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Chapter 10 of 30
Practice Quiz

Thermal Properties of Matter

ICSE · Class 11 · Physics

Practice quiz for Thermal Properties of Matter — ICSE Class 11 Physics. MCQs and questions with answers to test your preparation.

43 questions45 flashcards5 concepts

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Quick Quiz: Thermal Properties of Matter

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1

A steel rod of length 1 m and cross-sectional area 1 cm² is heated from 0°C to 200°C. If the rod is not allowed to expand (rigidly clamped at both ends), what is the compressive stress developed in the rod? (Given: α for steel = 1.2 × 10⁻⁵ /°C, Young's modulus Y = 2 × 10¹¹ Pa)

2

A platinum resistance thermometer records a resistance of 5 Ω at 0°C and 7 Ω at 100°C. When immersed in a hot liquid, it records 6.5 Ω. What is the temperature of the hot liquid?

3

A glass flask of volume 200 cm³ is completely filled with mercury at 20°C. How much mercury will overflow when the temperature is raised to 100°C? (Given: γ_mercury = 1.82 × 10⁻⁴ /°C, γ_glass = 2.7 × 10⁻⁵ /°C)

4

A pendulum clock keeps correct time at 20°C. The length of the pendulum is 1 m and α = 1.2 × 10⁻⁵ /°C. How many seconds will the clock lose or gain per day when the temperature rises to 40°C?

43 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

100 g of ice at 0°C is mixed with 100 g of steam at 100°C. What is the final temperature and composition of the mixture? (Latent heat of ice = 80 cal/g, Latent heat of steam = 540 cal/g, specific heat of water = 1 cal/g°C)

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100°C; mixture contains only water

Step 1: Heat required to convert 100 g ice at 0°C to water at 100°C = 100×80 + 100×1×100 = 8000 + 10000 = 18000 cal. Step 2: Heat available from condensing 100 g steam at 100°C = 100 × 540 = 54000 cal. Step 3: Since heat available (54000 cal) > heat required (18000 cal), all ice converts to water at 100°C. Remaining heat = 54000 − 18000 = 36000 cal after condensing some steam. Let x g of steam condense: x × 540 = 18000, x = 33.3 g. So 33.3 g steam condenses and all ice melts to water at 100°C. Final state: 100°C, all water. Step 4: All steam (100 g) gives 54000 cal but only 18000 cal is needed

2multiple choice
1 marks

A copper calorimeter of mass 100 g contains 200 g of water at 20°C. A piece of metal of mass 150 g at 80°C is dropped into the calorimeter. The final temperature is 25°C. If specific heat of copper = 0.1 cal/g°C and water = 1 cal/g°C, find the specific heat of the metal.

Show answer

0.1273 cal/g°C

Step 1: Heat gained by water = 200 × 1 × (25 − 20) = 1000 cal. Step 2: Heat gained by calorimeter = 100 × 0.1 × (25 − 20) = 50 cal. Step 3: Total heat gained = 1000 + 50 = 1050 cal. Step 4: By principle of calorimetry, heat lost by metal = heat gained. 150 × c × (80 − 25) = 1050. 150 × c × 55 = 1050. c = 1050 / 8250 = 0.1273 cal/g°C. Option B doubles the answer by using ΔT = 27.5 for metal. Option C halves the answer. Always account for calorimeter's heat gain.

3multiple choice
1 marks

In an experiment, 10 g of a solid at 100°C is dropped into a calorimeter containing 50 g of water at 20°C. The water equivalent of the calorimeter is 5 g. The final temperature is 30°C. What is the specific heat of the solid?

Show answer

0.786 cal/g°C

Step 1: Heat gained = (mass of water + water equivalent of calorimeter) × specific heat of water × rise in temp = (50 + 5) × 1 × (30 − 20) = 55 × 10 = 550 cal. Step 2: Heat lost by solid = m × c × ΔT = 10 × c × (100 − 30) = 700c. Step 3: By principle of calorimetry: 700c = 550. c = 550/700 = 0.786 cal/g°C. Option B forgets the water equivalent. Option C uses (100−20) instead of (100−30) for solid's temperature drop. Option D uses ΔT = 140 instead of 70.

4multiple choice
1 marks

The coefficient of volume expansion of an ideal gas at constant pressure is γ_P and at constant volume is γ_V. Which of the following statements is correct regarding an ideal gas at room temperature (300 K)?

Show answer

γ_P = γ_V = 1/T = 3.33 × 10⁻³ K⁻¹

Step 1: From ideal gas equation PV = μRT. At constant pressure: ΔV/V = ΔT/T, so γ_P = ΔV/(VΔT) = 1/T. Step 2: At constant volume: ΔP/P = ΔT/T, so γ_V = ΔP/(PΔT) = 1/T. Step 3: Both are equal to 1/T. At T = 300 K: γ_P = γ_V = 1/300 = 3.33 × 10⁻³ K⁻¹. Step 4: This shows that for ideal gases, both expansion coefficients are equal and DEPEND on temperature (they decrease as T increases). Option B is wrong — there is no reason for γ_P > γ_V. Option D is wrong because γ depends on T.

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What are the important topics in Thermal Properties of Matter for ICSE Class 11 Physics?
Key topics in Thermal Properties of Matter include Thermal Properties of Matter — Complete Chapter Overview, Step-by-step process showing how liquid level changes during heating in a container, Comparison of three temperature scales with conversion formulas. These are the concepts ICSE Class 11 examiners draw on most — study them first, then practise related questions.
How to score full marks in Thermal Properties of Matter — ICSE Class 11 Physics?
Understand the core concepts first, then work through the 43 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

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