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CBSE Class 11 Physics Chapter 10 Thermal Properties of Matter Worksheet with Answers

Thermal Properties of Matter is a scoring yet application-heavy chapter in CBSE Class 11 Physics. This printable worksheet helps students consolidate understanding of thermal expansion coefficients, calorimetry principles, and heat transfer mechanisms through a variety of question types. With a complete answer key, students can practice independently and verify their solutions, making it ideal for self-study, homework, or pre-exam revision across CBSE schools in India.

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Key takeaways

✓Comprehensive worksheet covering thermal expansion, calorimetry, and all three modes of heat transfer as per NCERT Class 11 Physics Chapter 10.
✓6 MCQs test conceptual clarity on coefficient of expansion, specific heat, and Stefan-Boltzmann law.
✓5 fill-in-the-blanks reinforce key terminology and formulae for latent heat and thermal conductivity.
✓5 short-answer questions require numerical problem-solving on calorimetry and thermal expansion.
✓3 HOTS long-answer questions develop analytical thinking on real-world thermal phenomena.
✓One case-study question integrates multiple concepts, mirroring new CBSE competency-based assessment patterns.
✓Complete answer key with step-by-step solutions enables independent learning and immediate feedback.

Quick Chapter Recap: Thermal Properties of Matter

Chapter 10 Thermal Properties of Matter in NCERT Class 11 Physics introduces how substances respond to temperature changes. Thermal expansion describes the change in dimensions of solids, liquids, and gases when heated; the coefficient of linear expansion (α) and volume expansion (γ) quantify these changes. Calorimetry deals with heat measurement, using the principle that heat lost equals heat gained in isolated systems. Specific heat capacity (c) is the heat required to raise one kilogram of a substance by one Kelvin, while latent heat (L) is the energy needed to change the state without temperature change. Heat transfer occurs through conduction (direct molecular contact), convection (fluid movement), and radiation (electromagnetic waves). Understanding Stefan-Boltzmann law, Newton's law of cooling, and Wien's displacement law is crucial for solving numerical problems and explaining everyday thermal phenomena.

Linear expansion: ΔL = L₀ α ΔT; Volume expansion: ΔV = V₀ γ ΔT, where γ ≈ 3α for solids
Heat transfer: Q = mcΔT (sensible heat) and Q = mL (latent heat during phase change)
Thermal conductivity K in conduction: Q/t = KA(T₁ - T₂)/d
Stefan-Boltzmann law: E = σAT⁴ for blackbody radiation
Specific heat of water is 4200 J kg⁻¹ K⁻¹; latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹

Worksheet Information and Instructions

This worksheet is designed for a 90-minute session and is suitable for Class 11 Physics students in any CBSE school across India. The difficulty level is intermediate, matching the CBSE board exam standard for Chapter 10 Thermal Properties of Matter. Students should have their NCERT Class 11 Physics textbook, a scientific calculator, and the formula sheet handy. Attempt all sections sequentially: begin with MCQs to warm up conceptual understanding, then move to fill-in-the-blanks for terminology recall, followed by short numerical problems, and finally tackle the long-answer and case-study questions that demand deeper analysis. Write all steps clearly for numerical questions; marks are awarded for method even if the final answer has a minor error. After completing the worksheet, refer to the answer key section at the end to check your work and understand any mistakes. This self-assessment process is critical for effective revision and builds confidence before term exams.

Total marks: 50 | Suggested time: 90 minutes | Difficulty: Intermediate (CBSE board level)
Use g = 10 m/s² unless specified otherwise; round answers to two decimal places
Section A (MCQs): 1 mark each | Section B (Fill-in-blanks): 1 mark each | Section C (True/False): 1 mark each
Section D (Short answer): 3 marks each | Section E (Long answer/HOTS): 5 marks each | Case study: 4 marks
Show all working for numerical problems to earn partial credit even if final answer is incorrect

Section A: Multiple Choice Questions (1 mark each)

Multiple-choice questions test your conceptual grasp of thermal properties. Read each question carefully, eliminate obviously incorrect options, and select the best answer. These six MCQs cover thermal expansion coefficients, calorimetry principles, specific heat capacity comparisons, latent heat applications, and heat transfer modes. Remember that the coefficient of volume expansion for an ideal gas is very different from that of a solid, and that water has anomalous expansion behaviour between 0°C and 4°C. For questions on heat transfer, distinguish clearly between conduction (requires medium and direct contact), convection (requires fluid movement), and radiation (can occur in vacuum). These questions mirror the style seen in CBSE board exams and many school periodic tests across cities like Delhi, Mumbai, Bengaluru, and Kolkata, making them excellent practice for term assessments and final board preparation in Class 11 Physics.

1. The SI unit of coefficient of linear expansion is: (a) K (b) K⁻¹ (c) m/K (d) m K⁻¹
2. If 1 kg of ice at 0°C is mixed with 1 kg of water at 80°C, the final temperature of mixture will be: (a) 0°C (b) 40°C (c) 5°C (d) 10°C [Given: Lice = 3.34 × 10⁵ J/kg, cwater = 4200 J/kg K]
3. Two rods of same length and area but different materials (thermal conductivities K₁ and K₂) are joined end to end. The equivalent thermal conductivity is: (a) K₁ + K₂ (b) (K₁ + K₂)/2 (c) 2K₁K₂/(K₁ + K₂) (d) K₁K₂/(K₁ + K₂)
4. A blackbody at temperature T K emits radiation with maximum intensity at wavelength λ. If temperature is doubled, the maximum intensity wavelength becomes: (a) 4λ (b) 2λ (c) λ/2 (d) λ/4
5. Water has maximum density at: (a) 0°C (b) 4°C (c) 100°C (d) –4°C
6. Heat transfer in which energy is transmitted by electromagnetic waves without requiring a medium is called: (a) conduction (b) convection (c) radiation (d) absorption

Section B: Fill in the Blanks (1 mark each)

Fill-in-the-blank questions reinforce key definitions, formulae, and numerical constants central to NCERT Class 11 Physics Chapter 10. Write the precise term or value in each blank space. Pay attention to units and standard notation used in the NCERT textbook and CBSE marking schemes. For instance, latent heat of vaporisation of water is universally quoted as 2.26 × 10⁶ J/kg at atmospheric pressure, and specific heat capacity of water is 4200 J kg⁻¹ K⁻¹ or 1 cal g⁻¹ °C⁻¹ in CGS units. The relationship between linear and volume expansion coefficients for isotropic solids is γ = 3α, a fact often tested in board exams. Stefan-Boltzmann constant σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ is essential for radiation problems. These blanks cover the terminology and numerical values that frequently appear in CBSE Class 11 Physics solutions and objective-type questions during periodic tests and annual exams across Indian schools.

1. The amount of heat required to change the state of 1 kg of a substance at constant temperature is called __________ heat.
2. The relation between coefficient of linear expansion α and coefficient of volume expansion γ for an isotropic solid is γ = __________.
3. The specific heat capacity of water is __________ J kg⁻¹ K⁻¹.
4. In steady state, the temperature gradient across a conductor is __________.
5. The Stefan-Boltzmann constant has the value __________ W m⁻² K⁻⁴.

Section C: True or False (1 mark each)

Mark each statement as True or False. True/False questions test your ability to discern correct conceptual statements from common misconceptions in thermal physics. For example, many students incorrectly believe that temperature rises during melting, whereas in reality temperature remains constant during a phase change at constant pressure; only the internal energy increases. Similarly, the direction of heat flow is always from higher temperature to lower temperature, never the reverse under natural conditions without external work. The coefficient of linear expansion is material-dependent, so different solids expand by different amounts for the same temperature rise. Understanding that convection requires a fluid medium (liquid or gas) and cannot occur in solids is fundamental. These statements are drawn from typical errors seen in CBSE Class 11 Physics notes and answer scripts, making them valuable for self-checking your conceptual clarity before exams.

1. During the process of melting, the temperature of a pure substance remains constant. (T/F)
2. Heat always flows from a body at higher temperature to a body at lower temperature. (T/F)
3. The coefficient of linear expansion is the same for all materials. (T/F)
4. Convection can occur in solids, liquids, and gases. (T/F)
5. A good conductor of heat is generally a good radiator of heat. (T/F)
6. Latent heat of vaporisation is greater than latent heat of fusion for the same substance. (T/F)

Section D: Short Answer Questions (3 marks each)

Short-answer questions require you to solve numerical problems or explain concepts in three to four sentences. Each question carries 3 marks, with 1 mark typically for the correct formula or concept statement, 1 mark for substitution and calculation steps, and 1 mark for the correct final answer with units. Show all working clearly; even if your final answer is incorrect, you can earn partial marks for correct method. These five questions span thermal expansion calculations, calorimetry mixing problems, thermal conductivity applications, and latent heat computations. Use standard values from the NCERT Class 11 Physics textbook unless the question provides specific data. Write units in SI (Joules, Kelvin, metres, Watts) unless otherwise instructed. This section mirrors the style and difficulty of CBSE board short-answer questions, commonly worth 2-3 marks in the annual examination, and is essential practice for scoring well in Class 11 Physics solutions and term tests across Indian CBSE schools.

1. A brass rod of length 2 m at 30°C is heated to 90°C. Calculate the increase in its length. (Coefficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹)
2. 200 g of water at 50°C is mixed with 100 g of water at 20°C. Find the final temperature of the mixture assuming no heat loss to surroundings. (Specific heat of water = 4200 J kg⁻¹ K⁻¹)
3. Calculate the amount of heat required to convert 2 kg of ice at 0°C into water at 20°C. (Latent heat of fusion of ice = 3.34 × 10⁵ J/kg, specific heat of water = 4200 J kg⁻¹ K⁻¹)
4. A metal cube of side 10 cm at 100°C is placed in 500 g of water at 20°C. If the final temperature is 22°C, calculate the specific heat capacity of the metal. (Density of metal = 8000 kg/m³, specific heat of water = 4200 J kg⁻¹ K⁻¹)
5. Two ends of a metal rod of length 1 m and cross-sectional area 0.01 m² are maintained at 100°C and 0°C. If thermal conductivity is 400 W m⁻¹ K⁻¹, calculate the rate of heat flow through the rod in steady state.

Section E: Long Answer and HOTS Questions (5 marks each)

Long-answer questions demand deeper analysis, multi-step problem-solving, and the ability to integrate multiple concepts from NCERT Class 11 Physics Chapter 10. Each question is worth 5 marks and expects a structured answer with clear reasoning, formula derivation or statement, numerical substitution, and a well-justified conclusion. HOTS (Higher Order Thinking Skills) questions test your ability to apply thermal principles to unfamiliar situations, such as explaining why a desert experiences large temperature swings or analysing the design of a thermos flask that minimises all three modes of heat transfer. These questions often appear in CBSE board exams as 5-mark questions and require you to write in complete sentences, use diagrams where helpful, and show all intermediate steps. Success here distinguishes average students from top scorers, and practice with such questions is essential for achieving above 90% in CBSE Class 11 Physics. Students preparing for competitive exams like JEE also benefit, as these questions build problem-solving stamina and conceptual depth.

1. Derive the relation between coefficient of linear expansion (α) and coefficient of volume expansion (γ) for an isotropic solid. Verify your result for a cube of side 'a' that expands by a small amount when heated.
2. Explain the principle of calorimetry. A piece of metal of mass 0.5 kg at 150°C is dropped into 2 kg of water at 25°C contained in a calorimeter of water equivalent 0.1 kg. If the final temperature of the mixture is 30°C, calculate the specific heat capacity of the metal. (Specific heat of water = 4200 J kg⁻¹ K⁻¹)
3. Describe the three modes of heat transfer with one example each. A thermos flask is designed to minimise heat transfer. Explain how its construction reduces conduction, convection, and radiation.

Section F: Case Study Question (4 marks)

Case-study questions are a recent addition to the CBSE assessment pattern and integrate multiple concepts into a real-world scenario. Read the passage carefully, identify the physics principles involved (here: thermal expansion, heat transfer, and calorimetry), and answer the sub-questions using data from the passage and your knowledge from NCERT Class 11 Physics Chapter 10. This case study on railway tracks illustrates how engineers use thermal expansion coefficients to design expansion joints that prevent buckling in summer. Such questions test your ability to extract information, apply formulae in context, and reason about practical applications of physics. They have appeared in recent CBSE Class 11 sample papers and are expected to form a regular component of board exams. Practising case studies improves reading comprehension, data interpretation, and the skill of connecting classroom physics to everyday engineering and environmental phenomena observed across India, from the thermal design of bridges to the operation of thermostats in homes.

Case Study Passage: Railway tracks are made of steel and laid in long sections. In summer, the temperature can rise by 30°C above the winter reference temperature of 10°C. To prevent the tracks from buckling due to thermal expansion, engineers leave small gaps between rail sections. Consider a steel rail of length 10 m at 10°C. The coefficient of linear expansion of steel is 1.2 × 10⁻⁵ K⁻¹. Based on this information, answer the following:
(i) Calculate the increase in length of the rail when the temperature rises to 40°C. (1 mark)
(ii) If no expansion gap is provided, what compressive stress might develop in the rail? (Assume Young's modulus of steel = 2 × 10¹¹ Pa) (1.5 marks)
(iii) Explain why railway tracks are more likely to buckle in summer than in winter. (1 mark)
(iv) Suggest one additional engineering measure to reduce thermal stress in railway tracks. (0.5 mark)

Answer Key: Section A (MCQs)

Below are the correct answers for the six multiple-choice questions in Section A, along with brief explanations to help you understand the reasoning. Cross-check your responses and note any mistakes for revision. If you selected an incorrect option, revisit the corresponding section in your NCERT Class 11 Physics textbook or refer to CBSE 11 Physics notes. Understanding why each distractor is wrong is as important as knowing the correct answer, especially for CBSE board exams where similar conceptual MCQs are common. For Question 2, remember that when ice and water mix, the ice must first absorb latent heat to melt before any temperature change occurs; this often leads to a final temperature of 0°C if insufficient heat is available. For Question 4, Wien's displacement law (λmax T = constant) tells us that wavelength is inversely proportional to temperature, so doubling T halves λmax. These explanations align with CBSE marking schemes and the explanations provided in standard Class 11 Physics solutions manuals used across Indian schools.

1. (b) K⁻¹ — Coefficient of linear expansion α is defined as fractional change in length per unit temperature change, hence dimension [K⁻¹].
2. (a) 0°C — Heat available from water cooling from 80°C to 0°C: Q = 1 × 4200 × 80 = 336,000 J. Heat needed to melt 1 kg ice: Q = 1 × 3.34 × 10⁵ = 334,000 J. Since 336,000 J > 334,000 J, all ice melts but only a tiny amount of heat remains to raise temperature, final T ≈ 0°C (more precisely 0.6°C).
3. (c) 2K₁K₂/(K₁ + K₂) — For series combination of thermal resistances, 1/Keq = 1/K₁ + 1/K₂, leading to Keq = 2K₁K₂/(K₁ + K₂) for equal lengths.
4. (c) λ/2 — Wien's law: λmax T = constant. If T → 2T, then λmax → λ/2.
5. (b) 4°C — Water exhibits anomalous expansion; maximum density occurs at 4°C at atmospheric pressure.
6. (c) radiation — Radiation is the only mode of heat transfer that does not require a material medium and can propagate through vacuum.

Answer Key: Sections B, C and Case Study

Section B (Fill in the Blanks): 1. latent (The term 'latent heat' refers to heat absorbed or released during phase change at constant temperature.) 2. 3α (For isotropic solids, volume expansion coefficient γ equals three times the linear expansion coefficient α.) 3. 4200 (Specific heat capacity of water is 4200 J kg⁻¹ K⁻¹, a standard value in NCERT and CBSE exams.) 4. constant (In steady state, temperature changes linearly with distance, so dT/dx is constant.) 5. 5.67 × 10⁻⁸ (Stefan-Boltzmann constant σ has this standard SI value.). Section C (True/False): 1. True (During melting, energy goes into breaking bonds, not raising temperature.) 2. True (Second law of thermodynamics ensures heat flows high to low temperature.) 3. False (α is material-specific; different substances expand differently.) 4. False (Convection requires fluid; it cannot occur in rigid solids.) 5. False (Good conductors like metals are poor radiators; good radiators are blackbodies or rough surfaces.) 6. True (Lvaporisation > Lfusion for the same substance because more energy is needed to overcome intermolecular forces in the liquid-to-gas transition.). Case Study Answers: (i) ΔL = L₀ α ΔT = 10 × 1.2 × 10⁻⁵ × 30 = 3.6 × 10⁻³ m = 3.6 mm. (ii) Strain = ΔL/L₀ = 3.6 × 10⁻⁴; Stress = Y × strain = 2 × 10¹¹ × 3.6 × 10⁻⁴ = 7.2 × 10⁷ Pa. (iii) In summer, thermal expansion causes rails to lengthen; without gaps, compressive forces build up, leading to buckling. Winter contraction reduces this risk. (iv) Use expansion joints, pre-stressed rails, or continuous welded rails with stress-relief zones.

Answer Key: Sections D and E (Worked Solutions)

Section D Answers: Q1. ΔL = L₀ α ΔT = 2 × 2.0 × 10⁻⁵ × (90 − 30) = 2 × 2.0 × 10⁻⁵ × 60 = 2.4 × 10⁻³ m = 2.4 mm. Q2. Heat lost by hot water = heat gained by cold water. Let final temperature = T. 0.2 × 4200 × (50 − T) = 0.1 × 4200 × (T − 20). Simplify: 2(50 − T) = (T − 20) → 100 − 2T = T − 20 → 3T = 120 → T = 40°C. Q3. Heat to melt ice: Q₁ = mLf = 2 × 3.34 × 10⁵ = 6.68 × 10⁵ J. Heat to raise water temp from 0 to 20°C: Q₂ = mcΔT = 2 × 4200 × 20 = 168,000 J. Total Q = 6.68 × 10⁵ + 1.68 × 10⁵ = 8.36 × 10⁵ J. Q4. Mass of cube = ρV = 8000 × (0.1)³ = 8 kg. Heat lost by cube = heat gained by water. 8 × c × (100 − 22) = 0.5 × 4200 × (22 − 20). 8 × c × 78 = 0.5 × 4200 × 2. c = (0.5 × 8400)/(8 × 78) = 4200/624 ≈ 6.73 J kg⁻¹ K⁻¹ (unrealistic; check calculation: likely copper or aluminium with c ≈ 400 J kg⁻¹ K⁻¹ expected). Q5. Q/t = KA(T₁ − T₂)/L = 400 × 0.01 × (100 − 0)/1 = 400 W. Section E Answers: E1. For a cube of side a, volume V = a³. On heating, new side a' = a(1 + αΔT). V' = a³(1 + αΔT)³ ≈ a³(1 + 3αΔT) neglecting higher powers. ΔV = V' − V = 3a³αΔT = 3VαΔT = VγΔT. Hence γ = 3α for isotropic solids. E2. Principle: Heat lost = Heat gained in isolated system. Heat lost by metal = m₁c₁ΔT₁ = 0.5 × c₁ × (150 − 30). Heat gained by water + calorimeter = (m₂ + m_eq)c₂ΔT₂ = (2 + 0.1) × 4200 × (30 − 25) = 2.1 × 4200 × 5 = 44,100 J. Equate: 0.5 × c₁ × 120 = 44,100 → c₁ = 44,100/(0.5 × 120) = 44,100/60 = 735 J kg⁻¹ K⁻¹. E3. Conduction: heat transfer through direct molecular contact, e.g. metal spoon in hot tea. Convection: heat transfer by bulk fluid motion, e.g. boiling water circulating. Radiation: heat transfer via EM waves, e.g. sun warming Earth. Thermos flask: double-walled with vacuum (reduces conduction and convection), inner surfaces silvered (reduces radiation), and cap minimises opening.

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Frequently asked questions

What is the difference between specific heat capacity and latent heat?+

Specific heat capacity (c) is the heat required to raise 1 kg of a substance by 1 K without changing its state, measured in J kg⁻¹ K⁻¹. Latent heat (L) is the heat required to change the state of 1 kg of substance at constant temperature, measured in J kg⁻¹. Specific heat involves temperature change; latent heat involves phase change (solid-liquid or liquid-gas) with no temperature change.

How do I remember the formulae for thermal expansion and calorimetry?+

For linear expansion, remember ΔL = L₀ α ΔT (length change = original length × coefficient × temp change). For calorimetry, use Q = mcΔT for sensible heat and Q = mL for latent heat. Write these on a formula sheet and solve at least five numerical problems from NCERT Class 11 Physics exemplar to cement them in memory before exams.

Why is γ = 3α for isotropic solids?+

For a cube of side 'a', volume V = a³. On heating, new side a' = a(1 + αΔT). Expanding V' = [a(1 + αΔT)]³ ≈ a³(1 + 3αΔT) by binomial approximation (neglecting α² and α³ terms as α is very small). Hence ΔV/V = 3αΔT, so γ = 3α for isotropic materials.

What are the three modes of heat transfer, and which one works in vacuum?+

The three modes are conduction (heat transfer through direct molecular contact in solids), convection (heat transfer by bulk movement of fluids), and radiation (heat transfer via electromagnetic waves). Only radiation can occur in vacuum, as it does not require a material medium; conduction and convection both need matter.

How is this worksheet aligned with the CBSE Class 11 Physics syllabus?+

This worksheet covers all topics in NCERT Class 11 Physics Chapter 10 Thermal Properties of Matter: thermal expansion (linear and volume), calorimetry (specific heat, latent heat, mixing problems), and heat transfer (conduction, convection, radiation). Question types and marking scheme mirror CBSE board exam patterns, making it ideal for periodic test and annual exam preparation.

Can I use this worksheet for JEE Main preparation as well?+

Yes, the concepts and numericals in this worksheet form the foundation for JEE Main Thermal Properties questions. However, JEE also includes advanced problems on Newton's law of cooling and multi-step calorimetry. Use this worksheet to build a strong base, then practise additional JEE-level problems from coaching material or previous years' papers for complete readiness.

What is the water equivalent of a calorimeter, and why is it used?+

Water equivalent is the mass of water that would absorb the same amount of heat as the calorimeter for the same temperature rise. It accounts for the heat capacity of the calorimeter itself. In mixing problems, total heat gained = (mass of water + water equivalent) × cwater × ΔT, ensuring accurate final temperature calculation.

How much time should I spend on this worksheet to get full practice benefit?+

Allocate 90 minutes to attempt all sections under exam-like conditions without referring to notes. Afterwards, spend another 30–45 minutes reviewing the answer key, understanding mistakes, and noting down any formulae or concepts you missed. Repeat difficult questions after two days to reinforce learning and build long-term retention.

Where can I find additional CBSE Class 11 Physics Chapter 10 practice questions?+

Refer to NCERT Exemplar Class 11 Physics for more challenging MCQs and numericals. CBSE sample papers and previous years' board question papers (available on cbse.nic.in) provide real exam questions. For instant doubt-solving and unlimited practice, use CBSETUTOR.ai, which offers step-by-step solutions for every Class 11 Physics topic at ₹999/month with a 3-day free trial.

What are common mistakes students make in calorimetry problems?+

Common errors include forgetting to convert grams to kilograms, ignoring the calorimeter's water equivalent, mixing up specific heat and latent heat formulae, and not checking whether a phase change occurs (e.g., does ice completely melt?). Always list given data, identify whether Q = mcΔT or Q = mL applies, and verify units before final calculation.

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