Convection & Radiation

Difficulty:
Convection

Definition: Heat transfer by actual movement of heated material (fluids - liquids & gases).

Types of Convection

  • Natural Convection: Due to density differences (hot air rises)
  • Forced Convection: Using external means (fans, pumps)

Examples: Sea breeze, Land breeze, Boiling water, Heating rooms

Radiation

Definition: Heat transfer through electromagnetic waves without requiring a medium.

Stefan-Boltzmann Law

P = εσAT⁴

Where: P = Power radiated (W), ε = Emissivity (0-1), σ = Stefan's constant (5.67×10⁻⁸ W/m²K⁴), A = Surface area (m²), T = Temperature (K)

Convection Currents

Hot
Cold
Convection Current
Hot air rises
Cold air sinks

Hot fluid rises → Cools down → Sinks back → Heats again → Cycle continues

Properties of Heat Transfer Modes

Property Conduction Convection Radiation
Medium Required Solid (best) Fluid (liquid/gas) No medium (vacuum OK)
Mechanism Molecular collisions Bulk fluid motion EM waves
Speed Slow in non-metals Moderate Speed of light
Temperature Dependence Linear Complex T⁴ (Stefan's Law)
Examples Metal spoon in hot tea Sea breeze, Heating room Sun's heat, Heat from fire

Laws of Thermodynamics

Difficulty:
Zeroth Law

Statement: If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.

Importance: Defines temperature and allows temperature measurement.

If A = B and B = C, then A = C
First Law

Statement: Energy cannot be created or destroyed, only converted from one form to another.

First Law Equation

ΔU = Q - W

Where: ΔU = Change in internal energy, Q = Heat added to system, W = Work done by system

For cyclic process: ΔU = 0 ⇒ Q = W
Second Law

Kelvin-Planck Statement: No heat engine can convert all heat into work.

Clausius Statement: Heat cannot flow from colder to hotter body spontaneously.

Efficiency η = 1 - (T₂/T₁) ≤ 1

NEET Important: Thermodynamic Processes

Isothermal (ΔT = 0)

• Temperature constant
• ΔU = 0
• Q = W
• Boyle's Law: PV = constant

Adiabatic (Q = 0)

• No heat exchange
• ΔU = -W
• PV^γ = constant
• Fast processes

Isochoric (ΔV = 0)

• Volume constant
• W = 0
• ΔU = Q
• Gay-Lussac's Law

Isobaric (ΔP = 0)

• Pressure constant
• W = PΔV
• Charles' Law: V/T = constant

PV Diagram for Different Processes

Volume (V)
Pressure (P)
Isothermal
Adiabatic
Isobaric
Isochoric

Different thermodynamic processes on Pressure-Volume diagram

Gas Laws & Kinetic Theory

Difficulty:
Ideal Gas Law

Equation of State

PV = nRT = NkT

Where: P = Pressure (Pa), V = Volume (m³), n = Number of moles, R = Universal gas constant (8.314 J/mol·K), T = Temperature (K), N = Number of molecules, k = Boltzmann constant (1.38×10⁻²³ J/K)

Assumptions of Kinetic Theory:

  1. Gas molecules are point masses
  2. No intermolecular forces except during collisions
  3. Collisions are perfectly elastic
  4. Molecular motion is random
Kinetic Energy & Speeds

Important Formulas

Average KE per molecule = (3/2)kT
RMS speed: v_rms = √(3RT/M) = √(3kT/m)
Average speed: v_avg = √(8RT/πM)
Most probable speed: v_mp = √(2RT/M)

Relation: v_mp : v_avg : v_rms = 1 : 1.128 : 1.224

Gas Laws Summary

Law Statement Equation Constant Parameter
Boyle's Law Pressure inversely proportional to volume at constant temperature P₁V₁ = P₂V₂ Temperature (T)
Charles' Law Volume directly proportional to temperature at constant pressure V₁/T₁ = V₂/T₂ Pressure (P)
Gay-Lussac's Law Pressure directly proportional to temperature at constant volume P₁/T₁ = P₂/T₂ Volume (V)
Avogadro's Law Volume directly proportional to number of moles at constant T & P V₁/n₁ = V₂/n₂ T & P

NEET Gas Laws Problem

Problem Statement

Q: Calculate the rms speed of oxygen molecules at 27°C. Given: Molar mass of O₂ = 32 g/mol, R = 8.314 J/mol·K

NEET Answer Writing Strategy

Step 1: Convert units
Temperature to Kelvin: T = 27 + 273 = 300 K
Molar mass to kg/mol: M = 32 g/mol = 0.032 kg/mol

Step 2: Write formula
v_rms = √(3RT/M)

Step 3: Substitute values
v_rms = √[(3 × 8.314 × 300) / 0.032]

Step 4: Calculate stepwise
3 × 8.314 = 24.942
24.942 × 300 = 7482.6
7482.6 ÷ 0.032 = 233831.25
√233831.25 ≈ 483.56 m/s

Step 5: Final answer with unit
v_rms ≈ 484 m/s

Black Body Radiation

Difficulty:
Black Body Definition

Black Body: An ideal body that absorbs all radiation incident on it and emits radiation at all wavelengths.

Properties

  • Perfect absorber (absorptivity = 1)
  • Perfect emitter (emissivity = 1)
  • Emissive power depends only on temperature
  • Real bodies: ε < 1 (gray bodies)
Radiation Laws

Wien's Displacement Law

λ_max T = b (constant)

Where: λ_max = wavelength at maximum emission, T = Temperature (K), b = Wien's constant (2.898×10⁻³ m·K)

Stefan-Boltzmann Law: P = σAT⁴

Application: Determines temperature of stars from their color

Black Body Radiation Spectrum

Wavelength (λ)
Intensity
T₁ (Hot)
T₂ (Cooler)
λ_max ∝ 1/T
(Wien's Law)

Hotter bodies emit more radiation at shorter wavelengths (blue shift)

Thermal Stress Applications

Difficulty:
Thermal Stress Concept

Thermal Stress: Stress developed when thermal expansion or contraction is constrained.

Thermal Stress Formula

Stress = Y × α × ΔT

Where: Y = Young's modulus (N/m²), α = Coefficient of linear expansion (/K), ΔT = Temperature change (K)

Force developed: F = A × Stress = A × Y × α × ΔT

Practical Applications

Engineering Applications

  1. Railway Tracks: Gaps left for expansion
  2. Bridges: Expansion joints provided
  3. Bimetallic Strips: Used in thermostats
  4. Power Lines: Sag allowed in summer
  5. Glassware: Pyrex has low α to prevent cracking

Coefficient of Linear Expansion (α) for Common Materials

Material α (×10⁻⁶ /K) Application Special Feature
Invar (Fe-Ni) 1.2 Pendulum clocks, Precision instruments Very low expansion
Glass (Pyrex) 3.3 Laboratory glassware Low expansion, thermal shock resistant
Steel 12 Construction, Railways Moderate expansion
Copper 17 Electrical wires, Plumbing High expansion
Aluminum 23 Aircraft, Cooking utensils Very high expansion

Thermometry & Instruments

Difficulty:
Temperature Scales

Scale Conversions

Celsius to Kelvin: K = °C + 273.15
Celsius to Fahrenheit: °F = (9/5)°C + 32
Fahrenheit to Celsius: °C = (5/9)(°F - 32)

Fixed Points:

  • Ice point: 0°C, 32°F, 273.15 K
  • Steam point: 100°C, 212°F, 373.15 K
Thermometers

Types of Thermometers

  1. Liquid-in-glass: Mercury/alcohol thermometer
  2. Bimetallic: Two metals with different α
  3. Gas Thermometer: Constant volume gas thermometer (most accurate)
  4. Resistance Thermometer: Platinum resistance thermometer
  5. Thermocouple: Based on Seebeck effect
  6. Pyrometer: For very high temperatures

Bihar Board Thermometry Question

5-Mark Question Pattern

Q: Describe the construction and working of a constant volume gas thermometer. Why is it considered the most accurate thermometer?

Bihar Board Answer Format (5 marks)

Introduction (1 mark)
A constant volume gas thermometer is based on the pressure-temperature relationship of gases at constant volume. According to Gay-Lussac's law, at constant volume, pressure of a gas is directly proportional to its absolute temperature.

Construction (2 marks)
It consists of:
1. A bulb containing gas (hydrogen/helium)
2. A mercury manometer to measure pressure
3. A capillary tube connecting bulb to manometer
4. A constant volume marker
5. A water bath for temperature control

Working (1 mark)
The bulb is placed in the substance whose temperature is to be measured. Mercury level is adjusted to keep gas volume constant. Pressure difference is measured using manometer. Temperature is calculated using: T = (P/P₀) × T₀, where P₀ and T₀ are pressure and temperature at ice point.

Accuracy Reason (1 mark)
It is most accurate because:
1. Gases have large expansion coefficients
2. They follow ideal gas law closely
3. Measurements are highly reproducible
4. Can measure very low and high temperatures

NEET Advanced Concepts

Difficulty:
Advanced Formulas

Must-Know Formulas for NEET

Mayer's Relation: C_p - C_v = R
Adiabatic Bulk Modulus: B_s = γP
Speed of sound in gas: v = √(γRT/M)
Mean free path: λ = 1/(√2 π d² n)
Degree of freedom: f = 3 (monatomic), 5 (diatomic), 6 (polyatomic)
Equipartition energy: U = (f/2) nRT
Degree of Freedom
Gas Type Translational Rotational Vibrational* Total (f) C_v C_p γ = C_p/C_v
Monatomic 3 0 0 3 (3/2)R (5/2)R 5/3 = 1.67
Diatomic 3 2 0* 5 (5/2)R (7/2)R 7/5 = 1.4
Polyatomic 3 3 0* 6 3R 4R 4/3 = 1.33

*Vibrational degrees activated at high temperatures

NEET High-Yield Topics

Heat Engine & Refrigerator

• Efficiency η = 1 - Q₂/Q₁ = 1 - T₂/T₁ (Carnot)
• Coefficient of performance (refrigerator): COP = Q₂/W = Q₂/(Q₁ - Q₂) = T₂/(T₁ - T₂)
• Real engines have η < Carnot efficiency

Thermal Conductivity Combination

• Series: R_total = R₁ + R₂, k_eq = L_total/(A·R_total)
• Parallel: 1/R_total = 1/R₁ + 1/R₂, k_eq = (k₁A₁ + k₂A₂)/(A₁ + A₂)
• Thermal resistance R = L/(kA)

Adiabatic Processes

• PV^γ = constant
• TV^(γ-1) = constant
• P^(1-γ)T^γ = constant
• Work done: W = (P₁V₁ - P₂V₂)/(γ-1) = nR(T₁ - T₂)/(γ-1)

Complete Question Bank

Practice Level:

Section A: Very Short Answer (1 mark each)

NEET Pattern

Q1: What is the SI unit of thermal conductivity?

Answer: W/m·K (Watt per meter per Kelvin)

Q2: State Wien's displacement law.

Answer: The product of wavelength at which maximum emission occurs and absolute temperature is constant. λ_max T = constant

Q3: What is the value of Stefan's constant?

Answer: σ = 5.67 × 10⁻⁸ W/m²K⁴

Section B: Short Answer (2 marks each)

Bihar Board Pattern

Q4: Differentiate between conduction and convection.

Expected Answer:
Conduction: Heat transfer through molecular collisions without bulk motion of material. Occurs in solids.
Convection: Heat transfer by actual movement of heated material. Occurs in fluids (liquids and gases).

Q5: Define coefficient of thermal conductivity. Give its SI unit.

Expected Answer:
Coefficient of thermal conductivity (k) is defined as the amount of heat flowing per second through a slab of unit cross-sectional area and unit thickness when temperature difference between opposite faces is 1 K.
SI Unit: W/m·K (Watt per meter per Kelvin)

Section C: Numerical Problems (3 marks each)

Calculation Based

Q6: Calculate the amount of heat radiated per second by a black body of surface area 0.1 m² at 127°C. (σ = 5.67 × 10⁻⁸ W/m²K⁴)

Solution:
Given: A = 0.1 m², T = 127°C = 400 K, σ = 5.67 × 10⁻⁸ W/m²K⁴
For black body: ε = 1
Using Stefan's law: P = εσAT⁴
= 1 × 5.67 × 10⁻⁸ × 0.1 × (400)⁴
= 5.67 × 10⁻⁹ × 256 × 10⁸
= 5.67 × 256 × 10⁻¹
= 1451.52 × 10⁻¹ = 145.152 W
Answer: 145.15 W

Section D: Long Answer (5 marks each)

Theory Questions

Q7: State and explain the first law of thermodynamics. Apply it to isothermal and adiabatic processes.

Model Answer (5 marks):

Statement (1 mark): The first law of thermodynamics states that energy can neither be created nor destroyed, but can only be converted from one form to another. Mathematically: ΔU = Q - W

Explanation (1 mark): Where ΔU is change in internal energy, Q is heat added to the system, and W is work done by the system. The law is essentially a statement of conservation of energy applied to thermal processes.

Application to Isothermal Process (1.5 marks): In isothermal process, temperature remains constant (ΔT = 0). For ideal gas, internal energy depends only on temperature, so ΔU = 0. From first law: 0 = Q - W ⇒ Q = W. Thus, all heat added is converted into work done by the gas.

Application to Adiabatic Process (1.5 marks): In adiabatic process, there is no heat exchange (Q = 0). From first law: ΔU = 0 - W ⇒ ΔU = -W. Thus, work is done at the expense of internal energy, causing temperature to drop. For expansion (W > 0), ΔU < 0 so temperature decreases.

Answer Writing Mastery

Importance:

Bihar Board Answer Writing Strategy

For 5-mark Theory Questions:

Component Marks What to Include
Introduction & Definition 1 Define key terms, state principle/law
Explanation/Derivation 2 Step-by-step explanation, mathematical derivation if applicable
Diagram 1 Neat, labeled diagram with proper labeling
Conclusion/Application 1 Summary, real-world applications, examples

For Numerical Problems (3 marks):

  1. Step 1 (0.5 mark): Write given data with proper units
  2. Step 2 (0.5 mark): Write relevant formula
  3. Step 3 (1.5 marks): Step-by-step calculation with units
  4. Step 4 (0.5 mark): Final answer with proper unit and boxed

Common Mistakes to Avoid:

Conceptual Errors

• Confusing heat and temperature
• Mixing up CGS and SI units
• Forgetting temperature in Kelvin for gas laws

Calculation Errors

• Wrong unit conversions
• Incorrect powers of 10
• Forgetting π in calculations
• Rounding off too early

Presentation Errors

• Missing units in final answer
• Unlabeled diagrams
• Messy handwriting
• No proper steps shown

Time Management in Exam:

NEET (180 mins)

• 1 minute per question maximum
• First pass: Solve known questions
• Second pass: Attempt difficult ones
• Last 15 mins: Review marked questions

Bihar Board (3 hours)

• Allocate time based on marks
• 1 mark = 1 minute approximately
• Start with Section A (easy)
• Leave time for diagram drawing
• Last 30 mins: Revision

Pro Tips for Maximum Marks

  1. Underline key terms in your answers
  2. Draw diagrams even if not asked - they help in explanation
  3. Write formulas before calculations
  4. Show all steps - examiners give marks for correct steps even if final answer is wrong
  5. Use proper units throughout the answer
  6. Box your final answer for easy identification
  7. Keep your handwriting neat and legible
  8. Leave margins for possible additions

संवहन और विकिरण

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