NCERT Plus One (Class 11) Physics • Chapter 10

Thermal Properties of Matter

Temperature scales, absolute zero, thermal expansion, water anomaly, calorimetry, latent heat, heat transfer & Newton cooling.

📄 Official NCERT Textbook PDF

Topic 10.2 – 10.4: Temperature Scales & Ideal-Gas Absolute Zero

When a clinic thermometer in Kochi reads 100.4 °F, you have a mild fever — yet on the Celsius weather report outside, 38 °C feels like a blazing afternoon! Meanwhile, scientists track liquid nitrogen at 77 K. Why do Celsius, Fahrenheit, and Kelvin all use different numbers for the exact same warmth?
HEAT vs TEMPERATUREHeat = energy flowing due to ΔT (Joule)  |  Temperature = degree of hotness!
CONVERSION FORMULAt_C100 = t_F − 32180 = T − 273.15100  ⇒  t_F = 95t_C + 32
ABSOLUTE ZERO & TRIPLE POINT0 K = −273.15 °C (universal gas zero!)  |  Water Triple Point = 273.16 K (0.01 °C)
🧠 Memory Trick: Step size of 1 °C = 1 K = 1.8 °F! And Celsius meets Fahrenheit at −40 °C = −40 °F!
💡 Simple Kid-Check: To invent any temperature ruler, pick two anchor points (ice melting and water boiling) and chop the gap into equal steps! Celsius cuts it into 100 steps (0° to 100°), Fahrenheit cuts it into 180 steps (32° to 212°), and Kelvin simply shifts the zero all the way down to Nature's coldest floor: −273.15 °C!
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    Ideal Gas Equation: For dilute gases, PV = μRT (R = 8.314 J mol⁻¹ K⁻¹). At constant volume, pressure rises linearly with temperature (P ∝ T).
  • 2
    Finding Absolute Zero: Extrapolating the straight P–T lines of Oxygen, Nitrogen, and Helium backward hits zero pressure at the EXACT same point for every gas: −273.15 °C = 0 K!
  • 3
    Modern Fixed Point: Instead of boiling/ice points that shift with air pressure, the Kelvin scale uses the unique Triple Point of Water (273.16 K) where ice, liquid water, and vapour coexist!
T = t_C + 273.15  |  t_F = 95 t_C + 32  |  General unknown scale: T − LFPUFP − LFP = constant
🧒 Super-Easy Example: Find the temperature where Celsius and Fahrenheit read the exact same number x:
Set x = 95 x + 32 ⇒ x − 95 x = 32 ⇒ −45 x = 32 ⇒ x = −40°!
Watch TWO classic 1-mark traps! (1) A temperature difference of 20 °C is equal to 20 K (NEVER add 273.15 to a change ΔT!), and in Fahrenheit Δt_F = (9/5) × 20 = 36 °F! (2) Never put a degree circle on Kelvin — write 300 K, NOT 300 °K!

Topic 10.5: Thermal Expansion (Linear, Area, Volume & Water Anomaly)

Look closely at steel railway tracks in Palakkad summer heat — tiny gaps are deliberately left between rails so they don't buckle! Even stranger: while almost everything expands on heating, chilled water between 0 °C and 4 °C actually SHRINKS, keeping winter fish alive under frozen lakes!
3 EXPANSION LAWSΔl/l = α_l ΔT  |  ΔA/A = α_a ΔT  |  ΔV/V = α_v ΔT
GOLDEN RATIO 1 : 2 : 3For isotropic solids: α_l : α_a : α_v = 1 : 2 : 3 (α_a = 2α_l, α_v = 3α_l)!
WATER ANOMALY (0–4 °C)Water contracts from 0 °C → 4 °C! Max density (1000 kg/m³) & Min volume at 4 °C!
🧠 Memory Trick: Heating expands EVERYTHING like a zoom-in photo — even a hole in a metal washer gets BIGGER! And remember Water hits peak density at 4 °C!
💡 Simple Kid-Check: Heating makes atoms jiggle harder and push their neighbours slightly farther apart. Glue a brass strip (expands more) to a steel strip (expands less) to build a bimetallic strip: when heated, it bends with brass on the outside curve — running automatic iron-box and toaster switches!
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    Why α_v = 3 α_l: A cube of side l expands to l(1 + α_l ΔT). Cubing gives l³(1 + 3α_l ΔT + ...) for small α_l ⇒ α_v = 3 α_l (and similarly α_a = 2 α_l).
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    Thermal Stress: If a rod is clamped rigidly at both ends so it cannot expand by Δl = l α_l ΔT, massive compressive stress builds up: Thermal Stress = Y α_l ΔT!
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    Anomalous Expansion of Water: Below 4 °C, ice-like open hydrogen-bonded cages form, so density drops below 4 °C! As lake surfaces freeze at 0 °C, lighter cold water & ice float on top while dense 4 °C water stays warm at the bottom, saving aquatic life!
Δl = l α_l ΔT  |  α_v = 3 α_l  |  Thermal Stress = Y α_l ΔT  |  Ideal Gas at const P: α_v = 1T
🧒 Super-Easy Example: A steel rail of length 20 m (α_l = 1.2 × 10⁻⁵ K⁻¹) warms from 20 °C to 45 °C (ΔT = 25 K).
Expansion Δl = 20 × (1.2 × 10⁻⁵) × 25 = 6 × 10⁻³ m = 6 mm!
Watch TWO favourite Kerala HSE traps! (1) When a flat metal plate with a circular hole is heated, does the hole shrink or grow? The hole EXPANDS at the exact same rate as a solid metal disc of the same material! (2) For an ideal gas at constant pressure, PV = μRT gives volume expansion coefficient α_v = 1 / T (in Kelvin!) — NOT a fixed constant!

Topic 10.6 – 10.7: Specific Heat Capacity & Calorimetry Principle

Place an empty aluminium pan and a pot full of water over identical gas flames in your kitchen — in 15 seconds the empty pan blisters your fingers while the water is barely warm! Water has one of Nature's highest specific heat capacities (4186 J kg⁻¹ K⁻¹), acting like a giant thermal sponge that keeps Kerala's coasts mild!
SPECIFIC HEAT (s)ΔQ = m s ΔT  |  s = ΔQm ΔT (SI: J kg⁻¹ K⁻¹)
WATER & CALORIEs_water = 4186 J kg⁻¹ K⁻¹ = 1 cal g⁻¹ °C⁻¹  (1 cal = 4.186 J)
CALORIMETRY RULEIn an insulated vessel: Heat Lost by Hot = Heat Gained by Cold!
🧠 Memory Trick: Specific heat s = "Stubbornness to warm up!" Water's huge s = 4186 makes it the king of coolants & coastal breezes!
💡 Simple Kid-Check: Think of temperature as water level in a bucket and Heat Q as poured water! A narrow bucket (small s, like copper 386) fills up fast with tiny heat; a super-wide swimming pool (huge s, water 4186) swallows tons of heat while its temperature barely nudges!
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    Three Heat Capacities:
    • Heat capacity of whole body: S = ΔQ / ΔT = m s (J/K).
    • Specific heat capacity (per kg): s = ΔQ / (m ΔT) (J kg⁻¹ K⁻¹).
    • Molar specific heat (per mole): C = ΔQ / (μ ΔT) (J mol⁻¹ K⁻¹).
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    Gas Specific Heats Hint: Gases have TWO molar specific heats: at constant volume (C_v) and at constant pressure (C_p > C_v, because expanding gas also does external work!).
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    Principle of Calorimetry: When hot body 1 (m₁, s₁, T₁) mixes with cold body 2 (m₂, s₂, T₂) inside an insulated calorimeter (m_c, s_c), heat stops flowing at equilibrium T_mix:
    m₁ s₁ (T₁ − T_mix) = m₂ s₂ (T_mix − T₂) + m_c s_c (T_mix − T₂)!
Q = m s ΔT  |  Mixture (neglecting cup): T_mix = m₁ s₁ T₁ + m₂ s₂ T₂m₁ s₁ + m₂ s₂
🧒 Super-Easy Example: Drop 0.2 kg of hot Iron (s = 450 J/kg·K) at 100 °C into 0.1 kg of Water (s = 4200 J/kg·K) at 25 °C.
0.2 × 450 × (100 − T) = 0.1 × 4200 × (T − 25) ⇒ 90(100 − T) = 420(T − 25) ⇒ 9000 − 90T = 420T − 10500 ⇒ 510T = 19500 ⇒ T = 38.2 °C!
Watch TWO classic numerical traps! (1) Always check if the question mentions the copper calorimeter's own mass and specific heat — if given, the calorimeter ALSO gains heat m_c s_c (T_mix − T₂) alongside the water! (2) Remember 1 calorie = 4.186 J ≈ 4.2 J, and watch grams vs kilograms when mixing cal/g·°C with J/kg·K!

Topic 10.8: Change of State, Latent Heat & Regelation

Drop ice cubes into a glass or steam idlis in a pressure cooker: while ice melts at 0 °C or water boils at 100 °C, the thermometer needle freezes completely still even though the burner keeps roaring! Where does all that extra burner heat vanish? It hides inside broken molecular bonds as Latent Heat!
LATENT HEAT (L)Q = m L at CONSTANT temperature! (Unit: J/kg or cal/g)
WATER CONSTANTSL_f (ice→water) = 3.35 × 10⁵ J/kg (80 cal/g)  |  L_v (water→steam) = 2.26 × 10⁶ J/kg (540 cal/g)
PRESSURE & REGELATIONHigh P lowers ice melting point (Regelation!), yet raises water boiling point (Pressure cooker!)
🧠 Memory Trick: Heating curve slants = Q = msΔT (speeding molecules); flat plateaus = Q = mL (breaking bonds, constant T)!
💡 Simple Kid-Check: "Latent" literally means hidden! During melting or boiling, incoming heat stops raising molecule speed (so thermometer stays flat!) and instead pays the "breakaway fee" to unlock solid ice crystals into free liquid, or liquid water into escaping steam!
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    5-Stage Heating Curve of Water: (1) Ice warms → 0 °C (ms_i ΔT); (2) Ice melts flat at 0 °C (m L_f); (3) Water warms 0 → 100 °C (ms_w ΔT); (4) Water boils flat at 100 °C (m L_v); (5) Steam heats above 100 °C. Notice L_v plateau is nearly 6.7× longer than L_f!
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    Regelation: Hang weights on a thin copper wire looped over a slab of ice. High pressure right under the wire lowers ice's melting point, melting it; once the wire passes through, water above refreezes — slicing cleanly without splitting the block in two! (Ice skating works similarly!)
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    Pressure & Boiling: At hills like Munnar/Ooty, lower air pressure drops water's boiling point below 100 °C (slowing cooking!). A pressure cooker traps steam, raising pressure and boiling point (~120 °C) to cook dal and rice super fast!
Phase Change: Q = m L  |  Steam vs Water at 100 °C: Steam carries +2.26 × 10⁶ J/kg extra latent heat!
🧒 Super-Easy Example: How much heat turns 10 g of ice at 0 °C completely into water at 50 °C? (Use L_f = 80 cal/g, s_w = 1 cal/g·°C)
• Melt ice at 0 °C: Q₁ = 10 × 80 = 800 cal.
• Warm water 0 → 50 °C: Q₂ = 10 × 1 × 50 = 500 cal.
• Total = 800 + 500 = 1300 cal (≈ 5442 J)!
Two classic Kerala HSE 2-mark questions! (1) Why are burns from steam at 100 °C much more severe than burns from boiling water at 100 °C? Because every gram of steam first condenses on skin, dumping an extra 2260 J (540 cal) of hidden latent heat of vaporisation BEFORE cooling! (2) On a T vs Q graph, smaller slope on a slanted line means larger specific heat (slope = 1/(ms)), and longer horizontal line means larger latent heat!

Topic 10.9: Conduction, Convection & Blackbody Radiation (Wien & Stefan)

Stir hot sambar with a steel spoon and heat climbs straight to your hand (conduction). Watch boiling rasam swirl from bottom to top (convection). Now stand outside in tropical sunlight — how does warmth cross 150 million km of empty space with zero air? Through electromagnetic radiation!
CONDUCTION (SOLIDS)H = dQdt = K A T₁ − T₂L  (No matter moves! Metals high K)
WIEN'S LAWλ_m T = b = 2.9 × 10⁻³ m·K  (Hotter star ⇒ shorter/bluer peak wavelength λ_m!)
STEFAN-BOLTZMANN LAWH = e σ A T⁴  |  Net: e σ A (T⁴ − T_s⁴)  (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)
🧠 Memory Trick: Conduction = Pass the parcel (atoms stay put); Convection = Hot rises, cold sinks (fluids move!); Radiation = Speed-of-light waves (needs NO medium, scales as T⁴ in Kelvin)!
💡 Simple Kid-Check: Look at glowing charcoal vs a welding arc vs stars: cool embers glow dull red (long wavelength λ_m), Sun glows yellow-white (λ_m ≈ 4753 Å → ~6000 K), and blazing hot stars shine fierce blue (tiny λ_m)! That is Wien's Law (λ_m T = constant)!
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    Steady-State Conduction: Across a rod of length L, area A, conductivity K: heat current H = K A (T₁ − T₂) / L. Thermal resistance R_th = L / (K A) adds in series just like electrical resistors!
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    Convection (Natural & Forced): Warm fluid expands, becomes less dense, and rises while cooler dense fluid sinks — creating Kerala's cooling afternoon sea breeze (land heats faster than sea!) and nighttime land breeze!
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    Blackbody Radiation: Every body above 0 K radiates EM waves! Perfect blackbody (emissivity e = 1) emits total power H = σ A T⁴; real bodies emit e σ A T⁴. (Black/rough surfaces are best emitters AND best absorbers!)
Conduction: H = K A ΔTL  |  Wien: λ_m T = 2.9 × 10⁻³ m·K  |  Stefan: H = e σ A T⁴
🧒 Super-Easy Example: If a glowing blackbody filament is heated from 1000 K to 2000 K (2× Kelvin temperature):
• By Wien's Law, peak wavelength λ_m becomes HALF (shifts toward blue!).
• By Stefan's Law, radiated power H ∝ T⁴ jumps 2⁴ = 16 TIMES!
Critical Stefan's Law trap! In H = e σ A T⁴, T MUST be in KELVIN, never Celsius! If an exam heats a body from 27 °C (300 K) to 327 °C (600 K), absolute temperature doubled — so radiation shoots up 2⁴ = 16× (NOT (327/27)⁴!). Also remember: good absorbers are always good emitters!

Topic 10.10: Newton's Law of Cooling

Pour a steaming cup of Kumbakonam filter coffee at 90 °C in a 25 °C classroom: in the first two minutes it cools rapidly by 10 °C, yet going from 35 °C to 25 °C feels like forever! Why does cooling speed slow down as coffee approaches room temperature?
NEWTON'S COOLING LAW−dTdt = K (T − T_s)  (Cooling rate ∝ excess temp over surroundings!)
EXPONENTIAL DECAYT(t) = T_s + (T₀ − T_s) e^(−Kt)  |  Plot ln(T − T_s) vs t = straight line!
EXAM NUMERICAL FORMULAT₁ − T₂t = K (T₁ + T₂2 − T_s)  (Use AVERAGE temp!)
🧠 Memory Trick: Huge temperature gap = steep cooling slide! Divide two cooling intervals using (T_avg − T_s) and constant K cancels straight out!
💡 Simple Kid-Check: Hot coffee has a huge temperature gap (90 − 25 = 65 °C) above the room, so heat escapes furiously! Lukewarm coffee only has a tiny gap (30 − 25 = 5 °C), so heat leaks lazily!
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    Differential Law: Rate of heat loss −dQ/dt = k(T − T_s). Since dQ = m s dT, rate of temperature drop is −dT/dt = K(T − T_s) where K = k/(ms).
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    Linear Graph Check: Taking natural logs gives ln(T − T_s) = −Kt + C, a downward straight line of slope −K used in lab verification!
  • 3
    Two-Interval Ratio Shortcut (NCERT Exam Favourite!): Always write (T₁ − T₂)/t = K [ (T₁ + T₂)/2 − T_s ] for both intervals and divide one equation by the other!
T₁ − T₂t = K (T₁ + T₂2 − T_s)  |  Valid when (T − T_s) is small!
🧒 NCERT Example 10.8: Hot food cools from 94 °C → 86 °C in 2 min in a 20 °C room. How long to cool from 71 °C → 69 °C?
• Interval 1: 8 / 2 = K (90 − 20) = 70 K.
• Interval 2: 2 / t = K (70 − 20) = 50 K.
• Dividing: (4) / (2/t) = 70 / 50 ⇒ 2t = 1.4 ⇒ t = 0.7 min = 42 s!
Watch TWO classic errors in cooling numericals! First, students plug the starting temperature T₁ instead of the average temperature (T₁ + T₂)/2 minus room temperature T_s! Second, Newton's Law of Cooling holds strictly when the temperature excess (T − T_s) is small (and airflow is steady) so T⁴ − T_s⁴ stays roughly linear!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): Convert −40 °C to Fahrenheit and Kelvin. What is meant by absolute zero on the ideal-gas scale?

• Fahrenheit: t_F = (9/5)(−40) + 32 = −72 + 32 = −40 °F.
• Kelvin: T = −40 + 273.15 = 233.15 K.
Absolute Zero (0 K = −273.15 °C): The universal lowest temperature obtained by extrapolating constant-volume ideal gas P–T lines to zero pressure, where substances have minimum possible molecular activity.

Q2 (Kerala HSE March • 3 Marks): Define coefficients of linear expansion (α_l) and volume expansion (α_v) of a solid. Prove that α_v = 3 α_l.

α_l = Δl / (l ΔT) and α_v = ΔV / (V ΔT).
Proof: Take a cube of side l, volume V = l³. Heating by ΔT increases each side to l' = l(1 + α_l ΔT).
New volume: V + ΔV = l³(1 + α_l ΔT)³ ≈ l³(1 + 3 α_l ΔT) (neglecting higher powers of small α_l).
Thus ΔV/V = 3 α_l ΔT ⇒ α_v = 3 α_l.

Q3 (Kerala HSE Say • 2 Marks): Draw a density vs temperature graph for water between 0 °C and 10 °C. Explain how this anomalous behaviour protects aquatic life in cold countries.

Density rises from 0 °C to a sharp peak at 4 °C (1000 kg/m³), then falls above 4 °C.
In winter, surface water cools to 4 °C and sinks; further cooling below 4 °C stays lighter at the top and freezes into floating ice (0 °C), insulating the 4 °C liquid below where fish survive.

Q4 (Kerala HSE Model • 2 Marks): Why does temperature remain constant during melting or boiling even though heat is continuously supplied? Why are steam burns worse than boiling water burns?

1. During phase change, supplied heat breaks intermolecular bonds (increasing potential energy) rather than increasing average molecular kinetic energy, so temperature remains fixed.
2. Steam at 100 °C carries extra latent heat of vaporisation (2.26 × 10⁶ J/kg = 540 cal/g), which it releases onto skin during condensation before cooling.

Q5 (NCERT / DHSE Model • 3 Marks): A 0.047 kg aluminium sphere is heated in boiling water (100 °C) and quickly transferred into a 0.14 kg copper calorimeter containing 0.25 kg of water at 20 °C. The final steady temperature is 23 °C. Find the specific heat capacity of aluminium. (s_water = 4186, s_Cu = 386 J/kg·K)

Heat lost by Al = 0.047 × s_Al × (100 − 23) = 3.619 s_Al.
Heat gained by (Water + Cu cup) = [0.25 × 4186 + 0.14 × 386] × (23 − 20) = (1046.5 + 54.04) × 3 = 3301.6 J.
Equating: s_Al = 3301.6 / 3.619 ≈ 912 J kg⁻¹ K⁻¹.

Q6 (Kerala HSE March • 2 Marks): What is regelation? Mention one daily-life or physics experiment illustrating it.

Regelation: Melting of ice under increased pressure below 0 °C and refreezing once pressure is removed.
Illustration: Looping a weighted copper wire over an ice slab causes the wire to melt through beneath it and refreeze above, passing completely through without splitting the ice block.

Q7 (Kerala HSE March • 2 Marks): State Wien's displacement law and Stefan-Boltzmann law for blackbody radiation. A blackbody emits peak wavelength 5000 Å at 6000 K; find its peak wavelength at 3000 K.

• Wien's Law: λ_m T = b = 2.9 × 10⁻³ m·K.
• Stefan-Boltzmann Law: Radiant power emitted per unit area E = σ T⁴ (H = e σ A T⁴).
• Numerical: λ₁ T₁ = λ₂ T₂ ⇒ λ₂ = 5000 × (6000 / 3000) = 10,000 Å (1 μm, Infrared).

Q8 (Kerala HSE Say • 3 Marks): State Newton's law of cooling. A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when room temperature is 20 °C. How long will it take to cool from 71 °C to 69 °C?

Law: Rate of heat loss is proportional to excess temperature above surroundings (−dQ/dt = k(T − T_s)) for small differences.
Using (T₁ − T₂)/t = K[(T₁ + T₂)/2 − T_s]:
• Case 1: (94 − 86)/2 = K(90 − 20) ⇒ 4 = 70 K.
• Case 2: (71 − 69)/t = K(70 − 20) ⇒ 2/t = 50 K.
Dividing: 2t = 70/50 = 1.4 ⇒ t = 0.7 min = 42 seconds.