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Thermal Properties of Matter NCERT Solutions, Complete PDF Content and Important Questions
NCERT Solutions for Class 11 Physics Chapter 10 PDF Download (2026) + Important Questions: Thermal Properties of Matter Welcome to Class 11 Physics! Have you ever wondered why there are small gaps left between railway tracks? Or why a hot cup of tea eventually cools down to room temperature, but never gets colder than the room itself? These everyday phenomena are governed by the concepts you will learn in CBSE Class 11 Physics Chapter 10: Thermal Properties of Matter . Whether you're gearing up for your 2026 board exams or preparing for highly competitive entrance tests like JEE Main and NEET, this chapter is incredibly important. It forms the foundation of Thermodynamics and introduces you to heat, temperature, specific heat capacity, and the various mechanisms of heat transfer.
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ChapterThermal Properties of Matter
SubjectPhysics
Class11
BoardCBSE
DifficultyModerate
Exam Weightage5 - 7 Marks
Learning Objectives
- Understand the fundamental difference between heat and temperature. Convert temperatures accurately between Celsius, Fahrenheit, and Kelvin scales.
- Calculate linear, area, and volume thermal expansion for different materials.
- Apply the principle of calorimetry to.
- Solve complex heat mixture problems.
- Differentiate between conduction, convection, and radiation.
- Apply Newton's Law of Cooling, Wien's Displacement Law, and the Stefan-Boltzmann Law to real-world scenarios.
Key Concepts, Definitions and Formulas
- Temperature and Heat: Heat is a form of energy transferring between two bodies due to a temperature difference. Temperature is the degree of hotness or coldness of a body
- Temperature Scales Conversion: $$\frac{T_C}{100} = \frac{T_F - 32}{180} = \frac{T_K - 273.15}{100}$$
- Thermal Expansion: The increase in the dimensions of a body due to an increase in temperature
- Linear Expansion: $\Delta L = \alpha L \Delta T$
- Area Expansion: $\Delta A = \beta A \Delta T$ (where $\beta = 2\alpha$)
- Volume Expansion: $\Delta V = \gamma V \Delta T$ (where $\gamma = 3\alpha$) Specific Heat Capacity ($s$): The amount of heat required to raise the temperature of a unit mass of a substance by 1C
- Heat Transfer (Conduction): The rate of heat flow through a solid
- Wien's Displacement Law: The wavelength corresponding to maximum energy emission is inversely proportional to the absolute temperature
- Stefan-Boltzmann Law: The total heat energy emitted by a perfect black body per second per unit area is directly proportional to the fourth power of its absolute temperature
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Full NCERT Solutions
NCERT Q10.1: The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Answer: Step 1: Convert Kelvin to Celsius Formula: $T_C = T_K - 273.15$ For Neon: $T_C = 24.57 - 273.15 = -248.58^\circ\text{C}$ For $CO_2$: $T_C = 216.55 - 273.15 = -56.60^\circ\text{C}$ Step 2: Convert Celsius to Fahrenheit Formula: $T_F = \frac{9}{5}T_C + 32$ For Neon: $T_F = \frac{9}{5}(-248.58) + 32 = -447.44 + 32 = -415.44^\circ\text{F}$ For $CO_2$: $T_F = \frac{9}{5}(-56.60) + 32 = -101.88 + 32 = -69.88^\circ\text{F}$
NCERT Q10.2: Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between $T_A$ and $T_B$?
Answer: The triple point of water on the standard absolute (Kelvin) scale is $T_{tr} = 273.16 \text{ K}$. Given: Triple point on scale A = 200 A Triple point on scale B = 350 B Since both scales are absolute scales, the size of 1 degree on scale A is $\frac{273.16}{200}$ K, and on scale B is $\frac{273.16}{350}$ K. To find a relation between $T_A$ and $T_B$, we equate the corresponding temperatures: $$\frac{T_A}{200} = \frac{T_B}{350}$$ $$\frac{T_A}{T_B} = \frac{200}{350} = \frac{4}{7}$$ Result: $7 T_A = 4 T_B$
NCERT Q10.3: The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: $R = R_0 [1 + \alpha (T - T_0)]$. The resistance is $101.6 \text{ \Omega}$ at the triple-point of water 273.16 K, and $165.5 \text{ \Omega}$ at the normal melting point of lead (600.5 K). What is the temperature when the resistance is $123.4 \text{ \Omega}$?
Answer: Given: $R_0 = 101.6 \text{ \Omega}$ at $T_0 = 273.16 \text{ K}$ $R_1 = 165.5 \text{ \Omega}$ at $T_1 = 600.5 \text{ K}$ We need to find $T_2$ when $R_2 = 123.4 \text{ \Omega}$. Step 1: Find $\alpha$ Using $R_1 = R_0 [1 + \alpha (T_1 - T_0)]$: $165.5 = 101.6 [1 + \alpha (600.5 - 273.16)]$ $165.5 = 101.6 [1 + \alpha (327.34)]$ $\frac{165.5}{101.6} - 1 = \alpha (327.34)$ $1.6289 - 1 = \alpha (327.34)$ $\alpha = \frac{0.6289}{327.34} \approx 1.92 \times 10^{-3} \text{ K}^{-1}$ Step 2: Find $T_2$ Using $R_2 = R_0 [1 + \alpha (T_2 - T_0)]$: $123.4 = 101.6 [1 + (1.92 \times 10^{-3}) (T_2 - 273.16)]$ $\frac{123.4}{101.6} - 1 = (1.92 \times 10^{-3}) (T_2 - 273.16)$ $1.2145 - 1 = (1.92 \times 10^{-3}) (T_2 - 273.16)$ $0.2145 = (1.92 \times 10^{-3}) (T_2 - 273.16)$ $T_2 - 273.16 = \frac{0.2145}{1.92 \times 10^{-3}} \approx 111.7$ $T_2 = 111.7 + 273.16 = 384.86 \text{ K}$
NCERT Q10.5: A steel tape 1m long is correctly calibrated for a temperature of 27.0 C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 C? Coefficient of linear expansion of steel = $1.20 \times 10^{-5} \text{ K}^{-1}$.
Answer: Given: Calibrated temperature $T_1 = 27^\circ\text{C}$ Hot day temperature $T_2 = 45^\circ\text{C}$ Measured length $L = 63.0 \text{ cm}$ $\alpha_{steel} = 1.20 \times 10^{-5} \text{ K}^{-1}$ 1. Actual length of the steel rod on the hot day (45 C): Because the tape is made of steel, it expands on the hot day. Therefore, the distance between the centimeter marks on the tape increases. The tape reads less than the actual length. Actual length = Measured length + Expansion of that length of the tape Expansion $\Delta L = \alpha L \Delta T = (1.20 \times 10^{-5}) \times 63.0 \times (45 - 27)$ $\Delta L = 1.20 \times 10^{-5} \times 63.0 \times 18 \approx 0.0136 \text{ cm}$ Actual length at $45^\circ\text{C} = 63.0 + 0.0136 = 63.0136 \text{ cm}$. 2. Length of the steel rod on a day when temperature is 27.0 C: Let $L_0$ be the length at 27 C. $L_{45} = L_0 [1 + \alpha (45 - 27)]$ $63.0136 = L_0 [1 + (1.2 \times 10^{-5}) \times 18]$ $63.0136 = L_0 [1 + 0.000216]$ $L_0 = \frac{63.0136}{1.000216} \approx 63.0 \text{ cm}$.
NCERT Q10.6: A large steel wheel is to be fitted on to a shaft of the same material. At 27 C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant and equal to $1.20 \times 10^{-5} \text{ K}^{-1}$.
Answer: Given: Initial temperature $T_1 = 27^\circ\text{C}$ Diameter of shaft $d_1 = 8.70 \text{ cm}$ Diameter of hole $d_2 = 8.69 \text{ cm}$ (This is the target diameter the shaft must shrink to) $\alpha = 1.20 \times 10^{-5} \text{ K}^{-1}$ Using the linear expansion formula for diameter: $d_2 = d_1 [1 + \alpha (T_2 - T_1)]$ $8.69 = 8.70 [1 + (1.20 \times 10^{-5}) (T_2 - 27)]$ $\frac{8.69}{8.70} - 1 = (1.20 \times 10^{-5}) (T_2 - 27)$ $-0.001149 = 1.20 \times 10^{-5} (T_2 - 27)$ $T_2 - 27 = \frac{-0.001149}{1.20 \times 10^{-5}} = -95.75^\circ\text{C}$ $T_2 = 27 - 95.75 = -68.75^\circ\text{C}$. Result: The shaft must be cooled to $-68.75^\circ\text{C}$ .
Extra Important Questions for 2026 Exams
Important Q1: The temperature of a body on Kelvin scale is found to be x K. When it is measured in Fahrenheit scale, it is found to be x F. Then the value of x is: A) 301.25 B) 574.25 C) 313 D) 40
Answer: B) 574.25 Explanation: Using the conversion formula: $\frac{x - 273.15}{100} = \frac{x - 32}{180}$. Solving for x yields $x \approx 574.25$.
Important Q2: Which of the following materials has the highest specific heat capacity? A) Copper B) Ice C) Water D) Mercury
Answer: C) Water Explanation: Water has a specific heat capacity of $4186 \text{ J kg}^{-1}\text{K}^{-1}$, which is exceptionally high, making it an excellent coolant.
Important Q3: According to Wien's displacement law, the color of a star indicates its: A) Mass B) Distance C) Temperature D) Size
Answer: C) Temperature Explanation: $\lambda_m \propto \frac{1}{T}$. The wavelength corresponding to maximum intensity (color) is inversely proportional to its absolute temperature.
Important Q4: Why are gaps left between rails on a railway track?
Answer: Metals expand upon heating. During summers, the steel rails undergo linear thermal expansion due to the high temperature. If gaps are not left, the rails would bend or warp, leading to train derailments.
Important Q5: Distinguish between heat capacity and specific heat capacity.
Answer: Heat capacity is the amount of heat required to raise the temperature of the entire body by 1C (Unit: J/K). Specific heat capacity is the amount of heat required to raise the temperature of a unit mass (1 kg) of the substance by 1C (Unit: J/kgK).
Important Q6: State Newton's Law of Cooling.
Answer: It states that the rate of loss of heat (rate of cooling) of a body is directly proportional to the temperature difference between the body and its surroundings, provided the temperature difference is small.
Important Q7: Why do we feel warmer on a cloudy night than on a clear night?
Answer: Earth radiates heat back into space continuously. On a cloudy night, the clouds act as a blanket (opaque to infrared radiation) and reflect the radiated heat back to Earth, trapping the heat and keeping the atmosphere warm.
Important Q8: State Stefan-Boltzmann Law. If a black body emits radiation at a rate of E at temperature T, what will be the rate of emission if the temperature is doubled?
Answer: Statement: The total thermal radiated energy emitted per second per unit area of a perfect black body is directly proportional to the fourth power of its absolute temperature. $E = \sigma T^4$. Calculation: If the initial temperature is $T$, $E_1 = \sigma T^4$. If the temperature is doubled to $2T$, the new emission rate $E_2 = \sigma (2T)^4 = 16 \sigma T^4$. Therefore, $E_2 = 16 E_1$. The rate of emission becomes 16 times the original.
Important Q9: Explain the principle of Calorimetry. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J/gC; heat of fusion of water = 335 J/g).
Answer: Principle of Calorimetry: When two bodies at different temperatures are mixed in an isolated system, the heat lost by the hot body is equal to the heat gained by the cold body. Calculation: Mass of copper $m_c = 2.5 \text{ kg} = 2500 \text{ g}$. Initial temp of copper $T_1 = 500^\circ\text{C}$. Final temp $T_2 = 0^\circ\text{C}$ (since it rests on ice). Heat lost by copper $Q = m_c \cdot s_c \cdot \Delta T = 2500 \times 0.39 \times (500 - 0) = 487,500 \text{ J}$. Let mass of ice melted be $m_i$. Heat gained by ice to melt $Q = m_i \cdot L$. $m_i \times 335 = 487,500$ $m_i = \frac{487500}{335} \approx 1455 \text{ g} = 1.455 \text{ kg}$.
Important Q11: Which mode of heat transfer is minimized by the vacuum between the double walls?
Answer: Conduction and Convection. Both require a material medium to transfer heat, which the vacuum removes.
Important Q12: What is the purpose of silvering the inner walls of the flask?
Answer: The highly polished silvered walls act as mirrors. They reflect the thermal radiation (infrared waves) back into the liquid, minimizing heat loss by Radiation.
Important Q13: Why is cork specifically chosen for the stopper?
Answer: Cork is a bad conductor of heat. It prevents heat transfer via conduction through the mouth of the flask. (Directions: Select A if both are true and Reason explains Assertion. Select B if both are true but Reason does not explain Assertion. Select C if Assertion is true, Reason is false. Select D if Assertion is false, Reason is true.)
Important Q14: Assertion (A): Water is used as a coolant in automobile radiators. Reason (R): Water has a very high specific heat capacity.
Answer: A . Both are true, and the reason correctly explains the assertion. Because of its high specific heat, water can absorb a large amount of heat energy from the engine without a significant rise in its own temperature.
Important Q15: Assertion (A): The coefficient of volume expansion is approximately half the coefficient of linear expansion. Reason (R): Volume expansion occurs in three dimensions.
Answer: D . The assertion is false. The coefficient of volume expansion ($\gamma$) is approximately three times the coefficient of linear expansion ($\gamma = 3\alpha$). The reason is true.
FAQ Section
Is Thermal Properties of Matter important for Class 11 boards?
Yes, it carries a weightage of about 5-7 marks. It's also foundational for the next chapter (Thermodynamics), making it a must-study.
Where can I download the Class 11 Physics NCERT PDF?
You can download the official, updated NCERT Physics textbook for free directly from the official NCERT website (ncert.nic.in) under the 'Publications' section.
What is the anomalous expansion of water?
Unlike most liquids that expand upon heating, water contracts when heated from 0C to 4C. Its maximum density is at 4C. This unique property helps aquatic life survive in frozen lakes!
What is the difference between conduction and convection?
Conduction transfers heat through stationary matter via molecular collisions (mainly in solids). Convection transfers heat through the actual movement of heated fluid particles (in liquids and gases).
Which topics from Chapter 10 are most important for JEE/NEET?
Calorimetry (mixing ice and water), Stefan-Boltzmann Law ($T^4$ relationships), and Newton's Law of Cooling are heavily tested in competitive exams
More Class 11 Physics Chapters
Ch 1: Units and MeasurementsCh 2: Motion in a Straight LineCh 3: Motion in a PlaneCh 4: Laws of MotionCh 5: Work, Energy and PowerCh 6: System of Particles and Rotational MotionCh 7: GravitationCh 8: Mechanical Properties of SolidsCh 9: Mechanical Properties of FluidsCh 10: Thermal Properties of MatterCh 11: ThermodynamicsCh 12: Kinetic TheoryCh 13: OscillationsCh 14: Waves