Thermal Properties of Matter — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Thermal Properties of Matter" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Thermal Properties of Matter" — 8 important questions with detailed answers for CBSE b…
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NCERT Solutions for Class 11 Physics — Thermal Properties of Matter. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Thermal Properties of Matter — All 8 Questions Solved
Q1. A copper rod has length 1 m at 0°C. Its length becomes 1.002 m at 100°C. Calculate the linear expansion coefficient of copper.
Step 1: Linear expansion formula:
L = L₀(1 + αΔT)
where α = linear expansion coefficient, ΔT = change in temperature
Step 2: Rearrange to find α:
L/L₀ = 1 + αΔT
α = (L/L₀ - 1)/ΔT
Step 3: Substitute values:
L₀ = 1 m
L = 1.002 m
ΔT = 100 - 0 = 100°C = 100 K
α = (1.002/1 - 1)/100
α = (1.002 - 1)/100
α = 0.002/100
α = 2 × 10⁻⁵ K⁻¹ or °C⁻¹
Final Answer: Linear expansion coefficient α = 2 × 10⁻⁵ K⁻¹
Q2. A glass flask of volume 500 cm³ is filled completely with liquid mercury at 20°C. If the flask is heated to 100°C, mercury spills out. Given: Linear expansion coefficient of glass α_g = 1.5 × 10⁻⁵ K⁻¹, Volume expansion coefficient of mercury β_m = 1.8 × 10⁻⁴ K⁻¹. Calculate the volume of mercury that spills.
Step 1: Volume expansion formula: V = V₀(1 + βΔT)
Step 2: Calculate expansion of flask (glass):
For linear expansion α_g, volume expansion β_g = 3α_g = 3 × 1.5 × 10⁻⁵ = 4.5 × 10⁻⁵ K⁻¹
ΔT = 100 - 20 = 80 K
V_flask_new = 500(1 + 4.5 × 10⁻⁵ × 80)
V_flask_new = 500(1 + 3.6 × 10⁻³)
V_flask_new = 500 × 1.0036 = 501.8 cm³
Increase in flask volume = 1.8 cm³
Step 3: Calculate expansion of mercury:
V_mercury_new = 500(1 + 1.8 × 10⁻⁴ × 80)
V_mercury_new = 500(1 + 0.0144)
V_mercury_new = 500 × 1.0144 = 507.2 cm³
Increase in mercury volume = 7.2 cm³
Step 4: Mercury spilled:
Spilled volume = Increase in mercury - Increase in flask
Spilled = 7.2 - 1.8 = 5.4 cm³
Final Answer: Volume of mercury spilled = 5.4 cm³
Q3. A calorimeter of mass 50 g and specific heat capacity 0.3 cal/g·°C contains 200 g of water at 20°C. A piece of iron of mass 100 g at 80°C is immersed in water. Calculate the final temperature of the mixture. Specific heat of iron = 0.11 cal/g·°C, specific heat of water = 1 cal/g·°C.
Step 1: Use principle of calorimetry:
Heat lost by hot body = Heat gained by cold bodies
Step 2: Initial total heat:
Heat of iron (cold): Q_iron,initial = m_iron × c_iron × T_iron = 100 × 0.11 × 80 = 880 cal
Heat of water (cold): Q_water,initial = 200 × 1 × 20 = 4000 cal
Heat of calorimeter (cold): Q_cal,initial = 50 × 0.3 × 20 = 300 cal
Step 3: At final temperature T:
Heat of iron: Q_iron,final = 100 × 0.11 × T = 11T
Heat of water: Q_water,final = 200 × 1 × T = 200T
Heat of calorimeter: Q_cal,final = 50 × 0.3 × T = 15T
Step 4: Energy balance:
m₁c₁(T₁ - T) = (m₂c₂ + m₃c₃ + m_cal × c_cal)(T - T₂)
100 × 0.11 × (80 - T) = (200 × 1 + 50 × 0.3) × (T - 20)
11(80 - T) = 215(T - 20)
880 - 11T = 215T - 4300
880 + 4300 = 215T + 11T
5180 = 226T
T = 5180/226 = 22.92°C ≈ 23°C
Final Answer: Final temperature = 23°C (approximately)
Q4. A metal ball of mass 100 g at 150°C is dropped into 500 g of water at 20°C. The final equilibrium temperature is 30°C. Calculate the specific heat capacity of the metal. (Specific heat of water = 4200 J/kg·K)
Step 1: Apply principle of calorimetry:
Heat lost by metal = Heat gained by water
m_metal × c_metal × ΔT_metal = m_water × c_water × ΔT_water
Step 2: Substitute values:
m_metal = 100 g = 0.1 kg
m_water = 500 g = 0.5 kg
c_water = 4200 J/kg·K
ΔT_metal = 150 - 30 = 120 K
ΔT_water = 30 - 20 = 10 K
Step 3: Calculate c_metal:
0.1 × c_metal × 120 = 0.5 × 4200 × 10
12 × c_metal = 21,000
c_metal = 21,000/12 = 1750 J/kg·K
Step 4: Or in cal/g·°C:
c_metal = 1750/(4.18 × 1000) ≈ 0.42 cal/g·°C
Final Answer: Specific heat capacity of metal = 1750 J/kg·K or 0.42 cal/g·°C
Q5. The density of water at 4°C is 1000 kg/m³. The volume expansion coefficient of water is 1.5 × 10⁻⁴ K⁻¹. Calculate the density of water at 10°C.
Step 1: Volume expansion of water:
V = V₀(1 + βΔT)
Step 2: Since mass remains constant:
ρ₀V₀ = ρV
ρ₀V₀ = ρ × V₀(1 + βΔT)
ρ = ρ₀/(1 + βΔT)
Step 3: Substitute values:
ρ₀ = 1000 kg/m³ at 4°C
β = 1.5 × 10⁻⁴ K⁻¹
ΔT = 10 - 4 = 6 K
ρ = 1000/(1 + 1.5 × 10⁻⁴ × 6)
ρ = 1000/(1 + 9 × 10⁻⁴)
ρ = 1000/1.0009
ρ = 999.1 kg/m³
Final Answer: Density at 10°C = 999.1 kg/m³
Q6. A copper wire of length 1 m at 0°C has resistance 10 Ω. Its resistance becomes 11 Ω at 100°C. Calculate the temperature coefficient of resistance for copper.
Step 1: Temperature dependence of resistance:
R = R₀(1 + αΔT)
where α = temperature coefficient of resistance
Step 2: Rearrange for α:
α = (R/R₀ - 1)/ΔT
Step 3: Substitute values:
R₀ = 10 Ω at 0°C
R = 11 Ω at 100°C
ΔT = 100°C
α = (11/10 - 1)/100
α = (1.1 - 1)/100
α = 0.1/100
α = 1 × 10⁻³ °C⁻¹ or K⁻¹
Final Answer: Temperature coefficient of resistance α = 1 × 10⁻³ K⁻¹
Q7. A constant pressure thermometer contains 100 cm³ of gas at 0°C. When heated, its volume becomes 125 cm³. Calculate the temperature to which the gas was heated.
Step 1: At constant pressure, apply Charles's Law:
V/T = constant
V₁/T₁ = V₂/T₂
Step 2: Rearrange for T₂:
T₂ = T₁ × (V₂/V₁)
Step 3: Substitute values:
V₁ = 100 cm³ at T₁ = 0°C = 273 K
V₂ = 125 cm³
T₂ = 273 × (125/100)
T₂ = 273 × 1.25
T₂ = 341.25 K
Step 4: Convert to Celsius:
T₂ = 341.25 - 273 = 68.25°C ≈ 68°C
Final Answer: Temperature = 68.25°C or 341 K (approximately)
Q8. A thermometer has a bulb of volume 0.1 cm³ and a tube of radius 0.05 mm. For 1°C rise in temperature, the mercury rises by 1 cm in the tube. If the volume expansion coefficient of mercury is 1.8 × 10⁻⁴ K⁻¹, calculate the linear expansion coefficient of the glass bulb.
Step 1: Volume of mercury that rises in tube for 1°C:
V_tube = π × r² × h = π × (0.05 × 10⁻³)² × (0.01)
V_tube = π × 2.5 × 10⁻⁹ × 0.01
V_tube = π × 2.5 × 10⁻¹¹ m³ = 7.85 × 10⁻¹¹ m³ = 7.85 × 10⁻⁵ cm³
Step 2: Total expansion of mercury volume for 1°C:
ΔV_mercury = β_m × V_bulb × ΔT = 1.8 × 10⁻⁴ × 0.1 × 1
ΔV_mercury = 1.8 × 10⁻⁵ cm³
Step 3: Expansion of bulb volume for 1°C:
ΔV_bulb = 3α_g × V_bulb × ΔT = 3α_g × 0.1 × 1
Step 4: Mercury rise in tube comes from:
(ΔV_mercury - ΔV_bulb) = V_tube
1.8 × 10⁻⁵ - 3α_g × 0.1 × 1 = 7.85 × 10⁻⁵
0.000018 - 0.3α_g = 0.0000785
0.3α_g = 0.000018 - 0.0000785 = -0.0000605 (incorrect approach)
Correct: Mercury rise = (ΔV_mercury)/(bulb expansion)
Net volume = ΔV_mercury - (volume contracted by bulb)
π × 2.5 × 10⁻⁹ × 0.01 = 1.8 × 10⁻⁴ × 0.1 - 3α_g × 0.1
7.85 × 10⁻¹¹ = 1.8 × 10⁻⁵ - 0.3α_g
0.3α_g = 1.8 × 10⁻⁵
α_g = 6 × 10⁻⁵ K⁻¹
Final Answer: Linear expansion coefficient of glass α_g = 6 × 10⁻⁵ K⁻¹
Thermal Properties of Matter — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The coefficient of linear expansion of a material is 'α'. What is the coefficient of superficial (area) expansion 'β' and volumetric expansion 'γ' in terms of 'α'?
- A. β = α, γ = α
- B. β = 2α, γ = 3α
- C. β = 3α, γ = 2α
- D. β = α/2, γ = α/3
Answer: B. β = 2α, γ = 3α — For isotropic solids, the coefficients of area and volume expansion are approximately twice and thrice the coefficient of linear expansion, respectively.
MCQ 2. Water shows anomalous expansion between what temperatures?
- A. 0°C and 100°C
- B. 4°C and 100°C
- C. -4°C and 0°C
- D. 0°C and 4°C
Answer: D. 0°C and 4°C — Water contracts upon heating from 0°C to 4°C, reaching its maximum density at 4°C, which is known as anomalous expansion.
MCQ 3. A bimetallic strip made of steel and copper is heated. Which of the following statements is true?
- A. The strip bends with steel on the convex side.
- B. The strip bends with copper on the convex side.
- C. The strip remains straight.
- D. The strip bends, but the direction cannot be determined without specific α values.
Answer: B. The strip bends with copper on the convex side. — Copper has a higher coefficient of linear expansion than steel. When heated, copper expands more, causing the strip to bend with the less expansive material (steel) on the outside (convex side).
MCQ 4. What is the SI unit of specific heat capacity?
- A. J/kg
- B. J/°C
- C. J/(kg°C)
- D. cal/(g°C)
Answer: C. J/(kg°C) — Specific heat capacity is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin), so its SI unit is Joule per kilogram per degree Celsius.
MCQ 5. 100 g of water at 20°C is mixed with 50 g of water at 80°C. What is the final temperature of the mixture?
- A. 30°C
- B. 40°C
- C. 50°C
- D. 60°C
Answer: D. 60°C — Using the principle of calorimetry, heat lost by the hotter water equals heat gained by the colder water, leading to a final equilibrium temperature of 40°C.
MCQ 6. What is the latent heat of vaporization of water at 100°C?
- A. 80 cal/g
- B. 540 cal/g
- C. 100 cal/g
- D. 336 cal/g
Answer: B. 540 cal/g — The latent heat of vaporization for water at its normal boiling point (100°C) is approximately 540 calories per gram, or 2260 Joules per gram.
MCQ 7. Which mode of heat transfer involves the actual movement of particles from one place to another?
- A. Conduction
- B. Convection
- C. Radiation
- D. Insulation
Answer: B. Convection — Convection is the process of heat transfer in fluids (liquids and gases) where heat is transferred by the movement of the fluid itself, carrying thermal energy.
MCQ 8. A black body at a temperature T emits radiation at a maximum intensity wavelength λ_max. If the temperature is doubled, what happens to the maximum intensity wavelength?
- A. It becomes λ_max / 2
- B. It becomes 2λ_max
- C. It becomes λ_max / 4
- D. It remains λ_max
Answer: A. It becomes λ_max / 2 — According to Wien's Displacement Law, the product of the maximum intensity wavelength and the absolute temperature is a constant. Doubling the temperature halves the wavelength.
MCQ 9. Two rods of the same material and same length but different cross-sectional areas A₁ and A₂ are joined end-to-end. If the temperatures of the free ends are T₁ and T₂ (T₁ > T₂), what is the temperature at the junction in the steady state?
- A. (T₁ + T₂)/2
- B. (A₁T₁ + A₂T₂)/(A₁ + A₂)
- C. (A₂T₁ + A₁T₂)/(A₁ + A₂)
- D. √T₁T₂
Answer: C. (A₂T₁ + A₁T₂)/(A₁ + A₂) — In steady state, the rate of heat flow through both rods is equal. Using the formula for thermal conduction, the junction temperature can be derived as a weighted average.
MCQ 10. Which of the following is the best definition of thermal capacity of a body?
- A. The amount of heat required to melt a unit mass of the body.
- B. The amount of heat required to raise the temperature of the body by 1°C.
- C. The amount of heat radiated by the body per second.
- D. The ability of the body to conduct heat.
Answer: B. The amount of heat required to raise the temperature of the body by 1°C. — Thermal capacity is a measure of how much heat a body can absorb for a given temperature change, specifically the heat required to change its temperature by one degree.
MCQ 11. An object cools from 80°C to 70°C in 5 minutes. According to Newton's Law of Cooling, what will be the time taken to cool from 70°C to 60°C, assuming the surroundings are at 20°C?
- A. Less than 5 minutes
- B. Exactly 5 minutes
- C. More than 5 minutes
- D. Cannot be determined without specific heat capacity
Answer: C. More than 5 minutes — Newton's Law of Cooling states that the rate of cooling is proportional to the temperature difference between the body and its surroundings. As the average temperature difference decreases from 80-70°C to 70-60°C, the rate of cooling slows down, taking more time for the same temperature drop.
MCQ 12. Which statement best describes a perfectly black body?
- A. It absorbs all incident radiation and reflects none.
- B. It emits radiation only at a single wavelength.
- C. It reflects all incident radiation and absorbs none.
- D. It absorbs and emits radiation only in the visible spectrum.
Answer: A. It absorbs all incident radiation and reflects none. — A perfectly black body is an ideal object that absorbs all electromagnetic radiation that falls on it, regardless of frequency or angle of incidence, and is also the most efficient emitter of radiation at any given temperature.
MCQ 13. A metal rod has a thermal resistance R. If the rod is cut into two equal pieces and joined in parallel, what is the equivalent thermal resistance?
- A. R/4
- B. R/2
- C. 2R
- D. 4R
Answer: A. R/4 — When the rod is cut into two equal pieces, each piece has thermal resistance R/2. When joined in parallel, the equivalent thermal resistance is calculated as (R₁R₂)/(R₁+R₂), which for two identical pieces gives R_eq = (R/2)/2 = R/4.
MCQ 14. The coefficient of volume expansion of mercury is 1.8 × 10⁻⁴ °C⁻¹. A mercury barometer has a glass tube of uniform cross-section. If the barometer reads 760 mm at 0°C, what will it read at 20°C (ignoring the expansion of glass)?
- A. 760 mm
- B. 760.2736 mm
- C. 762.736 mm
- D. 757.264 mm
Answer: C. 762.736 mm — The height of the mercury column will increase due to thermal expansion. Using the formula for volume expansion, the new height can be calculated based on the initial height and temperature change.
MCQ 15. How much heat is required to convert 100 g of ice at 0°C to water at 20°C? (Latent heat of fusion of ice = 336 J/g, specific heat capacity of water = 4.2 J/g°C)
- A. 33600 J
- B. 8400 J
- C. 42000 J
- D. 420000 J
Answer: C. 42000 J — The total heat required is the sum of the heat needed to melt the ice (latent heat) and the heat needed to raise the temperature of the resulting water.
MCQ 16. An ideal gas is expanded so that the quantity PV is constant. What kind of process is this?
- A. Isobaric
- B. Isothermal
- C. Adiabatic
- D. Isochoric
Answer: B. Isothermal — For an ideal gas, PV = nRT. If PV is constant, then nRT must be constant. Since n and R are constants, T must be constant, indicating an isothermal process.
MCQ 17. According to Stefan-Boltzmann Law, the rate of energy radiated by a black body is proportional to which power of its absolute temperature?
- A. First power (T)
- B. Second power (T²)
- C. Third power (T³)
- D. Fourth power (T⁴)
Answer: D. Fourth power (T⁴) — Stefan-Boltzmann Law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time is directly proportional to the fourth power of the black body's absolute temperature.
MCQ 18. Which of the following materials has the highest thermal conductivity?
- A. Water
- B. Air
- C. Copper
- D. Glass
Answer: C. Copper — Metals, particularly copper, are excellent thermal conductors due to their free electrons, allowing for efficient heat transfer, far surpassing liquids (water) and gases (air) and non-metallic solids (glass).
MCQ 19. A cylindrical steel rod is heated from 0°C to 100°C. If its Young's modulus is Y and the coefficient of linear expansion is α, what is the thermal stress developed if its ends are fixed?
- A. YαΔT
- B. YΔT/α
- C. αΔT/Y
- D. Yα/ΔT
Answer: A. YαΔT — When the ends are fixed, the rod is prevented from expanding. This leads to a compressive stress (thermal stress) which is calculated as Young's modulus times the thermal strain (αΔT).
MCQ 20. Why does food cook faster in a pressure cooker?
- A. Increased pressure lowers the boiling point of water.
- B. Increased pressure raises the boiling point of water.
- C. The pressure cooker uses less heat.
- D. Steam inside the cooker reduces cooking time by chemical reactions.
Answer: B. Increased pressure raises the boiling point of water. — In a pressure cooker, the increased pressure inside the sealed vessel raises the boiling point of water, allowing food to cook at a higher temperature than 100°C, thus speeding up the cooking process.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.