Class 11 Physics Chapter 10 Thermal Properties of Matter – Extra Questions with Answers

Temperature scales, specific heat, and modes of heat transfer are the recurring themes tested from this chapter, starting with conversions like 100°C into Kelvin. The questions below work through calorimetry, expansion coefficients, and radiation in short, exam-style answers.

Last Updated: September 23, 2026

Very Short Answer Questions (1 mark)

Q1. Convert 100°C to Kelvin.
Ans: K=100+273.15=373.15 K.

Q2. What is the SI unit of specific heat capacity?
Ans: J/(kg·K).

Q3. Which mode of heat transfer does not require a medium?
Ans: Radiation.

Q4. Write the relation between linear (α) and volume (γ) expansion coefficients.
Ans: γ=3α.

Q5. What is latent heat?
Ans: The heat absorbed or released per unit mass during a phase change at constant temperature.

Short Answer Questions (2–3 marks)

Q6. A metal rod of length 2 m and α=1.2×10⁻⁵/°C is heated through 50°C. Find its increase in length.
Ans: ΔL=LαΔT=2(1.2×10⁻⁵)(50)=1.2×10⁻³ m=1.2 mm.

Q7. How much heat is needed to raise the temperature of 2 kg of water by 10°C (c_water=4200 J/kg·K)?
Ans: Q=mcΔT=2(4200)(10)=84000 J.

Q8. Find the heat needed to melt 0.5 kg of ice at 0°C (L_fusion=3.34×10⁵ J/kg).
Ans: Q=mL=0.5(3.34×10⁵)=1.67×10⁵ J.

Higher-Order Thinking / Application Questions

Q9. 0.1 kg of steam at 100°C is passed into 1 kg of water at 20°C. Using the principle of calorimetry (and given L_vaporisation=2.26×10⁶ J/kg, c_water=4200 J/kg·K), set up (without fully solving) the heat balance equation, explaining each term.
Ans: Heat lost by steam condensing and then cooling = Heat gained by water in warming up. Heat lost = m_steamL_vap + m_steamc_water(100−T_f), where the first term is the heat released when steam condenses to water at 100°C, and the second term is heat released as this condensed water cools from 100°C to the final temperature T_f. Heat gained = m_waterc_water(T_f−20), the heat absorbed by the original water as it warms from 20°C to T_f. Setting heat lost = heat gained gives an equation that can be solved for T_f.

Q10. Explain why a metal spoon left in a hot cup of tea feels hot at the top end (outside the tea) after some time, and why the same effect is not observed with a wooden spoon, in terms of thermal conductivity.
Ans: The metal spoon has a high thermal conductivity, meaning heat is efficiently conducted through its molecular/electronic structure from the hot end (in contact with the tea) to the cooler end (outside the cup), causing the top of the spoon to become noticeably warm within a short time due to conduction. A wooden spoon has very low thermal conductivity (wood is a good thermal insulator), so heat conducted from the hot end to the top is extremely slow and negligible over the same time period, meaning the top of a wooden spoon remains close to room temperature even after prolonged contact with hot tea.

Quick visual: a worked diagram from the full Solutions page, for reference.

Frequently Asked Questions

How would you convert a temperature of 37 degrees Celsius to Kelvin?
Add 273 to the Celsius value, giving 310 K.

If 500 g of water is heated from 20 to 80 degrees Celsius, how would you find the heat required?
Using Q = mc times delta T = 0.5 times 4200 times 60 = 126000 joules.

Written by Satish

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