Thermal Properties of Matter | AAI ATC Physics Ch.10

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Section 10.2 & 10.3

Temperature, Heat & Measurement

Heat is the form of energy transferred between two systems by virtue of a temperature difference. SI unit of heat is Joule (J); SI unit of temperature is Kelvin (K).

Temperature scales: Celsius (°C) and Fahrenheit (°F) are related by:

(t_F − 32) / 180 = t_C / 100

Fahrenheit-Celsius conversion | Ice point: 0°C = 32°F | Steam point: 100°C = 212°F

Thermometers use physical properties (like liquid volume) that change with temperature. Two fixed reference points needed to define a temperature scale.

🖼️ Insert Diagram: Fahrenheit vs Celsius temperature graph (linear relationship) Recommended: Fig 10.1 from NCERT — Plot of tF versus tC

🎯 Practice MCQs — Temperature & Heat

AAI ATC Style

1 The temperature 37°C (normal human body temperature) on the Fahrenheit scale is:

A 96.8°F

B 98.6°F

C 100°F

D 37°F

✅ t_F = (9/5)×t_C + 32 = (9/5)×37 + 32 = 66.6 + 32 = 98.6°F. Always use: t_F = 1.8×t_C + 32 for quick conversion.

2 At what temperature do the Celsius and Fahrenheit scales show the same numerical reading?

A 0°

B 100°

C −40°

D −273°

✅ Set t_F = t_C = x: x = (9/5)x + 32 → x − (9/5)x = 32 → −(4/5)x = 32 → x = −40. Both scales read −40° at this unique crossover point.

3 Temperature 300 K on the Celsius scale is:

A 300°C

B 573°C

C 26.85°C

D −273.15°C

✅ t_C = T − 273.15 = 300 − 273.15 = 26.85°C. Kelvin = Celsius + 273.15. Room temperature (~27°C) = 300 K is a very commonly used value in problems.

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Section 10.4

Ideal Gas Equation & Absolute Temperature

The Ideal Gas Equation combines Boyle's Law (PV = constant at fixed T) and Charles' Law (V/T = constant at fixed P):

PV = μRT

P = Pressure | V = Volume | μ = Number of moles | R = 8.31 J mol⁻¹ K⁻¹ | T = Absolute temperature (K)

Absolute Zero = −273.15°C = 0 K — the temperature at which an ideal gas would have zero pressure and volume (theoretically). This is the foundation of the Kelvin scale.

T (K) = t_C + 273.15

Kelvin and Celsius scales have the same size of unit but different origins

🖼️ Insert Diagram: Comparison of Kelvin, Celsius, and Fahrenheit scales with fixed points Recommended: Fig 10.4 from NCERT — Temperature scale comparison

🎯 Practice MCQs — Ideal Gas & Absolute Temperature

AAI ATC Style

1 2 moles of an ideal gas occupy 44.8 L at 0°C. If temperature is raised to 273°C at constant pressure, the new volume is:

A 44.8 L

B 89.6 L

C 22.4 L

D 67.2 L

✅ V₁/T₁ = V₂/T₂ → 44.8/273 = V₂/546 → V₂ = 44.8 × 2 = 89.6 L. T₁ = 273 K, T₂ = 273 + 273 = 546 K. Temperature doubles → volume doubles (Charles' Law).

2 The universal gas constant R = 8.31 J mol⁻¹ K⁻¹. The pressure exerted by 1 mole of an ideal gas at 27°C in a volume of 24.9 L is approximately:

A 0.5 × 10⁵ Pa

B 1.0 × 10⁵ Pa

C 2.0 × 10⁵ Pa

D 8.31 × 10³ Pa

✅ P = μRT/V = 1 × 8.31 × 300 / (24.9 × 10⁻³) = 2493 / 0.0249 ≈ 1.0 × 10⁵ Pa ≈ 1 atm. At 27°C (300 K), 1 mole of ideal gas in ~24.9 L exerts approximately 1 atmosphere of pressure.

3 Absolute zero temperature on the Fahrenheit scale is:

A −273°F

B 0°F

C −459.67°F

D −212°F

✅ t_F = (9/5)(−273.15) + 32 = −491.67 + 32 = −459.67°F. Absolute zero = 0 K = −273.15°C = −459.67°F. This is shown in Fig 10.4 of NCERT.

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Section 10.5

Thermal Expansion

Most substances expand on heating. Three types of thermal expansion:

Δl/l = αₗ ΔT | ΔA/A = 2αₗ ΔT | ΔV/V = αᵥ ΔT

αₗ = Coefficient of linear expansion | αᵥ = Coefficient of volume expansion | αᵥ = 3αₗ

Anomalous expansion of water: Water contracts on heating from 0°C to 4°C. Maximum density of water = at 4°C (1000 kg/m³). This is why lakes freeze from the top — crucial for aquatic life.

Material	αₗ (×10⁻⁵ K⁻¹)	Material	αₗ (×10⁻⁵ K⁻¹)
Aluminium	2.5	Gold	1.4
Brass	1.8	Glass (pyrex)	0.32
Iron	1.2	Lead	0.29
Copper	1.7	Silver	1.9

      🔑 AAI ATC Exam Tip: Thermal stress in a constrained rod = Y × αₗ × ΔT (where Y = Young's modulus). Railway tracks have gaps to allow thermal expansion. At high altitudes, metal parts in aircraft experience significant temperature changes — thermal expansion is a critical design consideration.
    

🖼️ Insert Diagram: Linear, Area, and Volume Expansion comparison diagram Recommended: Fig 10.5 from NCERT — Thermal Expansion types

🎯 Practice MCQs — Thermal Expansion

AAI ATC Style

1 A blacksmith heats an iron ring (L = 5.231 m at 27°C) to fit a wooden wheel of diameter 5.243 m. The coefficient of linear expansion of iron = 1.2 × 10⁻⁵ K⁻¹. The required temperature is:

A 100°C

B 150°C

C 218°C

D 300°C

✅ L_T2 = L_T1[1 + αₗ(T₂−T₁)] → 5.243 = 5.231[1 + 1.2×10⁻⁵(T₂−27)] → T₂ ≈ 218°C. This classic NCERT example shows real-world application of linear expansion in metalworking.

2 The coefficient of volume expansion (αᵥ) is related to coefficient of linear expansion (αₗ) by:

A αᵥ = αₗ

B αᵥ = 2αₗ

C αᵥ = 3αₗ

D αᵥ = αₗ/3

✅ αᵥ = 3αₗ. This comes from the geometry of a cube expanding equally in all three dimensions. Similarly, area expansion coefficient = 2αₗ. These are fundamental relations used in all expansion problems.

3 A steel rail (L = 5 m, A = 40 cm², αₗ = 1.2×10⁻⁵ K⁻¹, Y = 2×10¹¹ N/m²) is prevented from expanding while temperature rises by 10°C. The thermal stress developed is:

A 1.2 × 10⁶ N/m²

B 2.4 × 10⁷ N/m²

C 2.4 × 10⁹ N/m²

D 1.2 × 10⁴ N/m²

✅ Thermal stress = Y × αₗ × ΔT = 2×10¹¹ × 1.2×10⁻⁵ × 10 = 2.4 × 10⁷ N/m². This is why railway tracks are laid with expansion gaps — otherwise this enormous stress would buckle the rails on hot days!

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Section 10.6 & 10.7

Specific Heat Capacity & Calorimetry

Specific heat capacity (s) is the amount of heat per unit mass required to change temperature by 1 K:

s = ΔQ / (m × ΔT)

s = Specific heat (J kg⁻¹ K⁻¹) | ΔQ = Heat added (J) | m = Mass (kg) | ΔT = Temperature change (K)

Calorimetry principle: Heat lost by hot body = Heat gained by cold body (in an isolated system).

m₁s₁ΔT₁ = m₂s₂ΔT₂

Heat lost = Heat gained | Calorimeter principle (no heat escapes)

Substance	s (J kg⁻¹ K⁻¹)	Substance	s (J kg⁻¹ K⁻¹)
Water	4186 (highest)	Ice	2060
Aluminium	900	Iron	450
Copper	386.4	Mercury	140
Silver	236.1	Lead	127.7

      🔑 Why Water is Special: Water has the highest specific heat capacity (4186 J/kg/K) — so it heats and cools slowly. This is why it is used as automobile radiator coolant, why sea breeze exists, and why coastal areas have moderate climates.
    

🎯 Practice MCQs — Specific Heat & Calorimetry

AAI ATC Style

1 A 0.047 kg aluminium sphere at 100°C is transferred to 0.14 kg copper calorimeter containing 0.25 kg water at 20°C. Final temp = 23°C. (s_water = 4186, s_Cu = 386 J/kg/K). The specific heat of aluminium is approximately:

A 386 J/kg/K

B 500 J/kg/K

C 911 J/kg/K

D 4186 J/kg/K

✅ Heat lost = 0.047 × sAl × 77. Heat gained = (0.25×4186 + 0.14×386)×3. Setting equal and solving: sAl ≈ 911 J/kg/K. This is the standard NCERT calorimetry problem. Always set heat lost = heat gained.

2 A 10 kW drilling machine drills an aluminium block of mass 8 kg for 2.5 minutes. If 50% power heats the block (s_Al = 910 J/kg/K), the temperature rise is:

A 50°C

B 75°C

C 103°C

D 206°C

✅ Useful heat = 0.5 × 10000 × (2.5×60) = 750,000 J. ΔT = Q/(ms) = 750000/(8×910) = 750000/7280 ≈ 103°C. Power × time = energy; only 50% is useful.

3 Which substance is used as coolant in automobile radiators because of its high specific heat capacity?

A Mercury

B Copper

C Water

D Glycerine

✅ Water has the highest specific heat capacity (4186 J/kg/K) among common substances. It can absorb a large amount of heat for a small temperature rise, making it ideal as a coolant. It is also used in hot water bags as a heater.

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Section 10.8

Change of State & Latent Heat

During a change of state (melting, vaporisation), temperature remains constant even though heat is being supplied. This heat is called Latent Heat:

Q = mL

L = Latent heat (J kg⁻¹) | Q = Heat required (J) | m = Mass (kg)

For water: Latent heat of fusion (Lf) = 3.33 × 10⁵ J/kg (ice→water at 0°C). Latent heat of vaporisation (Lv) = 22.6 × 10⁵ J/kg (water→steam at 100°C).

Why burns from steam are more severe than boiling water — steam at 100°C carries an extra 22.6 × 10⁵ J/kg compared to water at the same temperature.

📌 Key Facts — Change of State

Boiling point increases with pressure (pressure cooker) and decreases with altitude (cooking harder on hills).
Sublimation: Direct solid→vapour without passing through liquid. Example: dry ice (CO₂), iodine.
Triple point of water: 273.16 K and 6.11×10⁻³ Pa — all three phases coexist.
Regelation: Ice melts under high pressure and refreezes when pressure is removed. This is why ice skating is possible.
Latent heat of vaporisation >> latent heat of fusion for all substances.

🖼️ Insert Diagram: Temperature vs Heat graph for water (showing flat regions during phase change) Recommended: Fig 10.12 from NCERT — Temperature vs heat for water at 1 atm

🎯 Practice MCQs — Change of State & Latent Heat

AAI ATC Style

1 Heat required to convert 3 kg of ice at −12°C to steam at 100°C is: (s_ice = 2100, s_water = 4186 J/kg/K, Lf = 3.35×10⁵, Lv = 2.256×10⁶ J/kg)

A 3.35 × 10⁵ J

B 6.78 × 10⁶ J

C 9.1 × 10⁶ J

D 1.5 × 10⁷ J

✅ Q = Q₁+Q₂+Q₃+Q₄ = m×s_ice×12 + m×Lf + m×s_water×100 + m×Lv = 75600 + 1005000 + 1255800 + 6768000 = 9.1×10⁶ J. Four stages: warm ice → melt ice → warm water → vaporise water.

2 When 0.15 kg of ice at 0°C mixes with 0.30 kg of water at 50°C, final temp = 6.7°C. The latent heat of fusion of ice is: (s_water = 4186 J/kg/K)

A 2.26 × 10⁵ J/kg

B 3.34 × 10⁵ J/kg

C 4.18 × 10⁵ J/kg

D 2.0 × 10⁵ J/kg

✅ Heat lost by water = 0.30×4186×43.3 = 54376 J. Heat gained = 0.15×Lf + 0.15×4186×6.7 = 0.15Lf + 4207. Solving: Lf = (54376−4207)/0.15 = 3.34×10⁵ J/kg.

3 Why does cooking take longer at high altitudes?

A Higher atmospheric pressure raises boiling point

B Lower atmospheric pressure reduces boiling point of water

C Higher altitude means less oxygen for burning fuel

D Latent heat of vaporisation increases at high altitude

✅ At high altitudes, atmospheric pressure is lower → boiling point of water decreases below 100°C. Food cooks at a lower temperature, requiring more time. A pressure cooker works on the opposite principle — higher pressure raises boiling point → food cooks faster.

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Section 10.9

Heat Transfer — Conduction, Convection & Radiation

Three modes of heat transfer:

1. Conduction — heat transfer through molecular collisions without flow of matter (mainly in solids). Rate of heat flow:

H = KA(T_C − T_D) / L

H = Heat current (W) | K = Thermal conductivity (W m⁻¹ K⁻¹) | A = Area | L = Length | ΔT = Temp difference

2. Convection — heat transfer by actual bulk movement of fluid matter. Can be natural (buoyancy-driven) or forced (pump-driven). Sea breeze, land breeze, trade winds are all convection phenomena.

3. Radiation — heat transfer via electromagnetic waves, requires no medium. Stefan-Boltzmann Law for a perfect radiator:

H = AeσT⁴ | σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

H = Power radiated | A = Surface area | e = Emissivity (0–1) | T = Absolute temperature

Wien's Displacement Law: λ_m × T = 2.9 × 10⁻³ m·K. The wavelength of peak radiation decreases as temperature increases (iron glows red → orange → white hot).

Material	K (W m⁻¹ K⁻¹)	Material	K (W m⁻¹ K⁻¹)
Silver	406	Glass wool	0.04
Copper	385	Wood	0.12
Aluminium	205	Air	0.024
Steel	50.2	Water	0.8

🖼️ Insert Diagram: Day and night sea breeze/land breeze convection cycle Recommended: Fig 10.17 from NCERT — Convection cycles (day: sea breeze, night: land breeze)

🎯 Practice MCQs — Heat Transfer

AAI ATC Style

1 A steel rod (L = 15 cm, A_steel = 2×A_copper) and copper rod (L = 10 cm) are connected in series. One end is at 300°C, other at 0°C. K_steel = 50.2, K_copper = 385 W/m/K. The junction temperature is approximately:

A 150°C

B 100°C

C 44.4°C

D 200°C

✅ In steady state, H is same through both rods: K₁A₁(300−T)/L₁ = K₂A₂(T−0)/L₂. With A₁=2A₂: 50.2×2×(300−T)/15 = 385×T/10. Solving gives T ≈ 44.4°C. Copper being much better conductor, the junction stays close to the cold end.

2 A body at 3000 K has emissivity e = 0.4 and surface area 0.3 cm². The power radiated by it is: (σ = 5.67 × 10⁻⁸ W/m²/K⁴)

A 150 W

B 30 W

C 60 W

D 6 W

✅ H = AeσT⁴ = 0.3×10⁻⁴ × 0.4 × 5.67×10⁻⁸ × (3000)⁴ = 0.3×10⁻⁴ × 0.4 × 5.67×10⁻⁸ × 8.1×10¹³ ≈ 60 W. This is the NCERT example of a tungsten lamp filament.

3 The surface temperature of the sun corresponds to peak radiation wavelength λ_m = 4753 Å. Using Wien's law (λ_m×T = 2.9×10⁻³ m·K), the surface temperature is approximately:

A 4000 K

B 5000 K

C 6060 K

D 10000 K

✅ T = 2.9×10⁻³ / λ_m = 2.9×10⁻³ / (4753×10⁻¹⁰) = 2.9×10⁻³ / 4.753×10⁻⁷ ≈ 6060 K. Wien's law allows us to estimate temperature of stars from their peak emission wavelength — a key tool in astrophysics.

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Section 10.10

Newton's Law of Cooling

The rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings (valid for small temperature differences):

−dQ/dt = k(T₂ − T₁)

T₂ = Body temperature | T₁ = Surroundings temperature | k = constant (depends on surface area, nature of surface)

This gives an exponential decay: T₂ = T₁ + C′e^(−Kt). A plot of ln(T₂−T₁) vs time is a straight line with negative slope.

For approximate calculations, we use: Average cooling rate = K × (average temperature excess above surroundings).

🖼️ Insert Diagram: Cooling curve of hot water — temperature vs time (exponential decay) Recommended: Fig 10.19 from NCERT — Curve showing cooling of hot water with time

🎯 Practice MCQs — Newton's Law of Cooling

AAI ATC Style

1 A pan cools from 94°C to 86°C in 2 min when room temp is 20°C. How long will it take to cool from 71°C to 69°C?

A 2 min

B 1 min

C 42 s

D 28 s

✅ Rate₁ = 8°C/2min at avg excess 70°C. Rate₂ = 2°C/time at avg excess 50°C. Using K×ΔT = rate: (8/2)/70 = (2/t)/50 → t = 2×(50/70)/4 × 2 = 0.7 min = 42 s. Classic NCERT problem — use ratio method.

2 A body cools from 80°C to 50°C in 5 minutes. Room temperature is 20°C. Time to cool from 60°C to 30°C is:

A 5 min

B 7 min

C 9 min

D 15 min

✅ Case 1: 30°C drop in 5 min, avg excess = 65−20 = 45°C → K = 30/(5×45) = 2/15. Case 2: 30°C drop, avg excess = 45−20 = 25°C → t = 30/(K×25) = 30×15/(2×25) = 9 min.

3 According to Newton's law of cooling, the graph of ln(T₂ − T₁) vs time t is:

A A parabola

B A curve concave upward

C A straight line with negative slope

D A horizontal line

✅ From T₂ = T₁ + C′e^(−Kt), taking log: ln(T₂−T₁) = −Kt + lnC′. This is of the form y = mx + c, a straight line with negative slope (−K). Experimental verification confirms this (Fig 10.20b NCERT).

Thermal Propertiesof Matter

Temperature, Heat & Measurement

🎯 Practice MCQs — Temperature & Heat

Ideal Gas Equation & Absolute Temperature

🎯 Practice MCQs — Ideal Gas & Absolute Temperature

Thermal Expansion

🎯 Practice MCQs — Thermal Expansion

Specific Heat Capacity & Calorimetry

🎯 Practice MCQs — Specific Heat & Calorimetry

Change of State & Latent Heat

📌 Key Facts — Change of State

🎯 Practice MCQs — Change of State & Latent Heat

Heat Transfer — Conduction, Convection & Radiation

🎯 Practice MCQs — Heat Transfer

Newton's Law of Cooling

🎯 Practice MCQs — Newton's Law of Cooling

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Thermal Properties
of Matter