Topics Covered in Part 1

  • Temperature & Heat
  • Thermal Expansion
  • Specific Heat Capacity
  • Calorimetry
  • Latent Heat
  • Heat Transfer Modes
  • Heat Conduction
  • Practice Questions

Quick Tips

• Focus on conceptual understanding
• Practice numerical problems daily
• Revise formulas regularly
• Draw diagrams for theory questions
• Time management in exams

Temperature and Heat

What is Temperature?

Temperature is a physical quantity that measures the degree of hotness or coldness of an object. It's a measure of the average kinetic energy of the molecules in a substance.

Key Points

  • Temperature determines the direction of heat flow (from higher to lower temperature)
  • It is measured using thermometers
  • SI unit: Kelvin (K)
  • Other units: Celsius (°C), Fahrenheit (°F)

What is Heat?

Heat is a form of energy that flows from a body at higher temperature to a body at lower temperature when they are brought into thermal contact.

Important Formulas

Q = m × c × ΔT

Where: Q = Heat energy (J), m = mass (kg), c = specific heat capacity (J/kg·K), ΔT = temperature change (K)

1 calorie = 4.186 Joules
1 kcal = 4186 Joules

Heat Transfer Direction

Hot Body
80°C
Higher Temperature
→
Heat Flow
Cold Body
20°C
Lower Temperature

Heat always flows from higher temperature to lower temperature

Temperature vs Heat Comparison

Aspect Temperature Heat
Definition Measure of hotness/coldness Energy in transit due to temperature difference
SI Unit Kelvin (K) Joule (J)
Scalar/Vector Scalar quantity Scalar quantity
Device Thermometer Calorimeter
Dependence Depends on average KE of molecules Depends on mass, specific heat, and ΔT

Thermal Expansion

What is Thermal Expansion?

Most substances expand when heated and contract when cooled. This phenomenon is called thermal expansion.

Types of Thermal Expansion

  1. Linear Expansion: Expansion in length
  2. Area Expansion: Expansion in area
  3. Volume Expansion: Expansion in volume

Formulas for Thermal Expansion

ΔL = α × L₀ × ΔT

Linear Expansion: ΔL = change in length, α = coefficient of linear expansion, L₀ = original length, ΔT = temperature change

ΔA = β × A₀ × ΔT (β = 2α)

Area Expansion: ΔA = change in area, β = coefficient of area expansion, A₀ = original area

ΔV = γ × V₀ × ΔT (γ = 3α)

Volume Expansion: ΔV = change in volume, γ = coefficient of volume expansion, V₀ = original volume

Thermal Expansion of Solids

Original Rod at T₁
Expanded Rod at T₂ (T₂ > T₁)
↓
ΔL = L₂ - L₁

When temperature increases from T₁ to T₂, length increases from L₁ to L₂

NEET Important Points

  • Water shows anomalous expansion between 0°C to 4°C
  • For isotropic solids: γ = 3α
  • Invar (Fe-Ni alloy) has very low α (used in pendulum clocks)
  • Bimetallic strip works on different α of two metals

Practice Question on Thermal Expansion

Q: A steel rail of length 5 m is clamped at both ends at 20°C. Calculate the thermal stress developed in the rail if the temperature rises to 50°C. (α for steel = 1.2 × 10⁻⁵ /°C, Young's modulus Y = 2 × 10¹¹ N/m²)

Answer Writing Guidelines

  1. Step 1: Write given data with proper units
  2. Step 2: Write relevant formula
  3. Step 3: Show step-by-step calculation
  4. Step 4: Write final answer with unit
  5. Step 5: Mention important points if any

Model Answer

Given:
L₀ = 5 m, T₁ = 20°C, T₂ = 50°C
α = 1.2 × 10⁻⁵ /°C, Y = 2 × 10¹¹ N/m²

Step 1: Calculate temperature change
ΔT = T₂ - T₁ = 50 - 20 = 30°C

Step 2: If expansion is prevented, thermal stress develops
Formula: Thermal stress = Y × α × ΔT

Step 3: Calculation
Thermal stress = (2 × 10¹¹) × (1.2 × 10⁻⁵) × 30
= 2 × 1.2 × 30 × 10¹¹⁻⁵
= 72 × 10⁶ N/m²
= 7.2 × 10⁷ N/m² (or Pa)

Step 4: Final Answer
Thermal stress developed = 7.2 × 10⁷ Pa

Note: Since the rail is clamped, expansion is prevented, leading to compressive stress.

Specific Heat Capacity

Definition

Specific Heat Capacity (c) of a substance is defined as the amount of heat required to raise the temperature of 1 kg of the substance by 1 K (or 1°C).

Formula

c = Q / (m × ΔT)

Where: c = specific heat capacity (J/kg·K), Q = heat supplied (J), m = mass (kg), ΔT = temperature change (K)

Q = m × c × ΔT

Specific Heat Capacity of Common Substances

Substance Specific Heat Capacity (J/kg·K) Special Feature
Water 4186 Very high - good coolant
Ice 2100 About half of liquid water
Aluminum 900 High for a metal
Iron 450 Moderate value
Copper 385 Low - heats up quickly
Lead 130 Very low - poor conductor of heat

Why Water has High Specific Heat?

Water has unusually high specific heat capacity because:

  1. Hydrogen bonding between water molecules requires more energy to break
  2. Water molecules can absorb heat without much temperature rise
  3. This property helps in temperature regulation of living organisms and climate moderation

Practice Question on Specific Heat

Q: Calculate the amount of heat required to raise the temperature of 2 kg of water from 20°C to 80°C. (Specific heat capacity of water = 4186 J/kg·K)

Model Answer

Given:
m = 2 kg, T₁ = 20°C, T₂ = 80°C, c = 4186 J/kg·K

Step 1: Temperature change
ΔT = T₂ - T₁ = 80 - 20 = 60°C = 60 K (since 1°C = 1 K change)

Step 2: Formula: Q = m × c × ΔT

Step 3: Calculation
Q = 2 × 4186 × 60
= 2 × 4186 × 60
= 502320 J
= 5.0232 × 10⁵ J

Step 4: In calories (optional)
Since 1 cal = 4.186 J
Q = 502320 / 4.186 ≈ 120000 cal = 120 kcal

Final Answer: Heat required = 5.02 × 10⁵ J or 120 kcal

Calorimetry

Principle of Calorimetry

When two bodies at different temperatures are brought into thermal contact, heat flows from the hotter body to the colder body until thermal equilibrium is reached.

Calorimetry Equation

Heat lost by hot body = Heat gained by cold body
m₁c₁(T₁ - T) = m₂c₂(T - T₂) + Heat lost to surroundings (if any)

Where: T = final equilibrium temperature

Calorimeter Setup

Hot Water
80°C
Mass = m₁
+ →
Cold Water
20°C
Mass = m₂
=
Mixture
T = ?
Final Temperature

Heat lost by hot water = Heat gained by cold water

Calorimeter Components

  • Outer jacket: Provides insulation
  • Inner vessel: Contains the liquid
  • Stirrer: For uniform temperature
  • Thermometer: Measures temperature
  • Lid: Prevents heat loss

Calorimetry Problem

Q: 100 g of water at 80°C is mixed with 50 g of water at 20°C in a calorimeter. Calculate the final temperature of the mixture. (Specific heat of water = 1 cal/g°C)

Bihar Board Answer Format

For 5-mark questions:

  1. Step 1 (0.5 mark): Write given data
  2. Step 2 (1 mark): State principle/formula
  3. Step 3 (2 marks): Detailed calculation
  4. Step 4 (1 mark): Final answer with unit
  5. Step 5 (0.5 mark): Conclusion/explanation

Model Answer (Bihar Board Style)

Step 1: Given Data
• Mass of hot water, m₁ = 100 g
• Temperature of hot water, T₁ = 80°C
• Mass of cold water, m₂ = 50 g
• Temperature of cold water, T₂ = 20°C
• Specific heat of water, c = 1 cal/g°C
• Let final temperature = T

Step 2: Principle
According to principle of calorimetry:
Heat lost by hot water = Heat gained by cold water

Step 3: Calculation
m₁ × c × (T₁ - T) = m₂ × c × (T - T₂)
100 × 1 × (80 - T) = 50 × 1 × (T - 20)
100 × (80 - T) = 50 × (T - 20)
8000 - 100T = 50T - 1000
8000 + 1000 = 50T + 100T
9000 = 150T
T = 9000 ÷ 150
T = 60°C

Step 4: Final Answer
Final temperature of mixture = 60°C

Step 5: Conclusion
The final temperature (60°C) lies between the initial temperatures (80°C and 20°C) as expected from calorimetry principle.

Latent Heat

What is Latent Heat?

Latent Heat is the heat energy absorbed or released during a phase change at constant temperature.

Types of Latent Heat

  1. Latent Heat of Fusion (L_f): Heat required to change 1 kg of solid to liquid at melting point
  2. Latent Heat of Vaporization (L_v): Heat required to change 1 kg of liquid to vapor at boiling point
  3. Latent Heat of Sublimation: Heat required to change 1 kg of solid directly to vapor

Latent Heat Formulas

Q = m × L

Where: Q = heat absorbed/released during phase change, m = mass, L = latent heat

For water: L_f = 336 J/g = 80 cal/g
For water: L_v = 2260 J/g = 540 cal/g

Heating Curve of Water

Temperature (°C)
Heat Added
Ice heating
(-10°C to 0°C)
Melting at 0°C
(Latent heat)
Water heating
(0°C to 100°C)
Boiling at 100°C
(Latent heat)
Steam heating
(>100°C)

During phase changes (melting & boiling), temperature remains constant despite heat addition

Latent Heat Problem

Q: Calculate the total heat required to convert 10 g of ice at -5°C to steam at 100°C.
Given: Specific heat of ice = 0.5 cal/g°C, Specific heat of water = 1 cal/g°C, Latent heat of fusion = 80 cal/g, Latent heat of vaporization = 540 cal/g

Model Answer

Step 1: Break into stages
1. Heating ice from -5°C to 0°C
2. Melting ice at 0°C to water at 0°C
3. Heating water from 0°C to 100°C
4. Vaporizing water at 100°C to steam at 100°C

Step 2: Calculations
m = 10 g

1. Q₁ = m × c_ice × ΔT = 10 × 0.5 × [0 - (-5)] = 10 × 0.5 × 5 = 25 cal

2. Q₂ = m × L_f = 10 × 80 = 800 cal

3. Q₃ = m × c_water × ΔT = 10 × 1 × (100 - 0) = 10 × 100 = 1000 cal

4. Q₄ = m × L_v = 10 × 540 = 5400 cal

Step 3: Total heat
Q_total = Q₁ + Q₂ + Q₃ + Q₄
= 25 + 800 + 1000 + 5400
= 7225 cal

Step 4: Convert to joules (if needed)
1 cal = 4.186 J
Q_total = 7225 × 4.186 ≈ 30244 J ≈ 3.02 × 10⁴ J

Final Answer: Total heat required = 7225 cal or 3.02 × 10⁴ J

Modes of Heat Transfer

Three Modes of Heat Transfer

Mode Mechanism Medium Required Examples
1. Conduction Molecular collisions without bulk motion Solid (best in metals) Heating metal rod at one end
2. Convection Bulk movement of heated material Fluids (liquids & gases) Boiling water, Sea breeze
3. Radiation Electromagnetic waves No medium required (works in vacuum) Heat from Sun, Heat from fire

Modes of Heat Transfer

Conduction
Through solids
Convection
Fluid motion
Through fluids
Radiation
Waves
Through vacuum

NEET Important Facts

  • Good conductors of heat are generally good conductors of electricity (metals)
  • Air is a poor conductor but good insulator (used in thermos flasks)
  • Black surfaces are good absorbers and emitters of radiation
  • White/silver surfaces are poor absorbers but good reflectors
  • Thermos flask prevents all three modes of heat transfer

Heat Conduction

Thermal Conductivity

Thermal conductivity (k) is a measure of a material's ability to conduct heat. Higher k means better conduction.

Fourier's Law of Heat Conduction

dQ/dt = -k × A × (dT/dx)

Where:
dQ/dt = rate of heat flow (W)
k = thermal conductivity (W/m·K)
A = cross-sectional area (m²)
dT/dx = temperature gradient (K/m)
Negative sign indicates heat flows from high to low temperature

Thermal Conductivity Values

Material Thermal Conductivity (W/m·K) Application
Silver 429 Best conductor
Copper 401 Cooking utensils
Aluminum 237 Heat sinks
Iron 80 General use
Glass 0.8 Windows
Water 0.6 Poor conductor
Air 0.026 Insulator
Wood 0.04-0.4 Insulation

Heat Conduction Problem

Q: A brass rod (k = 109 W/m·K) of length 0.5 m and cross-sectional area 0.01 m² has its ends maintained at 100°C and 0°C. Calculate the rate of heat flow through the rod.

NEET Answer Writing Strategy

  1. 1 minute: Read and understand the problem
  2. 2 minutes: Write formula and substitute values
  3. 1 minute: Calculate carefully
  4. 1 minute: Verify units and final answer
  5. Tip: Show steps clearly for partial marks

Model Answer (NEET Style)

Given:
k = 109 W/m·K
L = 0.5 m
A = 0.01 m²
T₁ = 100°C, T₂ = 0°C

Step 1: Temperature difference
ΔT = T₁ - T₂ = 100 - 0 = 100°C = 100 K (since ΔT in °C = ΔT in K)

Step 2: Temperature gradient
dT/dx = ΔT/L = 100 K / 0.5 m = 200 K/m

Step 3: Apply Fourier's law
dQ/dt = -k × A × (dT/dx)
= 109 × 0.01 × 200 (ignoring negative sign for magnitude)
= 109 × 0.01 × 200
= 109 × 2
= 218 W

Final Answer: Rate of heat flow = 218 W

Note: The negative sign indicates direction of heat flow from hot to cold end.

Practice Questions Set 1

Time Management for Exams

  • NEET: 180 questions in 180 minutes = 1 minute/question
  • Bihar Board: Plan time based on marks - 1 mark/minute
  • Start with easy questions
  • Don't spend more than 2 minutes on any question initially
  • Review marked questions at the end

Short Answer Questions (2 marks each)

Q1: Define specific heat capacity. Give its SI unit.

Expected Answer

Definition: Specific heat capacity of a substance is the amount of heat required to raise the temperature of 1 kg of the substance by 1 K (or 1°C).

SI Unit: Joule per kilogram per Kelvin (J/kg·K)

Scoring: Definition (1 mark) + Unit (1 mark)

Q2: State the principle of calorimetry.

Expected Answer

According to the principle of calorimetry, when two bodies at different temperatures are brought into thermal contact, heat flows from the hotter body to the colder body until thermal equilibrium is reached, provided no heat is lost to the surroundings.

Mathematically: Heat lost by hot body = Heat gained by cold body

Numerical Problems (3 marks each)

Q3: A copper sphere of radius 2 cm is heated from 20°C to 120°C. Calculate the increase in its surface area. (α for copper = 1.7 × 10⁻⁵ /°C)

Step-by-Step Solution

Given: r = 2 cm = 0.02 m, T₁ = 20°C, T₂ = 120°C, α = 1.7 × 10⁻⁵ /°C

Step 1: ΔT = 120 - 20 = 100°C

Step 2: Original surface area A₀ = 4πr² = 4 × 3.14 × (0.02)² = 4 × 3.14 × 0.0004 = 0.005024 m²

Step 3: Coefficient of area expansion β = 2α = 2 × 1.7 × 10⁻⁵ = 3.4 × 10⁻⁵ /°C

Step 4: ΔA = β × A₀ × ΔT = (3.4 × 10⁻⁵) × 0.005024 × 100 = 1.708 × 10⁻⁵ m²

Answer: Increase in surface area = 1.71 × 10⁻⁵ m²

Scoring: Formula (1 mark) + Calculation (1.5 marks) + Answer (0.5 mark)

Bihar Board Long Answer Guidelines (5 marks)

  1. Introduction (0.5 mark): Define key terms
  2. Principle/Law (1 mark): State relevant principle
  3. Derivation/Explanation (2 marks): Detailed explanation
  4. Diagram (1 mark): Neat labeled diagram
  5. Conclusion/Application (0.5 mark): Summary or real-world application

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