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.

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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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Key Questions Covered:

A copper rod has length 1 m at 0°C. Its length becomes 1.002 m at 100°C. Calc…
A glass flask of volume 500 cm³ is filled completely with liquid mercury at 2…
A calorimeter of mass 50 g and specific heat capacity 0.3 cal/g·°C contains 2…
A metal ball of mass 100 g at 150°C is dropped into 500 g of water at 20°C. T…
The density of water at 4°C is 1000 kg/m³. The volume expansion coefficient o…
A copper wire of length 1 m at 0°C has resistance 10 Ω. Its resistance become…
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
A copper rod has length 1 m at 0°C. Its length becomes 1.…	✓ Solved
A glass flask of volume 500 cm³ is filled completely with…	✓ Solved
A calorimeter of mass 50 g and specific heat capacity 0.3…	✓ Solved
A metal ball of mass 100 g at 150°C is dropped into 500 g…	✓ Solved
The density of water at 4°C is 1000 kg/m³. The volume exp…	✓ Solved
A copper wire of length 1 m at 0°C has resistance 10 Ω. I…	✓ Solved

Showing 6 of 8 questions

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 = 50...

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...

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 Answe...

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

Showing 6 of 8 questions. Visit the full page for complete solutions.