Physics - Mechanics and Properties of Matter - How to Solve: Thermal Properties of Matter (NEET UG Physics)

How to Solve: Thermal Properties of Matter (NEET UG Physics)

Complete Guide

Introduction

"Master thermal expansion, calorimetry, and phase changes, and you’ll solve 3–4 NEET Physics questions worth 12–16 marks—enough to boost your rank by thousands!

WHAT YOU NEED TO KNOW FIRST

1. Basic algebra (rearranging equations, solving for unknowns).
2. SI units (Joules, Kelvin, Celsius, kg, m³).
3. Concept of heat (energy transfer due to temperature difference).

KEY TERMS & FORMULAS

1. Thermal Expansion

Linear Expansion: Formula: ΔL = αL₀ΔT - ΔL = change in length (m) - α = coefficient of linear expansion (K⁻¹ or °C⁻¹) → MEMORISE THIS - L₀ = original length (m) - ΔT = change in temperature (°C or K)

Volume Expansion: Formula: ΔV = βV₀ΔT - ΔV = change in volume (m³) - β = coefficient of volume expansion (K⁻¹ or °C⁻¹) → MEMORISE THIS (β ≈ 3α for solids) - V₀ = original volume (m³)

Area Expansion: Formula: ΔA = γA₀ΔT - γ ≈ 2α (for isotropic materials)

2. Calorimetry (Heat Transfer)

Specific Heat Capacity (c): Formula: Q = mcΔT - Q = heat energy (J) - m = mass (kg) - c = specific heat capacity (J/kg·K) → MEMORISE COMMON VALUES (e.g., water = 4186 J/kg·K) - ΔT = temperature change (°C or K)

Latent Heat (L): Formula: Q = mL - L = latent heat (J/kg) → MEMORISE VALUES (e.g., water’s latent heat of fusion = 3.34 × 10⁵ J/kg)

Principle of Calorimetry: Heat lost by hot body = Heat gained by cold body → m₁c₁ΔT₁ = m₂c₂ΔT₂

3. Phase Change (Heating Curve)

• Solid → Liquid: Melting (uses latent heat of fusion, Lf)
• Liquid → Gas: Vaporization (uses latent heat of vaporization, Lv)
• Temperature remains constant during phase change!

4. Newton’s Law of Cooling

Formula: dT/dt = -k(T – Tₛ) - T = temperature of the body (°C or K) - Tₛ = surrounding temperature (°C or K) - k = cooling constant (s⁻¹) → given on exam sheet - Approximate form (for small ΔT): ΔT = T₀e⁻ᵏᵗ

STEP-BY-STEP METHOD

Step 1: Identify the Type of Problem

• Thermal expansion? → Use ΔL = αL₀ΔT or ΔV = βV₀ΔT
• Heat transfer (calorimetry)? → Use Q = mcΔT or Q = mL
• Phase change? → Use latent heat + specific heat
• Newton’s cooling? → Use dT/dt = -k(T – Tₛ) or ΔT = T₀e⁻ᵏᵗ

Step 2: List Given Data & Unknowns

• Write all given values (with units).
• Identify what you need to find.

Step 3: Choose the Right Formula

• Match the problem type to the correct formula.
• If phase change is involved, break into stages (e.g., heating ice → melting → heating water).

Step 4: Solve for the Unknown

• Rearrange the formula if needed.
• Plug in values with units.
• Calculate step-by-step.

Step 5: Check Units & Reasonableness

• Final answer should have correct units (e.g., J for heat, m for length).
• Does the answer make sense? (e.g., ΔL should be small for solids).

WORKED EXAMPLES

Example 1 – Basic (Linear Expansion)

Problem: A steel rod (α = 1.2 × 10⁻⁵ °C⁻¹) is 2 m long at 20°C. What is its length at 120°C?

Solution:
1. Identify: Thermal expansion (linear).
2. Given: - L₀ = 2 m - α = 1.2 × 10⁻⁵ °C⁻¹ - ΔT = 120°C – 20°C = 100°C
3. Formula: ΔL = αL₀ΔT
4. Calculate: ΔL = (1.2 × 10⁻⁵)(2)(100) = 2.4 × 10⁻³ m = 2.4 mm
5. Final length: L = L₀ + ΔL = 2 m + 0.0024 m = 2.0024 m

What we did and why: - Used linear expansion formula because the problem involved length change. - Added ΔL to original length to get final length.

Example 2 – Medium (Calorimetry with Phase Change)

Problem: 50 g of ice at -10°C is heated until it becomes water at 50°C. Calculate total heat required. Given: - c_ice = 2100 J/kg·K - c_water = 4186 J/kg·K - Lf = 3.34 × 10⁵ J/kg

Solution:
1. Break into stages: - Heating ice (-10°C → 0°C) - Melting ice (0°C → water at 0°C) - Heating water (0°C → 50°C)

1. Stage 1: Heating ice Q₁ = mcΔT = (0.05 kg)(2100 J/kg·K)(10 K) = 1050 J

2. Stage 2: Melting ice Q₂ = mLf = (0.05 kg)(3.34 × 10⁵ J/kg) = 16700 J

3. Stage 3: Heating water Q₃ = mcΔT = (0.05 kg)(4186 J/kg·K)(50 K) = 10465 J

4. Total heat: Q_total = Q₁ + Q₂ + Q₃ = 1050 + 16700 + 10465 = 28215 J

What we did and why: - Split the problem into stages because phase change occurs. - Used specific heat for temperature change and latent heat for phase change.

Example 3 – Exam-Style (Newton’s Cooling)

Problem: A cup of coffee cools from 90°C to 60°C in 10 minutes in a room at 20°C. How long will it take to cool to 40°C?

Solution:
1. Identify: Newton’s cooling (exponential decay).
2. Given: - T₀ = 90°C (initial temp) - T = 60°C at t = 10 min - Tₛ = 20°C (surroundings) - Find t when T = 40°C

1. Formula: T – Tₛ = (T₀ – Tₛ)e⁻ᵏᵗ

2. Step 1: Find k 60 – 20 = (90 – 20)e⁻ᵏ(10) 40 = 70e⁻¹⁰ᵏ e⁻¹⁰ᵏ = 40/70 = 4/7 -10k = ln(4/7) k = -ln(4/7)/10 ≈ 0.0559 min⁻¹

3. Step 2: Find t for T = 40°C 40 – 20 = (90 – 20)e⁻ᵏᵗ 20 = 70e⁻ᵏᵗ e⁻ᵏᵗ = 20/70 = 2/7 -kt = ln(2/7) t = -ln(2/7)/k ≈ 22.5 min

What we did and why: - Used Newton’s cooling formula because the problem involved temperature decay over time. - First found k using given data, then used it to find the required time.

COMMON MISTAKES

1. MISTAKE: Forgetting to convert mass to kg. WHY IT HAPPENS: Using grams instead of kg in Q = mcΔT. CORRECT APPROACH: Always convert mass to kg (1 g = 0.001 kg).

2. MISTAKE: Ignoring phase change in calorimetry. WHY IT HAPPENS: Only using Q = mcΔT and missing Q = mL. CORRECT APPROACH: Break the problem into stages (heating → phase change → heating).

3. MISTAKE: Using wrong ΔT in expansion problems. WHY IT HAPPENS: Using final temp instead of ΔT (T_final – T_initial). CORRECT APPROACH: Always calculate ΔT = T_final – T_initial.

4. MISTAKE: Confusing α and β in expansion. WHY IT HAPPENS: Using linear expansion formula for volume change. CORRECT APPROACH: Use ΔL = αL₀ΔT for length, ΔV = βV₀ΔT for volume.

5. MISTAKE: Misapplying Newton’s cooling formula. WHY IT HAPPENS: Using T instead of (T – Tₛ) in the formula. CORRECT APPROACH: Always use (T – Tₛ) in dT/dt = -k(T – Tₛ).

EXAM TRAPS

1. TRAP: Mixing up latent heat values (Lf vs. Lv). HOW TO SPOT IT: Problem mentions "melting" or "vaporization." HOW TO AVOID IT: Memorize:
2. Lf (fusion) = 3.34 × 10⁵ J/kg (ice → water)

3. Lv (vaporization) = 2.26 × 10⁶ J/kg (water → steam)

4. TRAP: Assuming temperature changes during phase change. HOW TO SPOT IT: Problem describes melting/boiling but asks for temperature. HOW TO AVOID IT: Remember: Temperature stays constant during phase change!

5. TRAP: Using wrong specific heat values. HOW TO SPOT IT: Problem gives no c value but expects you to know it. HOW TO AVOID IT: Memorize:

6. c_water = 4186 J/kg·K
7. c_ice = 2100 J/kg·K
8. c_steam = 2000 J/kg·K

1-MINUTE RECAP

"Listen up—this is your last-minute thermal properties cheat sheet!

1. Thermal expansion: Use ΔL = αL₀ΔT for length, ΔV = βV₀ΔT for volume. Remember β ≈ 3α for solids.
2. Calorimetry: Q = mcΔT for temperature change, Q = mL for phase change. Break problems into stages if both are involved.
3. Phase change: Temperature does not change while melting/boiling. Use latent heat here.
4. Newton’s cooling: dT/dt = -k(T – Tₛ). For small changes, use ΔT = T₀e⁻ᵏᵗ.
5. Units matter! Always convert mass to kg, temperature to Kelvin if needed, and check final units.

Now go crush those NEET questions!