📖 Chapter PN03 · NDA Class 11–12 Level🎯 NDA Level : High Priority
Heat and Thermodynamics covers how thermal energy is stored, transferred, and converted into work. This chapter is conceptually rich and very regularly tested in NDA — from basic temperature conversions to gas laws, calorimetry, and the First Law of Thermodynamics. Most questions are either formula-application or concept-identification, making this chapter highly rewarding for average students who prepare systematically.
📌 What to expect in NDA (based on 2022–2025 pattern): (1) Temperature scale conversions — Celsius, Fahrenheit, Kelvin; (2) Thermal expansion — linear, area, volume coefficients; anomalous expansion of water; (3) Heat transfer — conduction (metals), convection (fluids), radiation (vacuum); (4) Specific heat capacity and latent heat — calorimetry calculations; (5) Gas laws — Boyle's law (P–V), Charles's law (V–T), combined gas law; (6) Kinetic theory postulates, RMS speed, pressure of a gas; (7) First Law of Thermodynamics — ΔU, Q, W for isothermal/adiabatic/isochoric processes.
Topics at a Glance
① Temperature & Thermal Expansion
Scales, α, β, γ; anomalous expansion of water
② Heat Transfer
Conduction, convection, radiation; Newton's cooling; Stefan's law
③ Calorimetry
Specific heat, latent heat, principle of calorimetry
④ Kinetic Theory of Gases
Postulates, pressure, RMS speed, degrees of freedom
⑤ Gas Laws
Boyle's, Charles's, Gay-Lussac's, combined gas law, ideal gas
Celsius, Fahrenheit, and Kelvin — and how to convert between them
Temperature measures the average kinetic energy of molecules. Three scales are used: Celsius (°C) for everyday use, Fahrenheit (°F) used in the US and older medical contexts, and Kelvin (K) — the absolute thermodynamic scale used in all physics equations.
⚡ Temperature Scale Conversions
°C to K: K = °C + 273.15 ≈ °C + 273
K to °C: °C = K − 273
°C to °F: °F = (9/5)°C + 32 or °F = 1.8°C + 32
°F to °C: °C = (5/9)(°F − 32) or °C = (°F − 32)/1.8
Triple point of water: 273.16 K = 0.01°C
Absolute zero: 0 K = −273.15°C ≈ −273°C
Key fixed points:
Ice point: 0°C = 273 K = 32°F
Steam point: 100°C = 373 K = 212°F
Memory: The interval between ice and steam points is 100°C = 100 K = 180°F. So 1°C = 1 K = 1.8°F. Kelvin scale has no negative temperatures — it starts at absolute zero.
💡 Quick conversion trick: To convert 37°C (body temperature) to °F: (37 × 1.8) + 32 = 66.6 + 32 = 98.6°F. To convert 98.6°F back: (98.6 − 32)/1.8 = 66.6/1.8 = 37°C. NDA sometimes asks at what temperature Celsius and Fahrenheit readings are equal → answer: −40° (both scales read −40 at the same point).
1.2
Thermal Expansion of Solids, Liquids & Gases
Materials expand on heating — with different coefficients for different dimensions
When a body is heated, its particles vibrate more vigorously, increasing average separation. This causes the body to expand. Gases expand far more than liquids, and liquids more than solids for the same temperature rise.
⚡ Thermal Expansion Formulae
Linear expansion (solids — length):
ΔL = α L₀ ΔT → L = L₀(1 + αΔT)
α = coefficient of linear expansion (per °C or per K)
Area expansion (solids — surface area):
ΔA = β A₀ ΔT → β = 2α (area coefficient = 2 × linear)
Volume expansion (solids, liquids, gases):
ΔV = γ V₀ ΔT → γ = 3α (volume coefficient = 3 × linear)
For gases (at constant pressure):
γ_gas = 1/273 per °C ≈ 3.67 × 10⁻³ per °C (much larger than solids)
Apparent vs Real expansion of liquids:
γ_real = γ_apparent + γ_vessel
(a liquid expands more than its container — the net visible expansion is apparent)
The relationship β = 2α and γ = 3α holds because expansion in each dimension contributes independently. For isotropic materials, all three are related this simply.
🌊 Anomalous Expansion of Water
Water contracts on heating from 0°C to 4°C
Maximum density of water at 4°C (1000 kg/m³)
Above 4°C: water expands normally
Result: lakes freeze from top down, preserving aquatic life
Ice is less dense than water → ice floats
🔸 Practical Applications
Gaps in railway tracks: allow for thermal expansion
Bimetallic strips: different α values → bending on heating
Telephone wires: sag in summer (expansion)
Pyrex glass: very low α → thermal shock resistant
Invar: very low α → used in precision instruments
Fig. 1 — Density of water vs temperature. Density increases from 0°C to 4°C (anomalous), reaching a maximum at 4°C, then decreases. This causes lakes to freeze from the top down.
📝 TOPIC-WISE PYQ
Temperature & Thermal Expansion — NDA Pattern Questions
Q1. At what temperature do Celsius and Fahrenheit scales give the same reading?
(a) 0° (b) 40° (c) −40° (d) 100°
Answer: (c) −40°
Set °F = °C = x: x = (9/5)x + 32 → x − (9/5)x = 32 → −(4/5)x = 32 → x = −40. Both scales read −40 at the same temperature.
Q2. A steel rod of length 1 m has a coefficient of linear expansion 1.2×10⁻⁵ /°C. Its increase in length when heated by 100°C is:
(a) 0.12 mm (b) 1.2 mm (c) 0.012 mm (d) 12 mm
Answer: (b) 1.2 mm
ΔL = α L ΔT = 1.2×10⁻⁵ × 1 × 100 = 1.2×10⁻³ m = 1.2 mm.
Q3. The temperature at which water has maximum density is:
(a) 0°C (b) 100°C (c) −4°C (d) 4°C
Answer: (d) 4°C
Water behaves anomalously — it contracts on heating from 0°C to 4°C and expands beyond 4°C. Maximum density (1000 kg/m³) is at 4°C. This anomalous behaviour is vital for aquatic life in winter.
2. Heat Transfer
2.1
Conduction, Convection & Radiation
Three mechanisms of heat transfer — each with distinct conditions and examples
Heat always flows from a hotter body to a cooler body (Second Law of Thermodynamics). It does so by three mechanisms — only radiation can transfer heat through a vacuum.
🔥 Conduction
Heat transfers through direct contact
Medium required; no bulk movement
Best in solids (especially metals)
Mechanism: free electron transfer + vibration
Rate: Q/t = kA(T₁−T₂)/d
k = thermal conductivity (W/m·K)
🌊 Convection
Heat transfers through fluid movement
Requires a fluid (liquid or gas)
Bulk movement of molecules carries heat
Natural: hot air rises (less dense)
Forced: fan, pump-driven flow
Sea breeze, land breeze, trade winds
🌟 Radiation
Heat as electromagnetic waves (infrared)
No medium required — travels through vacuum
Speed = speed of light (3×10⁸ m/s)
Best absorbed by: black, rough surfaces
Best reflected by: white, shiny, polished surfaces
Stefan's law: E = σT⁴ (per unit area)
⚡ Heat Transfer Laws
Fourier's Law (Conduction):
Q/t = k A (T₁ − T₂) / d
k = thermal conductivity (W/m·K or W/m·°C)
Good conductors: silver (429), copper (401), aluminium (237) W/m·K
Bad conductors: wood, glass, air (insulators)
Newton's Law of Cooling:
dT/dt = −k(T − T_surroundings)
Rate of cooling ∝ temperature difference from surroundings
Stefan-Boltzmann Law (Radiation):
E = εσT⁴ (energy emitted per unit area per second)
σ = 5.67 × 10⁻⁸ W/m²·K⁴ (Stefan's constant)
ε = emissivity (0 to 1; black body: ε = 1)
Wien's Displacement Law:
λ_max × T = constant = 2.898 × 10⁻³ m·K
(Hotter body emits shorter wavelength radiation)
A perfect black body absorbs ALL incident radiation (ε = 1) and is also the best emitter. A white polished surface reflects most radiation — used in thermos flasks to minimise radiation loss.
Land heats faster than sea: specific heat of land < water
📝 TOPIC-WISE PYQ
Heat Transfer — NDA Pattern Questions
Q1. Heat from the Sun reaches the Earth primarily by:
(a) Conduction (b) Convection (c) Radiation (d) Both conduction and convection
Answer: (c) Radiation
Space between Earth and Sun is a near-perfect vacuum — no medium for conduction or convection. Heat is transferred as electromagnetic radiation (infrared and visible light) at the speed of light.
Q2. According to Stefan's law, the rate of heat radiation by a black body is proportional to:
(a) T (b) T² (c) T³ (d) T⁴
Answer: (d) T⁴
Stefan-Boltzmann law: E = εσT⁴. Rate of radiation is proportional to the fourth power of absolute temperature. Doubling T increases radiation by 2⁴ = 16 times.
Q3. A thermos flask keeps liquid hot by preventing heat loss through:
(a) Conduction only (b) Convection only (c) Radiation only (d) All three modes
Answer: (d) All three modes
A thermos (Dewar) flask: vacuum between walls prevents conduction and convection; silvered walls reflect radiation and minimise emission. Together they minimise all three modes of heat loss.
🤔 TRICKY QUESTIONS
Heat Transfer — Think Beyond the Formula
T1. Why do we feel warmer in woollen clothes even though wool itself is a poor conductor of heat?
Wool traps air — a very poor conductor.
The warmth comes from the tiny pockets of air trapped in the fibres, not from wool itself. Air has extremely low thermal conductivity (≈ 0.024 W/m·K). Wool acts as a framework to keep this insulating air layer still (prevents convection too). The air pockets prevent body heat from escaping. Same principle applies to double-pane windows and hollow walls.
T2. A black body and a white body are both at the same temperature. Which emits more radiation and which absorbs more radiation?
Black body: emits more AND absorbs more.
Emissivity ε = 1 for black body, ε < 1 for white. By Kirchhoff's law, a good absorber is also a good emitter. So black body emits more (E ∝ ε) and absorbs more (absorptivity = emissivity at thermal equilibrium). A white body reflects most radiation — so absorbs and emits less. In thermal equilibrium with surroundings, both maintain the same temperature despite different rates — they emit and absorb at equal rates (for each).
3. Calorimetry
3.1
Specific Heat & Latent Heat
How much heat changes temperature — and what happens at phase changes
When heat is added to a body, it either raises its temperature (sensible heat) or causes a phase change (latent heat) without changing temperature. The principle of calorimetry states that heat lost by the hot body equals heat gained by the cold body in an isolated system.
⚡ Calorimetry Formulae
Heat gained/lost (sensible heat):
Q = mcΔT
m = mass (kg), c = specific heat capacity (J/kg·K), ΔT = temperature change
Latent Heat (phase change — no temperature change):
Q = mL
L = specific latent heat (J/kg)
L_fusion (ice → water) = 3.36 × 10⁵ J/kg (= 80 cal/g)
L_vaporisation (water → steam) = 2.26 × 10⁶ J/kg (= 540 cal/g)
Specific Heat Values (approx.):
Water: 4200 J/kg·K (= 1 cal/g·°C — highest of common substances)
Ice: 2100 J/kg·K
Aluminium: 900 J/kg·K
Iron: 450 J/kg·K
Copper: 385 J/kg·K
Mercury: 140 J/kg·K (lowest liquid specific heat)
Principle of Calorimetry:
Heat lost by hot body = Heat gained by cold body
m₁c₁(T₁ − T_mix) = m₂c₂(T_mix − T₂)
Mechanical Equivalent of Heat:
1 calorie = 4.18 J ≈ 4.2 J
Water has the highest specific heat of all common substances — this is why coastal climates are mild (sea moderates temperature), and why water is used as a coolant in engines and radiators.
Fig. 2 — Heating curve of water. Sloped portions: temperature rises (Q = mcΔT). Flat portions: phase change at constant temperature (Q = mL). Boiling plateau is longer because L_vaporisation >> L_fusion.
⚠ NDA Exam Trap — Latent Heat of Vaporisation vs Fusion: L_vaporisation (2.26 MJ/kg) is about 7 times larger than L_fusion (0.336 MJ/kg). This is why steam burns are far more severe than boiling water burns — steam releases enormous extra heat as it condenses. Steam at 100°C causes more damage than water at 100°C.
📝 TOPIC-WISE PYQ
Calorimetry — NDA Pattern Questions
Q1. How much heat is required to convert 5 kg of ice at 0°C completely to water at 0°C? (L_fusion = 3.36×10⁵ J/kg)
Answer: (b) 1.68×10⁶ J
Q = mL = 5 × 3.36×10⁵ = 1.68×10⁶ J. Temperature does not change during melting — all heat goes into breaking bonds between ice molecules.
Q2. Equal masses of water at 10°C and 50°C are mixed. The final temperature of the mixture is: (c_water is same for both)
(a) 20°C (b) 25°C (c) 30°C (d) 60°C
Answer: (c) 30°C
By principle of calorimetry: m×c×(T − 10) = m×c×(50 − T). The m and c cancel: T − 10 = 50 − T → 2T = 60 → T = 30°C. Simple average only because masses and specific heats are equal.
Q3. Which substance has the highest specific heat capacity?
(a) Copper (b) Iron (c) Water (d) Mercury
Answer: (c) Water
Water: c = 4200 J/kg·K — the highest of all common substances. This is why water is used as a coolant and why coastal areas have milder climates. Mercury has the lowest specific heat of common liquids (≈ 140 J/kg·K).
🤔 TRICKY QUESTIONS
Calorimetry — Unexpected Answers
T1. 1 kg of ice at −10°C is mixed with 1 kg of water at 10°C. What is the final state? (c_ice = 2100, c_water = 4200, L_f = 3.36×10⁵ J/kg)
Final state: water at 0°C (not all ice melts).
Heat needed to warm ice to 0°C: Q₁ = 1×2100×10 = 21,000 J.
Heat released by water cooling to 0°C: Q₂ = 1×4200×10 = 42,000 J.
Surplus heat = 42,000 − 21,000 = 21,000 J → used to melt ice.
Ice that melts = 21,000 / 3.36×10⁵ = 0.0625 kg (only 62.5 g melts).
Final state: mixture of ice (937.5 g) + water (1062.5 g) at 0°C.
T2. Why does a burn from steam at 100°C cause more severe injury than a burn from boiling water at 100°C, even though both are at the same temperature?
Steam releases extra latent heat on condensing.
When steam at 100°C contacts skin, it first condenses to water at 100°C — releasing L_vaporisation = 2.26×10⁶ J/kg. Only then does the water cool further. Boiling water at 100°C only releases Q = mcΔT as it cools. The condensation step releases 5–6× more heat than the sensible cooling, causing far worse burns.
4. Kinetic Theory of Gases
4.1
Basic Postulates & Gas Pressure
A molecular model that explains all macroscopic gas behaviour
The kinetic theory models a gas as an enormous number of tiny molecules in random, continuous motion. From this simple picture, all gas laws emerge naturally, and temperature is understood as a measure of average kinetic energy.
📋 Key Postulates
Gas molecules are point masses (negligible volume)
Molecules move randomly in all directions
Collisions between molecules are perfectly elastic
No intermolecular forces except during collision
Time of collision much smaller than time between collisions
All directions of motion are equally probable
📈 Temperature & Kinetic Energy
Temperature is a measure of average KE of molecules
Average KE = (3/2)kT per molecule
k = Boltzmann constant = 1.38×10⁻²³ J/K
At 0 K: molecular motion stops (absolute zero)
All gases have same average KE at same T
⚡ Kinetic Theory Formulae
Pressure of gas: P = (1/3) ρ v_rms² = (1/3)(mN/V)v_rms²
RMS speed: v_rms = √(3RT/M) = √(3kT/m)
Average speed: v_avg = √(8RT/πM) ≈ 0.92 × v_rms
Most probable: v_mp = √(2RT/M) ≈ 0.82 × v_rms
Speed order: v_mp < v_avg < v_rms (always remember this order!)
Average KE per molecule: E_k = (3/2)kT
Average KE per mole: E_k = (3/2)RT
Degrees of freedom (f):
Monoatomic (He, Ar): f = 3 KE = (3/2)kT
Diatomic (H₂, O₂, N₂): f = 5 KE = (5/2)kT
Triatomic (H₂O, CO₂): f = 6 KE = (6/2)kT = 3kT
R = 8.314 J/mol·K (Universal gas constant)
k = 1.38 × 10⁻²³ J/K (Boltzmann constant)
R = N_A × k (N_A = Avogadro's number = 6.022 × 10²³ /mol)
v_rms > v_avg > v_mp. All three speeds are proportional to √T and proportional to 1/√M (lighter gas = faster). At same temperature, lighter gas molecules move faster.
5. Gas Laws
5.1
Boyle's, Charles's & Combined Gas Law
The macroscopic rules governing ideal gas behaviour — highly tested in NDA
⚡ Gas Laws — All-in-One
Boyle's Law (constant T):
PV = constant → P₁V₁ = P₂V₂
Pressure ∝ 1/Volume (at constant temperature)
Charles's Law (constant P):
V/T = constant → V₁/T₁ = V₂/T₂
Volume ∝ Temperature (T must be in Kelvin!)
Gay-Lussac's Law (constant V):
P/T = constant → P₁/T₁ = P₂/T₂
Pressure ∝ Temperature (at constant volume)
Combined Gas Law:
P₁V₁/T₁ = P₂V₂/T₂
Ideal Gas Equation:
PV = nRT (n = moles, R = 8.314 J/mol·K)
PV = NkT (N = number of molecules, k = 1.38×10⁻²³ J/K)
STP (Standard Temperature & Pressure):
T = 0°C = 273 K, P = 1 atm = 101.325 kPa
1 mole of ideal gas at STP occupies 22.4 litres
Critical rule: In all gas law calculations, temperature MUST be in Kelvin (K = °C + 273). Using Celsius in Charles's law or combined gas law gives wrong answers. This is the most common student error in NDA.
Fig. 3 — Left: Boyle's Law — P–V isotherms (hyperbolas) at different temperatures. Right: Charles's Law — V–T straight lines at different pressures. All lines extrapolate to V = 0 at 0 K.
⚠ NDA Exam Trap — Always use Kelvin! Charles's law and combined gas law require temperature in Kelvin. If a problem says "0°C", convert to 273 K before substituting. Substituting 0°C directly gives V = 0, which is nonsensical. This is the single most common error in gas law problems.
📝 TOPIC-WISE PYQ
Gas Laws & Kinetic Theory — NDA Pattern Questions
Q1. A gas at 27°C and 1 atm is compressed to half its volume at constant temperature. Its new pressure is:
(a) 0.5 atm (b) 1 atm (c) 2 atm (d) 4 atm
Answer: (c) 2 atm
Boyle's law (constant T): P₁V₁ = P₂V₂. 1 × V = P₂ × (V/2). P₂ = 2 atm. Halving volume doubles pressure at constant temperature.
Q2. At what temperature will the rms speed of gas molecules be double that at 27°C?
Q3. The kinetic energy of one mole of an ideal monoatomic gas at temperature T is:
(a) (1/2)RT (b) (3/2)RT (c) (5/2)RT (d) 3RT
Answer: (b) (3/2)RT
Monoatomic gas: 3 degrees of freedom. KE per mole = (f/2)RT = (3/2)RT. For diatomic gas (5 DOF): KE = (5/2)RT. This is a very frequent NDA question.
🤔 TRICKY QUESTIONS
Gas Laws — Beyond Substitution
T1. A balloon is filled with hydrogen and another with oxygen, both at the same temperature and pressure. Which gas molecules move faster and by how much? (M_H₂ = 2, M_O₂ = 32)
Hydrogen molecules move 4 times faster.
v_rms ∝ 1/√M (at same T). v_H₂/v_O₂ = √(M_O₂/M_H₂) = √(32/2) = √16 = 4. Lighter gas = faster molecules. Same temperature = same average KE, but KE = ½mv² → lighter mass needs higher v for same KE.
T2. An ideal gas is enclosed in a rigid container. If temperature is increased by 100%, what happens to pressure?
Pressure also increases by 100% (doubles).
Rigid container → constant volume. By Gay-Lussac's law: P/T = constant. If T increases by 100%, new T = 2T (in Kelvin). So P₂ = P₁ × (2T/T) = 2P₁. Pressure doubles. Note: 100% increase means the new value is 2 times the original.
6. First Law of Thermodynamics
6.1
ΔU = Q − W — Energy Conservation for Thermal Systems
Heat added to a system either raises internal energy or does work
The First Law of Thermodynamics is simply the law of conservation of energy applied to thermal systems. It connects the heat supplied to a system with the change in its internal energy and the work done by the system.
⚡ First Law & Thermodynamic Processes
First Law: ΔU = Q − W
ΔU = change in internal energy (J)
Q = heat added to system (positive if absorbed, negative if released)
W = work done BY the system (positive if system expands)
Work done by gas: W = P × ΔV (for constant pressure process)
Four Thermodynamic Processes:
1. Isothermal (constant T):
ΔU = 0 (ideal gas: internal energy depends only on T)
∴ Q = W (all heat absorbed = work done by gas)
PV = constant (Boyle's law holds)
2. Adiabatic (no heat exchange, Q = 0):
ΔU = −W (work done at expense of internal energy)
System cools when it expands, heats when compressed
PV^γ = constant (γ = C_p/C_v)
3. Isochoric (constant volume, ΔV = 0):
W = 0 (no expansion → no work done)
ΔU = Q (all heat goes into raising internal energy)
4. Isobaric (constant pressure):
W = PΔV (work done by expanding gas)
ΔU = Q − PΔV
Sign convention: Q positive = heat absorbed by system; W positive = work done by system. Internal energy U of an ideal gas depends only on temperature (not on P or V separately).
Process
Condition
Q
W
ΔU
Example
Isothermal
T = const
= W
≠ 0
= 0
Slow expansion in heat bath
Adiabatic
Q = 0
= 0
= −ΔU
= −W
Sudden compression, diesel engine
Isochoric
V = const
= ΔU
= 0
= Q
Heating gas in rigid container
Isobaric
P = const
= ΔU + PΔV
= PΔV
= Q − W
Heating gas in a piston at atm pressure
💡 Real-life examples to remember: Adiabatic compression occurs in a diesel engine — the rapid compression raises temperature enough to ignite fuel (no spark plug needed). A refrigerator uses adiabatic expansion to cool. Pumping a bicycle tyre quickly heats the pump cylinder (adiabatic compression).
📝 TOPIC-WISE PYQ
First Law of Thermodynamics — NDA Pattern Questions
Q1. In an isothermal process, the internal energy of an ideal gas:
(a) Increases (b) Decreases (c) Remains constant (d) First increases then decreases
Answer: (c) Remains constant
Isothermal means constant temperature. For an ideal gas, internal energy depends only on temperature. Since T doesn't change, ΔU = 0. All heat absorbed equals work done by the gas: Q = W.
Q2. In which process is no work done by or on the gas?
Answer: (d) Isochoric
Isochoric = constant volume. Work done W = PΔV = P×0 = 0. No volume change means no work done (no expansion or compression). All heat supplied goes directly into raising internal energy.
Q3. 500 J of heat is supplied to a gas, and the gas does 200 J of work. The change in internal energy is:
(a) 700 J (b) 300 J (c) 200 J (d) 500 J
Answer: (b) 300 J
First Law: ΔU = Q − W = 500 − 200 = 300 J. The internal energy increases by 300 J — the remaining 200 J was used to do work (expand against external pressure).
🤔 TRICKY QUESTIONS
Thermodynamics — Conceptual Challenges
T1. A gas is compressed adiabatically. Does its temperature increase, decrease, or remain the same? Why?
Temperature increases.
Adiabatic: Q = 0. First law: ΔU = Q − W = 0 − W = −W. During compression, work is done ON the gas (W is negative for work done by gas). So ΔU = −(negative) = positive. Internal energy increases → temperature increases. Example: diesel engine ignites fuel by rapid adiabatic compression alone.
T2. Can a gas do work without absorbing heat? Can heat be absorbed without doing work?
Yes to both — these are adiabatic and isochoric processes respectively. Adiabatic expansion: Q = 0, but gas does work W > 0. Energy comes from internal energy: ΔU = −W (gas cools). Example: air released from tyre becomes cold. Isochoric heating: W = 0 (V constant), but heat Q is absorbed entirely as ΔU. No work done at all. Example: heating gas in a sealed rigid container.
⚡ High-Yield Formula Sheet — PN03 Heat & Thermodynamics
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