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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Key Questions Covered:

  1. A copper rod has length 1 m at 0°C. Its length becomes 1.002 m at 100°C. Calc…
  2. A glass flask of volume 500 cm³ is filled completely with liquid mercury at 2…
  3. A calorimeter of mass 50 g and specific heat capacity 0.3 cal/g·°C contains 2…
  4. + 5 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

Showing 3 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...

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...

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):...

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