Problem Types & Solved Examples

🔬 Exam Insight

This is where most students lose marks. They know the formula but not the selection logic. Always ask: What type of thermal process is this? What conservation principle applies? Which formula subset is relevant? Answer these BEFORE writing equations.

📊 Type 1 — Direct Formula Problems

Straight application of a formula

These are your marks-guarantee questions. No concept twist — just formula selection and arithmetic. Practice until you can do these in 90 seconds.

P1.1

Heat required to melt 500g of ice at 0°C

Latent Heat Phase Change CBSE/NEET

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Given

Mass of ice: m = 500 g = 0.5 kg. Initial temperature = 0°C. Latent heat of fusion of ice: L_f = 3.34 × 10⁵ J/kg. Find: heat required to melt completely.

🧠 Thinking Step

Phase change → temperature stays at 0°C throughout. Use Q = mL only. No Q = mcΔT here.

Identify: This is a phase change (solid→liquid) at constant temperature. Use Q = mL_f

Q = m × L_f = 0.5 × 3.34 × 10⁵

Q = 1.67 × 10⁵ J

✓ Answer: Q = 1.67 × 10⁵ J = 167 kJ

⚡ Shortcut Insight

For 1 kg of ice: Q = 3.34 × 10⁵ J. For 500 g, exactly half: Q = 1.67 × 10⁵ J. Scale linearly with mass.

P1.2

Linear expansion of a steel rod heated from 20°C to 120°C

Linear Expansion JEE Main

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Given

Steel rod of length L₀ = 2 m at 20°C. Heated to 120°C. α_steel = 12 × 10⁻⁶ K⁻¹. Find: (a) increase in length, (b) new length.

ΔT = 120 − 20 = 100°C = 100 K (ΔT is same in Celsius and Kelvin)

ΔL = αL₀ΔT = 12 × 10⁻⁶ × 2 × 100 = 2400 × 10⁻⁶ = 2.4 × 10⁻³ m

New length = L₀ + ΔL = 2 + 0.0024 = 2.0024 m

✓ ΔL = 2.4 mm | New length = 2.0024 m

❌ Common Mistake Alert

Students often forget to convert to the correct temperature difference unit. Here ΔT = 100°C = 100 K — the same. But for absolute temperature in radiation formulas, 120°C = 393 K ≠ 120 K!

P1.3

Rate of heat conduction through composite wall

Fourier's Law JEE Main NEET

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Given

A copper rod (K = 400 W/mK, L = 50 cm, A = 2 cm²) connects a heat source at 100°C to a sink at 0°C. Find: rate of heat flow in steady state.

🧠 Thinking Step

"Steady state" is the key phrase. It means temperature distribution doesn't change with time. Use H = KA(T₁−T₂)/L directly — no time integration needed.

Convert: L = 0.5 m, A = 2 × 10⁻⁴ m², ΔT = 100 − 0 = 100 K

H = KA(ΔT)/L = 400 × 2×10⁻⁴ × 100 / 0.5

H = 400 × 2×10⁻⁴ × 200 = 16 W

✓ H = 16 W (rate of heat flow)

🧠 Type 2 — Conceptual Problems

What happens physically — no formula crunch

NEET specializes in these. You need to understand the mechanism, not just the math. If this step is wrong, the entire solution fails.

P2.1

Why does the rate of cooling decrease with time? (Newton's Cooling)

Newton's Cooling Conceptual NEET

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What Examiner Tests

Understanding of Newton's Law: dT/dt ∝ (T − T₀). Do students understand WHY the rate changes, not just the formula?

Newton's Law: dT/dt = −k(T − T₀). The rate is PROPORTIONAL to the temperature excess (T − T₀).

As the body cools, T decreases → (T − T₀) decreases → |dT/dt| decreases.

This creates an exponential approach: T → T₀ asymptotically, never exactly reaching T₀.

🎯 Strategy Tip

In an exam, if asked "what is the rate of cooling when body reaches surrounding temperature?", the answer is ZERO — because (T − T₀) = 0. This is often a trap option.

P2.2

A hole in a metal plate — does it expand or contract on heating?

Thermal Expansion Most Tricky NEET JEE

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What Examiner Tests

Conceptual trap: Students assume the hole contracts. This is the most repeated conceptual trap in thermal expansion.

🧠 Thinking Step

Think of the hole as a "ghost plate" made of the same material. If that plate would expand, the hole expands too. The boundaries of the hole move outward, just like any dimension of the solid.

When a solid expands, ALL dimensions increase — including any holes or cavities.

Reason: The molecules around the hole move apart from each other, increasing the hole's area.

A circular hole expands as ΔA_hole = β × A_hole_initial × ΔT.

✓ The hole EXPANDS on heating (same coefficient β as the solid)

❌ Common Mistake Alert

This is where most students lose marks. Intuition says "the hole will shrink" — it DOESN'T. The ghost plate analogy is the exam-reliable reasoning.

P2.3

Good absorbers are good emitters — explain Kirchhoff's Law

Kirchhoff's Law NEET CBSE

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What Examiner Tests

Physical reasoning behind Kirchhoff's Law. Not just statement recall — logical derivation from equilibrium conditions.

Consider two bodies at the same temperature T in an enclosure. For thermal equilibrium: heat absorbed = heat emitted.

If body A absorbs more, it would heat up → equilibrium violated. So absorption rate must equal emission rate for both.

Therefore: (emissivity/absorptivity) = constant for all bodies at same T = blackbody emissive power.

For any body at equilibrium: emissivity (e) = absorptivity (a). Good absorber (a → 1) → good emitter (e → 1).

✓ Kirchhoff's Law: e = a at thermal equilibrium (derived from 2nd law of thermodynamics)

⚙️ Type 3 — Multi-Step Problems

Multiple concepts in sequence

JEE Main specializes in these. Each step must be right — one wrong step cascades into a completely wrong answer.

P3.1

Mixing ice and water — find final temperature and state

Calorimetry + Phase Change JEE Main NEET

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Given

200 g of ice at −10°C is mixed with 400 g of water at 60°C. Find final temperature and state. (c_ice = 2100 J/kg·K, c_water = 4200 J/kg·K, L_f = 3.36 × 10⁵ J/kg)

🧠 Thinking Step

Always do a PHASE CHECK first. Step 1: Can ice reach 0°C? Step 2: Can all ice melt? Step 3: If yes, find T_f. This is the key methodology that differentiates rank 1000 from rank 100.

Heat available from water to cool from 60°C to 0°C: Q_available = 0.4 × 4200 × 60 = 100,800 J

Heat needed to warm ice from −10°C to 0°C: Q₁ = 0.2 × 2100 × 10 = 4,200 J ✓ (affordable)

Heat needed to melt all ice at 0°C: Q₂ = 0.2 × 3.36 × 10⁵ = 67,200 J. Q₁ + Q₂ = 71,400 J < 100,800 J ✓

All ice melts. Remaining heat = 100,800 − 71,400 = 29,400 J for heating water mixture.

Total water = 0.2 + 0.4 = 0.6 kg. 29,400 = 0.6 × 4200 × ΔT → ΔT = 11.67°C

T_final = 0 + 11.67 ≈ 11.7°C (all water)

✓ Final state: All water at ≈ 11.7°C

🎯 Strategy Tip

Always compute "available" and "required" heat before solving. If required > available, equilibrium is at 0°C with partial melting. If required < available, all ice melts and T_f > 0°C.

P3.2

Thermal stress when a rod is prevented from expanding

Thermal Stress + Young's Modulus JEE Main ADV

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Given

A steel rod (α = 12 × 10⁻⁶ K⁻¹, Y = 2 × 10¹¹ Pa) is rigidly fixed at both ends. Its temperature rises by 50 K. Find the thermal stress developed.

The rod would expand by: ΔL = αL₀ΔT, but is prevented. This creates a compressive strain = αΔT.

Thermal Strain = α × ΔT = 12 × 10⁻⁶ × 50 = 6 × 10⁻⁴

Thermal Stress = Y × Strain = 2 × 10¹¹ × 6 × 10⁻⁴ = 1.2 × 10⁸ Pa

✓ Thermal Stress = 1.2 × 10⁸ Pa = 120 MPa (compressive)

🔬 Exam Insight

JEE twists this by asking: "What force must be applied to prevent expansion?" Answer: F = Stress × A = YαΔT × A. The force is compressive (pushes inward on both ends).

📈 Type 4 — Graph-Based Problems

Reading, interpreting, and deriving from graphs

JEE Main and NEET both use graph questions. These test visual pattern recognition — different from numerical problems.

P4.1

From a heating curve, identify (a) melting point, (b) specific heat of liquid, (c) latent heat

Heating Curve NEET JEE

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Problem Setup

A heating curve shows: rise from −30°C to 20°C (steep), then flat at 20°C for some time, then rise from 20°C to 80°C (less steep), flat at 80°C, then rise above 80°C. Identify each phase.

Melting point = temperature of first horizontal segment = 20°C

Boiling point = temperature of second horizontal segment = 80°C

Specific heat of liquid = slope of rising segment between 20°C and 80°C: slope = 1/(mc_liquid) → c_liquid = 1/(m × slope)

Latent heat of fusion = (heat added during first flat region) / mass = Q₁/m

🎯 Strategy Tip

Slope of rising segment in a T vs Q graph = 1/(mc). Steeper slope → smaller c (less heat needed per degree). Longer flat region → larger latent heat.

P4.2

From Newton's cooling graph, find the cooling constant k

Newton's Cooling JEE

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Given

A body cools from 80°C to 40°C in surroundings at 20°C. A graph of ln(T − T₀) vs t is a straight line. The slope of this line is −0.02 min⁻¹. Find k.

From Newton's Law: T(t) = T₀ + (T_i − T₀)e^(−kt). Taking ln: ln(T − T₀) = ln(T_i − T₀) − kt

This is y = c − kx form. Slope of ln(T−T₀) vs t graph = −k

Slope = −0.02 min⁻¹ → k = 0.02 min⁻¹

✓ Cooling constant k = 0.02 min⁻¹

✅ Type 5 — Assertion & Reason (A&R)

Two statements — is each true? Does reason explain assertion?

NEET and CBSE boards extensively test A&R. The mistake zone: both statements are true but one doesn't explain the other.

📋

A&R Option Key (Standard)

Option A: Both A and R are true, and R is the correct explanation of A
Option B: Both A and R are true, but R is NOT the correct explanation of A
Option C: A is true, R is false
Option D: A is false, R is true

P5.1

A: A steel rod is heated — it expands. R: Molecules gain kinetic energy on heating

NEET

▼ Show Solution

Assertion: TRUE — solids expand on heating (linear expansion).

Reason: PARTIALLY correct but INCOMPLETE. Molecules do gain KE, but expansion is due to increase in AVERAGE INTERATOMIC DISTANCE — caused by asymmetry in potential energy curve, not directly KE.

The correct reason is: the potential energy well for atomic interaction is asymmetric — atoms oscillate about a center that moves away from equilibrium at higher energies.

✓ Option B: Both true, but R is not the CORRECT explanation of A

P5.2

A: Water at 4°C has maximum density. R: Water is anomalous between 0°C and 4°C

NEETCBSE

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Assertion: TRUE — density of water is maximum at 4°C.

Reason: TRUE — water contracts from 0°C to 4°C (anomalous expansion), hence density increases in this range.

R correctly explains A: Because of anomalous contraction from 0°C to 4°C, density increases to a maximum at 4°C, then normal expansion reduces density above 4°C.

✓ Option A: Both true, R IS the correct explanation

📋 Type 6 — Case-Based Problems

Paragraph + multiple questions based on scenario

CBSE 2020+ introduced case-based questions. JEE Advanced uses "comprehension" format. The skill: extract the physics model from the narrative.

📋

Case Study: Bimetallic Strip Thermostat

Passage

A bimetallic strip consists of two metals bonded together — Brass (α = 19×10⁻⁶ K⁻¹) and Invar (α = 1.5×10⁻⁶ K⁻¹). When temperature changes, one expands more than the other, causing the strip to bend. This principle is used in thermostats, circuit breakers, and thermometers. In a thermostat, the strip bends to break a circuit when temperature exceeds a set point, thereby cutting off power to a heater.

On heating, which metal is on the outer (convex) curve of the bent strip?

▼ Answer

Brass has a higher α (19 vs 1.5 × 10⁻⁶ K⁻¹) → expands more when heated.

The metal that expands more becomes the OUTER (convex) side of the bend.

✓ Brass is on the convex (outer) side when heated

What happens to the strip on cooling below room temperature?

▼ Answer

On cooling, Brass contracts more than Invar (reverse of heating).

The strip bends in the OPPOSITE direction (Invar becomes convex).

✓ Strip bends in reverse — Invar on convex side

Next: Interlinking Concepts

How thermal connects with KTG, thermodynamics, mechanics, and optics.

← Precision Interlinking →