Class XI · Chapter 10
Thermal Properties of Matter — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering temperature scales, specific heat, calorimetry, and heat transfer.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Why does a body at higher temperature radiate significantly more energy than one at lower temperature?
Answer
By Stefan’s Law, radiated energy per unit area per unit time is E = σT⁴ — proportional to the fourth power of absolute temperature. Even a modest temperature increase produces a dramatic rise in radiated energy.
By Stefan’s Law, radiated energy per unit area per unit time is E = σT⁴ — proportional to the fourth power of absolute temperature. Even a modest temperature increase produces a dramatic rise in radiated energy.
Distinguish between heat and specific heat capacity, and derive the expression for heat required to change a substance’s temperature.
Answer
Heat (Q) is energy transferred due to a temperature difference. Specific heat capacity (c) is a material property: the heat needed to raise 1 kg of the substance by 1°C.
Heat (Q) is energy transferred due to a temperature difference. Specific heat capacity (c) is a material property: the heat needed to raise 1 kg of the substance by 1°C.
For a mass m undergoing a temperature change ΔT, the total heat required is: Q = mcΔT, obtained by scaling the per-unit-mass, per-degree heat requirement (c) by both the actual mass and the actual temperature change.
Q3. (a) Explain the three modes of heat transfer — conduction, convection, and radiation — with one everyday example of each. (b) Find the heat required to convert 2 kg of ice at 0°C completely into water at 0°C. (Latent heat of fusion of ice = 336,000 J/kg)
Answer
(a) Conduction: heat passed through direct particle collisions with no bulk motion — a metal spoon heating up in hot soup. Convection: heat carried by bulk fluid movement — water circulating as it boils. Radiation: heat transferred as electromagnetic waves, needing no medium — sunlight warming the Earth across empty space.
(a) Conduction: heat passed through direct particle collisions with no bulk motion — a metal spoon heating up in hot soup. Convection: heat carried by bulk fluid movement — water circulating as it boils. Radiation: heat transferred as electromagnetic waves, needing no medium — sunlight warming the Earth across empty space.
(b) Q = mL = 2 × 336,000 = 672,000 J = 672 kJ.
2NEET Style (Single-Correct MCQ)
Q4. The SI unit of heat is:
Solution
Heat is a form of energy, so it shares the SI unit of energy — the joule.
Heat is a form of energy, so it shares the SI unit of energy — the joule.
Q5. The triple point of water occurs at:
Solution
The triple point — where solid, liquid, and gas phases coexist in equilibrium — occurs at one uniquely defined temperature and pressure for water, making it a reliable reference for defining temperature scales.
The triple point — where solid, liquid, and gas phases coexist in equilibrium — occurs at one uniquely defined temperature and pressure for water, making it a reliable reference for defining temperature scales.
Q6. Thermal conductivity is generally highest for:
Solution
Metals have abundant free electrons that efficiently transport thermal energy, giving them far higher thermal conductivity than insulators or gases.
Metals have abundant free electrons that efficiently transport thermal energy, giving them far higher thermal conductivity than insulators or gases.
Q7. Newton’s Law of Cooling is valid only for:
Solution
The linear approximation dT/dt ∝ (T−Ts) breaks down for large temperature differences, where the more general Stefan’s Law behaviour dominates instead.
The linear approximation dT/dt ∝ (T−Ts) breaks down for large temperature differences, where the more general Stefan’s Law behaviour dominates instead.
3JEE Main Style
Q8. 500 g of water at 30°C is mixed with 300 g of water at 80°C. Find the final equilibrium temperature (assume no heat loss to surroundings).
Solution
Heat lost by hot water = heat gained by cold water: 300(80−T) = 500(T−30).
Heat lost by hot water = heat gained by cold water: 300(80−T) = 500(T−30).
24000 − 300T = 500T − 15000 → 39000 = 800T → T = 48.75°C.
Q9. The SI unit of the coefficient of linear expansion (α) is:
Solution
From ΔL = αLΔT, α must have units of (length/length)/temperature = 1/°C or K⁻¹.
From ΔL = αLΔT, α must have units of (length/length)/temperature = 1/°C or K⁻¹.
Q10. A 2 m steel rod is heated from 20°C to 120°C. Find its final length. (αsteel = 12×10⁻⁶ /°C)
Solution
ΔL = αLΔT = 12×10⁻⁶ × 2 × 100 = 0.0024 m = 2.4 mm.
ΔL = αLΔT = 12×10⁻⁶ × 2 × 100 = 0.0024 m = 2.4 mm.
Final length = 2 + 0.0024 = 2.0024 m.
You must be logged in to post a comment.