps.
  3. Radiation:

    • Transfer of heat in the form of electromagnetic waves.
    • Does not require a medium.
    • Governing Law (Stefan-Boltzmann Law): P=σAT4P = \sigma A T^4P=σAT4
      • σ\sigmaσ: Stefan-Boltzmann constant (5.67×10−8 W/m2K45.67 \times 10^{-8} \, W/m^2 K^45.67×10−8W/m2K4).
      • AAA: Surface area.
      • TTT: Absolute temperature in Kelvin.

Specific Heat Capacity

  1. Definition:

    • The amount of heat required to raise the temperature of 1 kg1 \, kg1kg of a substance by 1 K1 \, K1K.
    • Formula: Q=mcΔTQ = mc\Delta TQ=mcΔT
      • mmm: Mass.
      • ccc: Specific heat capacity.
      • ΔT\Delta TΔT: Change in temperature.
  2. Units:

    • SI Unit: J/(kg K)J/(kg \, K)J/(kgK).
  3. Applications:

    • Water has a high specific heat capacity, making it ideal for cooling systems.

Latent Heat

  1. Definition:

    • The amount of heat required to change the state of a unit mass of a substance without changing its temperature.
    • Formula: Q=mLQ = mLQ=mL
      • mmm: Mass.
      • LLL: Latent heat.
  2. Types:

    • Latent Heat of Fusion: Heat required to convert a solid into a liquid.
    • Latent Heat of Vaporization: Heat required to convert a liquid into a gas.

Numerical Example

  1. Example 1: Calculate the heat required to raise the temperature of 2 kg2 \, kg2kg of water from 20∘C20^\circ C20∘C to 80∘C80^\circ C80∘C. (cwater=4200 J/kg Kc_{\text{water}} = 4200 \, J/kg \, Kcwater​=4200J/kgK)

    • Formula: Q=mcΔTQ = mc\Delta TQ=mcΔT
    • Substituting values: Q=2⋅4200⋅(80−20)Q = 2 \cdot 4200 \cdot (80 - 20)Q=2⋅4200⋅(80−20) Q=2⋅4200⋅60=504,000 JQ = 2 \cdot 4200 \cdot 60 = 504,000 \, JQ=2⋅4200⋅60=504,000J
  2. Example 2: Calculate the heat required to convert 1 kg1 \, kg1kg of ice at 0∘C0^\circ C0∘C to water at 0∘C0^\circ C0∘C. (Lf=334,000 J/kgL_f = 334,000 \, J/kgLf​=334,000J/kg)

    • Formula: Q=mLQ = mLQ=mL
    • Substituting values: Q=1⋅334,000=334,000 JQ = 1 \cdot 334,000 = 334,000 \, JQ=1⋅334,000=334,000J

Thermal Expansion

  1. Definition:

    • When a substance is heated, its dimensions increase due to an increase in the average kinetic energy of its particles.
  2. Types of Expansion:

    • Linear Expansion:
      • Change in length: ΔL=L0αΔT\Delta L = L_0 \alpha \Delta TΔL=L0​αΔT
        • α\alphaα: Coefficient of linear expansion (K−1K^{-1}K−1).
        • L0L_0L0​: Original length.
    • Area Expansion:
      • Change in area: ΔA=A0βΔT\Delta A = A_0 \beta \Delta TΔA=A0​βΔT
        • β=2α\beta = 2\alphaβ=2α: Coefficient of area expansion.
        • A0A_0A0​: Original area.
    • Volume Expansion:
      • Change in volume: ΔV=V0γΔT\Delta V = V_0 \gamma \Delta TΔV=V0​γΔT
        • γ=3α\gamma = 3\alphaγ=3α: Coefficient of volume expansion.
        • V0V_0V0​: Original volume.
  3. Applications:

    • Gaps in railway tracks to accommodate expansion.
    • Expansion joints in bridges.

Laws of Thermodynamics

  1. Zeroth Law of Thermodynamics:

    • If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
    • Forms the basis for temperature measurement.
  2. First Law of Thermodynamics:

    • Law of energy conservation: ΔQ=ΔU+W\Delta Q = \Delta U + WΔQ=ΔU+W
      • ΔQ\Delta QΔQ: Heat added to the system.
      • ΔU\Delta UΔU: Change in internal energy.
      • WWW: Work done by the system.
  3. Second Law of Thermodynamics:

    • Heat cannot spontaneously flow from a colder body to a hotter body without external work.
    • Clausius Statement: It is impossible to construct a device that operates in a cycle and transfers heat from a cold body to a hot body without external energy input.
    • Kelvin-Planck Statement: No engine can convert all the heat it absorbs into work.
  4. Third Law of Thermodynamics:

    • As the temperature of a system approaches absolute zero, the entropy of the system approaches a constant minimum.

Heat Engines

  1. Definition:

    • A heat engine is a device that converts heat energy into mechanical work.
  2. Working Principle:

    • Operates between two heat reservoirs: a hot reservoir and a cold reservoir.
    • Efficiency (η\etaη): η=WQH=1−QCQH\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}η=QH​W​=1−QH​QC​​
      • QHQ_HQH​: Heat absorbed from the hot reservoir.
      • QCQ_CQC​: Heat rejected to the cold reservoir.
  3. Carnot Engine:

    • An ideal heat engine with maximum efficiency.
    • Efficiency: ηCarnot=1−TCTH\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}ηCarnot​=1−TH​TC​​
      • THT_HTH​: Temperature of the hot reservoir.
      • TCT_CTC​: Temperature of the cold reservoir.

Specific Heat Capacity at Constant Pressure and Volume

  1. Molar Specific Heat:

    • At constant volume (CVC_VCV​): CV=(∂U∂T)VC_V = \left( \frac{\partial U}{\partial T} \right)_VCV​=(∂T∂U​)V​
    • At constant pressure (CPC_PCP​): CP=(∂U∂T)PC_P = \left( \frac{\partial U}{\partial T} \right)_PCP​=(∂T∂U​)P​
  2. Relation Between CPC_PCP​ and CVC_VCV​:

    • For an ideal gas: CP−CV=RC_P - C_V = RCP​−CV​=R
      • RRR: Universal gas constant.
  3. Applications:

    • Gases behave differently under constant volume and pressure.

Thermal Conductivity

  1. Thermal Resistance:

    • Reciprocal of thermal conductivity.
    • Formula for heat flow: Q=kAΔTd⋅tQ = \frac{kA\Delta T}{d} \cdot tQ=dkAΔT​⋅t
      • kkk: Thermal conductivity.
      • ddd: Thickness of the material.
  2. Composite Slabs:

    • Heat flow through multiple layers: ΔTRtotal=Q\frac{\Delta T}{R_{\text{total}}} = QRtotal​ΔT​=Q
      • RtotalR_{\text{total}}Rtotal​: Sum of thermal resistances.

Numerical Examples

  1. Example 1: A metal rod of length 2 m2 \, m2m expands by 1.6 mm1.6 \, mm1.6mm when heated from 20∘C20^\circ C20∘C to 100∘C100^\circ C100∘C. Find the coefficient of linear expansion (α\alphaα).

    • Formula: ΔL=L0αΔT\Delta L = L_0 \alpha \Delta TΔL=L0​αΔT
    • Substituting values: 1.6×10−3=2⋅α⋅(100−20)1.6 \times 10^{-3} = 2 \cdot \alpha \cdot (100 - 20)1.6×10−3=2⋅α⋅(100−20) α=1.6×10−32⋅80=1×10−5 K−1\alpha = \frac{1.6 \times 10^{-3}}{2 \cdot 80} = 1 \times 10^{-5} \, K^{-1}α=2⋅801.6×10−3​=1×10−5K−1
  2. Example 2: A Carnot engine operates between TH=500 KT_H = 500 \, KTH​=500K and TC=300 KT_C = 300 \, KTC​=300K. Find its efficiency.

    • Formula: η=1−TCTH\eta = 1 - \frac{T_C}{T_H}η=1−TH​TC​​
    • Substituting values: η=1−300500=1−0.6=0.4 or 40%\eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 \, \text{or} \, 40\%η=1−500300​=1−0.6=0.4or40%

Heat Capacity and Calorimetry

  1. Heat Capacity:

    • The amount of heat required to raise the temperature of a substance by 1 degree.
    • Formula: C=mcC = mcC=mc
      • CCC: Heat capacity.
      • mmm: Mass of the substance.
      • ccc: Specific heat capacity.
  2. Calorimetry:

    • The study of heat transfer during physical or chemical changes.

    • Principle of Calorimetry:

      • Heat lost by a hot body = Heat gained by a cold body. Qlost=QgainedQ_{\text{lost}} = Q_{\text{gained}}Qlost​=Qgained​
    • Formula for heat exchange: m1c1(T1−Tf)=m2c2(Tf−T2)m_1c_1(T_1 - T_f) = m_2c_2(T_f - T_2)m1​c1​(T1​−Tf​)=m2​c2​(Tf​−T2​)

      • T1T_1T1​, T2T_2T2​: Initial temperatures of bodies.
      • TfT_fTf​: Final equilibrium temperature.

Change of State and Phase Diagram

  1. Phase Diagram:

    • A graph showing different phases of a substance as a function of temperature and pressure.
    • Key Points:
      • Triple Point: All three phases coexist.
      • Critical Point: Above this, the substance exists as a supercritical fluid.
  2. Cooling Curve:

    • Represents how the temperature of a substance changes as it loses heat.
    • Plateaus indicate phase changes (e.g., melting, boiling).
  3. Supercooling and Superheating:

    • Supercooling: A liquid is cooled below its freezing point without solidifying.
    • Superheating: A liquid is heated above its boiling point without vaporizing.

Thermal Efficiency and Insulation

  1. Thermal Insulators:

    • Materials that resist the flow of heat.
    • Examples: Wood, glass wool, and Styrofoam.
  2. Thermal Efficiency:

    • Efficiency of a heat transfer process: η=Useful heat transferTotal heat input⋅100\eta = \frac{\text{Useful heat transfer}}{\text{Total heat input}} \cdot 100η=Total heat inputUseful heat transfer​⋅100
  3. Applications:

    • Thermos flasks use insulation to reduce heat exchange.
    • Double-glazed windows reduce heat loss in buildings.

Blackbody Radiation

  1. Blackbody:

    • An idealized object that absorbs all incident radiation and emits radiation based on its temperature.
  2. Planck’s Law:

    • Describes the spectral distribution of radiation emitted by a blackbody.
  3. Stefan-Boltzmann Law:

    • Total energy radiated per unit surface area is proportional to the fourth power of temperature: P=σAT4P = \sigma A T^4P=σAT4
  4. Wien’s Displacement Law:

    • Wavelength at which the intensity of radiation is maximum: λmaxT=b\lambda_{\text{max}} T = bλmax​T=b
      • bbb: Wien’s constant (2.897×10−3 m⋅K2.897 \times 10^{-3} \, m \cdot K2.897×10−3m⋅K).

Heat Engines and Refrigerators

  1. Heat Engines:

    • Convert heat energy into mechanical work.
    • Efficiency: η=WQH=1−QCQH\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}η=QH​W​=1−QH​QC​​
  2. Refrigerators:

    • Transfer heat from a colder region to a hotter region.
    • Coefficient of Performance (COP): COP=QCWCOP = \frac{Q_C}{W}COP=WQC​​
      • QCQ_CQC​: Heat extracted from the cold reservoir.
      • WWW: Work done.

Numerical Examples

  1. Example 1: A 500 g500 \, g500g aluminum block (cAl=900 J/kg⋅Kc_{\text{Al}} = 900 \, J/kg \cdot KcAl​=900J/kg⋅K) is heated from 20∘C20^\circ C20∘C to 100∘C100^\circ C100∘C. Calculate the heat absorbed.

    • Formula: Q=mcΔTQ = mc\Delta TQ=mcΔT
    • Substituting values: Q=0.5⋅900⋅(100−20)Q = 0.5 \cdot 900 \cdot (100 - 20)Q=0.5⋅900⋅(100−20) Q=0.5⋅900⋅80=36,000 JQ = 0.5 \cdot 900 \cdot 80 = 36,000 \, JQ=0.5⋅900⋅80=36,000J
  2. Example 2: A refrigerator extracts 500 J500 \, J500J of heat from its interior using 100 J100 \, J100J of work. Find its COP.

    • Formula: COP=QCWCOP = \frac{Q_C}{W}COP=WQC​​
    • Substituting values: COP=500100=5COP = \frac{500}{100} = 5COP=100500​=5
  3. Example 3: Calculate the peak wavelength of radiation emitted by a blackbody at T=6000 KT = 6000 \, KT=6000K.

    • Formula: λmaxT=b\lambda_{\text{max}} T = bλmax​T=b
      • Substituting values: λmax=bT=2.897×10−36000\lambda_{\text{max}} = \frac{b}{T} = \frac{2.897 \times 10^{-3}}{6000}λmax​=Tb​=60002.897×10−3​ λmax=4.83×10−7 m=483 nm\lambda_{\text{max}} = 4.83 \times 10^{-7} \, m = 483 \, nmλmax​=4.83×10−7m=483nm

Recap: Key Points to Remember

  • Heat transfer occurs via conduction, convection, or radiation.
  • Thermal expansion is significant in practical applications like bridges and railway tracks.
  • Laws of thermodynamics govern heat, energy, and work in physical processes.
  • Heat engines and refrigerators operate based on thermodynamic principles.

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