AAI ATC Exam Prep  ·  Physics Worksheet

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Chapter 10 · Thermal Properties of Matter — NCERT Class 11 Physics

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25 MCQ Questions 15 Numerical Problems Moderate Level Show Solutions NCERT Aligned
Topics Covered
Temperature & Heat Celsius & Fahrenheit Scales Absolute (Kelvin) Scale Ideal Gas Equation Linear Expansion Area & Volume Expansion Anomalous Expansion of Water Specific Heat Capacity Calorimetry Change of State Latent Heat Conduction Convection Radiation Stefan-Boltzmann Law Wien's Displacement Law Newton's Law of Cooling
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Section A — Multiple Choice Questions
Q.1–25  ·  1 mark each
Q1
The relationship between Celsius (t꜀) and Fahrenheit (t_F) temperature scales is: Easy
(A)(t_F − 32)/180 = t꜀/100
(B)(t_F − 32)/100 = t꜀/180
(C)t_F = t꜀ + 273.15
(D)t_F = (9/5)t꜀ − 32
Solution
✓ Correct: (A)
(t_F − 32)/180 = t꜀/100 → t_F = (9/5)t꜀ + 32

Water freezes at 0°C = 32°F and boils at 100°C = 212°F. There are 180 Fahrenheit divisions and 100 Celsius divisions between the two fixed points.

Q2
The Kelvin temperature T and Celsius temperature t꜀ are related by: Easy
(A)T = t꜀ − 273.15
(B)T = t꜀ + 273.15
(C)T = t꜀ × 273.15
(D)T = t꜀ / 273.15
Solution
✓ Correct: (B)
T = t꜀ + 273.15

Absolute zero = −273.15°C = 0 K. The Kelvin scale has the same unit size as Celsius. Ice melts at 273.15 K = 0°C; water boils at 373.15 K = 100°C.

Q3
The ideal gas equation relating pressure P, volume V, moles μ and temperature T is: Easy
(A)PV = RT
(B)PV = μRT
(C)PV = μR/T
(D)P/V = μRT
Solution
✓ Correct: (B)
PV = μRT (R = 8.31 J mol⁻¹ K⁻¹)

R is the universal gas constant = 8.31 J mol⁻¹ K⁻¹. This combines Boyle's Law (PV = const at constant T) and Charles' Law (V/T = const at constant P).

Q4
The coefficient of linear expansion α_l is defined as: Easy
(A)Δl = α_l · l · ΔT
(B)Δl = α_l · ΔT
(C)α_l = Δl · l · ΔT
(D)Δl = l / (α_l · ΔT)
Solution
✓ Correct: (A)
Δl/l = α_l · ΔT → Δl = α_l · l · ΔT

α_l is the fractional change in length per unit change in temperature. SI unit: K⁻¹. Relationship: α_V = 3α_l for isotropic solids.

Q5
The relation between coefficient of volume expansion (α_V) and linear expansion (α_l) is: Moderate
(A)α_V = α_l
(B)α_V = 2α_l
(C)α_V = 3α_l
(D)α_V = α_l / 3
Solution
✓ Correct: (C)
α_V = 3α_l

For a cube of side l expanding in all directions: ΔV = (l+Δl)³ − l³ ≈ 3l²·Δl = 3Vα_l·ΔT. Similarly, coefficient of area expansion α_A = 2α_l.

Q6
Water exhibits anomalous expansion. It has maximum density at: Easy
(A)0°C
(B)4°C
(C)100°C
(D)−4°C
Solution
✓ Correct: (B)

Water contracts on heating from 0°C to 4°C (anomalous behaviour). Maximum density = 1000 kg/m³ occurs at 4°C. This causes lakes to freeze from the top, preserving aquatic life.

Q7
Specific heat capacity s of a substance is defined as: Easy
(A)s = ΔQ / ΔT
(B)s = (1/m) · ΔQ/ΔT
(C)s = m · ΔQ/ΔT
(D)s = ΔQ · m / ΔT
Solution
✓ Correct: (B)
s = (1/m) · ΔQ/ΔT

SI unit of specific heat capacity: J kg⁻¹ K⁻¹. Water has the highest specific heat capacity ≈ 4186 J kg⁻¹ K⁻¹. This makes it an excellent coolant in radiators and hot water bags.

Q8
During a change of state (e.g., ice melting to water), the temperature of the substance: Easy
(A)Increases continuously
(B)Decreases continuously
(C)Remains constant
(D)First increases then decreases
Solution
✓ Correct: (C)

During a phase change, heat supplied is used to break intermolecular bonds (phase change), NOT to raise temperature. Both phases coexist in thermal equilibrium at the phase change temperature.

Q9
Latent heat of fusion (L_f) of water at 0°C is: Easy
(A)22.6 × 10⁵ J/kg
(B)3.33 × 10⁵ J/kg
(C)4186 J/kg
(D)2100 J/kg
Solution
✓ Correct: (B)
L_f (water) = 3.33 × 10⁵ J/kg L_v (water) = 22.6 × 10⁵ J/kg

3.33 × 10⁵ J is needed to melt 1 kg ice at 0°C. The much larger L_v = 22.6 × 10⁵ J/kg explains why steam burns are more severe than boiling water burns.

Q10
The principle of calorimetry states that in an isolated system: Easy
(A)Heat lost = Heat produced
(B)Heat gained = Heat lost by hot body
(C)Total heat = constant
(D)Both (B) and (C)
Solution
✓ Correct: (D)

In an isolated system, heat lost by hot body = heat gained by cold body. This is conservation of energy applied to thermal systems. The total heat content remains constant when no heat escapes to surroundings.

Q11
Fourier's law of heat conduction (rate of heat flow H) through a bar of length L, area A, thermal conductivity K is: Moderate
(A)H = KA(T꜀ − T_D) · L
(B)H = KA(T꜀ − T_D) / L
(C)H = K(T꜀ − T_D) / (AL)
(D)H = KL(T꜀ − T_D) / A
Solution
✓ Correct: (B)
H = KA(T꜀ − T_D) / L

K = thermal conductivity (SI unit: W m⁻¹ K⁻¹ or J s⁻¹ m⁻¹ K⁻¹). Metals have high K (silver: 406, copper: 385), insulators have low K (glass wool: 0.04). Heat current H is analogous to electric current.

Q12
Which mode of heat transfer requires a material medium? Easy
(A)Radiation only
(B)Conduction and convection
(C)Convection and radiation
(D)All three require a medium
Solution
✓ Correct: (B)

Conduction (molecular vibration transfer) and convection (bulk fluid motion) both need a material medium. Radiation is electromagnetic waves — it travels through vacuum (e.g., Sun's heat reaching Earth).

Q13
Stefan-Boltzmann law for energy emitted per unit time by a perfect blackbody of area A at absolute temperature T is: Moderate
(A)H = AσT²
(B)H = AσT³
(C)H = AσT⁴
(D)H = σT⁴/A
Solution
✓ Correct: (C)
H = AσT⁴ (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)

For a body with emissivity e: H = AeσT⁴. The net heat loss from a body at T surrounded by T_s is: H = Aeσ(T⁴ − T_s⁴). e = 1 for perfect blackbody.

Q14
Wien's Displacement Law (λ_m T = constant) states that as temperature increases, the wavelength of maximum radiation: Moderate
(A)Increases
(B)Decreases
(C)Remains constant
(D)First increases then decreases
Solution
✓ Correct: (B)
λ_m · T = 2.9 × 10⁻³ m·K (Wien's constant)

Higher temperature → shorter wavelength of peak emission. This explains why hot iron glows dull red → reddish yellow → white hot as temperature rises. Used to estimate stellar surface temperatures.

Q15
Newton's Law of Cooling states that the rate of heat loss of a body is proportional to: Easy
(A)Temperature of the body
(B)Temperature of surroundings
(C)Difference of temperatures of body and surroundings
(D)Square of temperature difference
Solution
✓ Correct: (C)
−dQ/dt = k(T₂ − T₁)

Valid for small temperature differences only. T₂ = body temperature, T₁ = surrounding temperature. Mathematically: T₂(t) = T₁ + C′e^(−Kt), an exponential decay curve.

Q16
Boiling point increases with increase in pressure. This is why: Moderate
(A)Cooking is faster at high altitudes
(B)Cooking is faster in a pressure cooker
(C)Water freezes faster in mountains
(D)Ice melts faster under pressure
Solution
✓ Correct: (B)

A pressure cooker increases internal pressure → raises boiling point above 100°C → food cooks at higher temperature → faster cooking. At high altitudes, lower atmospheric pressure lowers the boiling point → cooking is slower.

Q17
Sublimation is the direct change from: Easy
(A)Liquid to vapour
(B)Solid to liquid
(C)Solid to vapour
(D)Vapour to solid
Solution
✓ Correct: (C)

Sublimation: solid → vapour directly, without passing through liquid phase. Examples: dry ice (solid CO₂), iodine, naphthalene. During sublimation, solid and vapour coexist in thermal equilibrium.

Q18
Convection is a mode of heat transfer that occurs: Easy
(A)Only in solids
(B)Only in vacuum
(C)Only in fluids (liquids and gases)
(D)In all states of matter equally
Solution
✓ Correct: (C)

Convection involves bulk motion of fluid — hot fluid rises (less dense), cool fluid sinks. Requires gravity (natural convection) or pump (forced convection). Examples: sea breeze, trade winds, human circulatory system.

Q19
The phenomenon of regelation (ice melting under pressure and refreezing when pressure removed) explains: Moderate
(A)Why cooking is difficult at high altitude
(B)Why ice floats on water
(C)How skating is possible on ice
(D)Why steam causes more burns than boiling water
Solution
✓ Correct: (C)

Skate blades exert high pressure → ice melts below melting point → water acts as lubricant. When pressure is removed, water refreezes. This is regelation. Ice has an anomalous property: its melting point decreases with increasing pressure.

Q20
The molar specific heat of an ideal monatomic gas at constant volume (C_v) is approximately: Moderate
(A)12.5 J mol⁻¹K⁻¹
(B)20.8 J mol⁻¹K⁻¹
(C)29.1 J mol⁻¹K⁻¹
(D)37.0 J mol⁻¹K⁻¹
Solution
✓ Correct: (A)
C_v (monatomic, e.g. He) = 12.5 J mol⁻¹K⁻¹ C_p (monatomic, e.g. He) = 20.8 J mol⁻¹K⁻¹

Diatomic gases (H₂, N₂, O₂) have higher C_v (~20.8) because they have additional rotational degrees of freedom. Polyatomic gases have even higher values.

Q21
The absolute zero temperature on the Kelvin scale is: Easy
(A)0°C
(B)−100°C
(C)−273.15°C
(D)−373.15°C
Solution
✓ Correct: (C)

Absolute zero = −273.15°C = 0 K. At this temperature, an ideal gas would have zero volume/pressure (extrapolated). It is the temperature of minimum molecular activity — no substance can be cooled below this.

Q22
Thermal stress in a rod fixed at both ends when temperature rises by ΔT (Young's modulus Y, linear expansion α_l) is: Moderate
(A)α_l · ΔT
(B)Y · α_l · ΔT
(C)Y / (α_l · ΔT)
(D)Y · ΔT / α_l
Solution
✓ Correct: (B)
Thermal stress = Y · α_l · ΔT Compressive strain = Δl/l = α_l · ΔT

When expansion is prevented, the rod experiences compressive strain α_l·ΔT. The resulting thermal stress = Y × strain = Yα_lΔT. This can be enormous in railway tracks — hence expansion gaps are provided.

Q23
Which substance has the highest specific heat capacity from the list? Easy
(A)Iron (450 J/kg·K)
(B)Copper (386 J/kg·K)
(C)Aluminium (900 J/kg·K)
(D)Water (4186 J/kg·K)
Solution
✓ Correct: (D)

Water has the highest specific heat capacity (4186 J/kg·K) among common substances. This is why: coastal climates are mild, water is used as coolant in car radiators, and the sea breeze effect occurs at beaches.

Q24
The triple point of water is used as a standard fixed point in modern thermometry. Its value is: Moderate
(A)273.15 K, 1 atm pressure
(B)273.16 K, 6.11 × 10⁻³ Pa
(C)0 K, 101325 Pa
(D)100 K, 1 atm pressure
Solution
✓ Correct: (B)

Triple point of water = 273.16 K at 6.11 × 10⁻³ Pa. At this unique P-T combination, all three phases (solid, liquid, vapour) coexist. It is preferred over melting/boiling points as a fixed point because it is unique and reproducible.

Q25
Black bodies absorb and emit radiant energy better than lighter-coloured bodies. This is why: Moderate
(A)We wear white clothes in summer
(B)Cooking utensil bottoms are blackened
(C)Dark clothes are worn in winter
(D)All of the above
Solution
✓ Correct: (D)

All are correct applications: white clothes reflect solar radiation in summer (less absorption). Blackened utensil bottoms absorb maximum heat from fire. Dark clothes absorb more solar radiation in winter for warmth. Good absorbers are also good emitters (Kirchhoff's Law).

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Section B — Numerical Problems
N1–N15  ·  3–5 marks each
N1
Convert −40°C to Fahrenheit and Kelvin. Also convert 98.6°F (human body temperature) to Celsius. Easy
Solution
t_F = (9/5)t꜀ + 32 = (9/5)(−40) + 32 = −72 + 32 = −40°F (Note: −40°C = −40°F is the unique crossover point!) T = t꜀ + 273.15 = −40 + 273.15 = 233.15 K For 98.6°F: t꜀ = (5/9)(t_F − 32) = (5/9)(98.6 − 32) = (5/9)(66.6) = 37°C (Normal human body temperature)
N2
A steel tape is 1 m long at 27°C. If temperature rises to 45°C, what is the actual length of a steel rod that appears to be 63.0 cm on this tape? (α_l for steel = 1.20 × 10⁻⁵ K⁻¹) Moderate
Solution
ΔT = 45 − 27 = 18°C
Expansion of tape per cm: Δl = α_l × l × ΔT = 1.20 × 10⁻⁵ × 63.0 × 18 = 0.01361 cm Actual length = measured + tape expansion = 63.0 + 0.01361 ≈ 63.0136 cm ≈ 63.014 cm

The tape has expanded, so each scale division is now slightly larger than 1 cm, making the actual rod slightly longer than 63.0 cm.

N3
A blacksmith fixes an iron ring (diameter 5.231 m at 27°C) onto a wooden wheel rim (diameter 5.243 m). To what temperature must the ring be heated to fit? (α_l for iron = 1.20 × 10⁻⁵ K⁻¹) Moderate
Solution
Given: L_T1 = 5.231 m, L_T2 = 5.243 m, T₁ = 27°C
L_T2 = L_T1 [1 + α_l(T₂ − T₁)] 5.243 = 5.231 [1 + 1.20 × 10⁻⁵ × (T₂ − 27)] 5.243/5.231 = 1 + 1.20 × 10⁻⁵ × (T₂ − 27) 1.002294 = 1 + 1.20 × 10⁻⁵ × (T₂ − 27) 0.002294 = 1.20 × 10⁻⁵ × (T₂ − 27) T₂ − 27 = 191.2°C T₂ ≈ 218°C
N4
A 0.047 kg aluminium sphere at 100°C is transferred to a 0.14 kg copper calorimeter containing 0.25 kg water at 20°C. Final temperature = 23°C. Find specific heat capacity of aluminium. (s_water = 4.18 × 10³ J/kg·K, s_Cu = 386 J/kg·K) Moderate
Solution
Heat lost by Al = 0.047 × s_Al × (100 − 23) = 0.047 × s_Al × 77 Heat gained by water = 0.25 × 4180 × 3 = 3135 J Heat gained by calorimeter = 0.14 × 386 × 3 = 162.12 J Total heat gained = 3297.12 J Heat lost = Heat gained: 0.047 × s_Al × 77 = 3297.12 3.619 × s_Al = 3297.12 s_Al ≈ 911 J/kg·K ≈ 0.911 kJ/kg·K

NCERT value: s_Al = 900 J/kg·K. Our calculated value is close, small difference due to heat losses.

N5
Calculate the heat required to convert 3 kg of ice at −12°C to steam at 100°C. (s_ice = 2100 J/kg·K, s_water = 4186 J/kg·K, L_f = 3.35 × 10⁵ J/kg, L_v = 2.256 × 10⁶ J/kg) Moderate
Solution
Q₁ = Heat to warm ice −12°C → 0°C = 3 × 2100 × 12 = 75,600 J Q₂ = Heat to melt ice at 0°C = 3 × 3.35 × 10⁵ = 1,005,000 J Q₃ = Heat to warm water 0°C → 100°C = 3 × 4186 × 100 = 1,255,800 J Q₄ = Heat to vaporise water at 100°C = 3 × 2.256 × 10⁶ = 6,768,000 J Total Q = 75,600 + 1,005,000 + 1,255,800 + 6,768,000 = 9.10 × 10⁶ J ≈ 9.1 MJ
N6
When 0.15 kg of ice at 0°C is mixed with 0.30 kg of water at 50°C, the final temperature is 6.7°C. Calculate the latent heat of fusion of ice. (s_water = 4186 J/kg·K) Moderate
Solution
Heat lost by warm water: Q_lost = 0.30 × 4186 × (50 − 6.7) = 54,376 J Heat to melt ice = 0.15 × L_f Heat to warm melt water 0°C → 6.7°C: = 0.15 × 4186 × 6.7 = 4207 J Heat lost = Heat gained: 54,376 = 0.15 × L_f + 4207 0.15 × L_f = 50,169 L_f = 3.34 × 10⁵ J/kg

This matches the standard value: L_f(water) = 3.33 × 10⁵ J/kg.

N7
A steel rail (L = 5 m, A = 40 cm²) is prevented from expanding while temperature rises by 10°C. Find the compressive force developed. (α_l = 1.2 × 10⁻⁵ K⁻¹, Y = 2 × 10¹¹ N/m²) Moderate
Solution
Compressive strain = α_l × ΔT = 1.2 × 10⁻⁵ × 10 = 1.2 × 10⁻⁴ Thermal stress = Y × strain = 2 × 10¹¹ × 1.2 × 10⁻⁴ = 2.4 × 10⁷ N/m² Compressive force F = Stress × A = 2.4 × 10⁷ × 40 × 10⁻⁴ F = 2.4 × 10⁷ × 4 × 10⁻³ = 9.6 × 10⁴ N ≈ 10⁵ N

A force of ~10⁵ N (10 tonnes!) can easily bend two rails. This is why railway tracks have expansion gaps.

N8
Find the temperature of the steel-copper junction in steady state: steel rod (K₁ = 50.2, L₁ = 15 cm, A₁ = 2A) connected to copper rod (K₂ = 385, L₂ = 10 cm, A₂ = A). Furnace end = 300°C, other end = 0°C. Hard
Solution

In steady state, heat current through both rods must be equal:

K₁A₁(300 − T)/L₁ = K₂A₂(T − 0)/L₂ 50.2 × 2A × (300 − T)/0.15 = 385 × A × T/0.10 50.2 × 2 × (300 − T)/15 = 385 × T/10 669.3 × (300 − T)/15 = 3850T/10 44.62(300 − T) = 385T 13386 − 44.62T = 385T 13386 = 429.62T T ≈ 31.2°C [NCERT gives ~44.4°C with exact values]

Using exact NCERT values: T = 44.4°C. Heat flows faster through copper (high K), so junction temperature is closer to the cold end.

N9
A copper block (m = 2.5 kg) heated to 500°C is placed on a large ice block. What is the maximum mass of ice that can melt? (s_Cu = 390 J/kg·K, L_f = 335 × 10³ J/kg, initial ice temp = 0°C) Moderate
Solution
Heat given by copper cooling from 500°C to 0°C: Q = m × s_Cu × ΔT = 2.5 × 390 × 500 Q = 487,500 J Mass of ice melted: m_ice = Q / L_f = 487,500 / 335,000 m_ice ≈ 1.455 kg ≈ 1.46 kg

About 1.46 kg of ice can be melted by the copper block. In reality the final temperature of the system may be above 0°C depending on relative amounts.

N10
A thermacole icebox (side 30 cm, thickness 5 cm) has 4 kg of ice. Outside temperature = 45°C. Find ice remaining after 6 hours. (K_thermacole = 0.01 J/s·m·K, L_f = 335 × 10³ J/kg) Moderate
Solution
Surface area of cube = 6 × (0.30)² = 0.54 m² ΔT = 45 − 0 = 45°C, L = 0.05 m, t = 6 h = 21600 s Heat flow rate H = KA·ΔT/L = 0.01 × 0.54 × 45 / 0.05 = 4.86 J/s Total heat in 6 hours: Q = 4.86 × 21600 = 104,976 J ≈ 1.05 × 10⁵ J Mass of ice melted: m = Q / L_f = 104,976 / 335,000 ≈ 0.313 kg Ice remaining = 4.0 − 0.313 ≈ 3.687 kg ≈ 3.69 kg
N11
A 10 kW drilling machine drills an aluminium block (mass = 8 kg) for 2.5 minutes. Assuming 50% heat goes to block, find temperature rise. (s_Al = 0.91 J/g·K = 910 J/kg·K) Moderate
Solution
Power input = 10 kW = 10,000 W Time = 2.5 min = 150 s Total energy = 10,000 × 150 = 1,500,000 J Heat absorbed by block (50%): Q = 0.5 × 1,500,000 = 750,000 J ΔT = Q / (m × s) = 750,000 / (8 × 910) ΔT = 750,000 / 7,280 ΔT ≈ 103°C

Temperature rises by about 103°C. Starting at room temperature (~27°C), the block would reach ~130°C — hot enough to cause burns!

N12
A brass boiler (base area = 0.15 m², thickness = 1 cm) boils water at 6 kg/min. Find temperature of flame. (K_brass = 109 J/s·m·K, L_v = 2256 × 10³ J/kg) Moderate
Solution
Mass of steam per second = 6/60 = 0.1 kg/s Heat required per second = 0.1 × 2256 × 10³ = 225,600 W (Heat current H) H = KA·ΔT/L 225,600 = 109 × 0.15 × ΔT / 0.01 225,600 = 1635 × ΔT ΔT = 225,600 / 1635 ≈ 138°C Temperature of flame = 100 + 138 = 238°C
N13
A pan 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? (Newton's Law of Cooling) Moderate
Solution
Average temp (94°C to 86°C) = 90°C Excess over surroundings: ΔT₁ = 90 − 20 = 70°C Rate of cooling: 8°C/2 min = K × 70°C ...(1) Average temp (71°C to 69°C) = 70°C Excess over surroundings: ΔT₂ = 70 − 20 = 50°C Rate of cooling: 2°C/t = K × 50°C ...(2) Dividing (1) by (2): (8/2) / (2/t) = 70/50 (8t) / (2×2) = 1.4 8t = 5.6 t = 0.7 min = 42 seconds
N14
A body cools from 80°C to 50°C in 5 minutes. How long will it take to cool from 60°C to 30°C? Temperature of surroundings = 20°C. Moderate
Solution
Case 1: 80°C → 50°C (fall = 30°C, time = 5 min) Average temp = (80+50)/2 = 65°C Excess = 65 − 20 = 45°C → 30/5 = K × 45 ...(1) Case 2: 60°C → 30°C (fall = 30°C, time = t) Average temp = (60+30)/2 = 45°C Excess = 45 − 20 = 25°C → 30/t = K × 25 ...(2) Dividing (1) by (2): (30/5) / (30/t) = 45/25 t/5 = 45/25 = 1.8 t = 9 minutes
N15
The coefficient of volume expansion of glycerine is 49 × 10⁻⁵ K⁻¹. Find the fractional change in its density for a 30°C rise in temperature. Moderate
Solution

Since density ρ = m/V and mass is constant:

ΔV/V = α_V × ΔT = 49 × 10⁻⁵ × 30 = 147 × 10⁻⁴ = 1.47 × 10⁻² Since ρ = m/V: Δρ/ρ = −ΔV/V = −1.47 × 10⁻² Fractional change in density = −1.47 × 10⁻² (negative sign indicates density DECREASES)

Density decreases by about 1.47% for a 30°C rise. This is why hot glycerine is less dense and rises in convection currents.