NCERT Solutions for Class 11 Physics Chapter 10 – Thermal Properties of Matter

Thermal Properties of Matter is a conceptual and scoring chapter in Class 11 Physics that explains how materials respond to changes in temperature and heat. This chapter covers key topics such as temperature scales, thermal expansion (linear, area, volume), heat transfer (conduction, convection, radiation), specific heat capacity, calorimetry, change of state, and Newton’s law of cooling, which are important for school exams and competitive exams like JEE and NEET.

NCERT Solutions for Class 11 Physics Chapter 10 – Thermal Properties of Matter are prepared strictly according to the latest CBSE syllabus and exam pattern. The solutions are written in simple, step-by-step language with clear formulas, graphs, and solved numericals, helping students build strong conceptual clarity and score well in Class 11 examinations.

NCERT Solutions for Class 11 Physics Chapter 10 – Thermal Properties of Matter

Thermal Properties of Matter
Medium
Q.
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.
Thermal Properties of Matter
Medium
Q.
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g–1K–1.
Thermal Properties of Matter
Medium
Q.
A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1m–1K–1. [Heat of fusion of water = 335 × 103 J kg–1]
Thermal Properties of Matter
Medium
Q.
A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1.
Thermal Properties of Matter
Medium
Q.
Answer the following questions based on the P–T phase diagram of CO2:
(a) CO2 at 1 atm pressure and temperature –60°C is compressed isothermally. Does it go through a liquid phase?
(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature – 65°C as it is heated up to room temperature at constant pressure.
(d) CO2 is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?
Thermal Properties of Matter
Medium
Q.
Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO2?
(c) What are the critical temperature and pressure for CO2? What is their significance?
(d) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm?
Thermal Properties of Matter
Medium
Q.
Given below are observations on molar specific heats at room temperature of some common gases.
Gas
Molar specific heat (Cv)
(cal mol–1k–1)
Hydrogen
4.87
Nitrogen
4.97
Oxygen
5.02
Nitric oxide
4.99
Carbon monoxide
5.01
Chlorine
6.17
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
Thermal Properties of Matter
Medium
Q.
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Thermal Properties of Matter
Medium
Q.
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 Jg–1K–1; heat of fusion of water = 335 J g–1).
Thermal Properties of Matter
Medium
Q.
The coefficient of volume expansion of glycerin is 49 × 10–5 K–1. What is the fractional change in its density for a 30°C rise in temperature?
Thermal Properties of Matter
Medium
Q.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?
Thermal Properties of Matter
Medium
Q.
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10-5 K-1).
Thermal Properties of Matter
Medium
Q.
A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.
Thermal Properties of Matter
Medium
Q.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1.
Thermal Properties of Matter
Medium
Q.
A large steel wheel is to be lifted on 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 over the required temperature range:
αsteel = 1.20×10-5  K-1.
Thermal Properties of Matter
Medium
Q.
A steel tape 1 m long is correctly calibrated for a temperature of 27.0 oC. 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 45.0oC. What is the actual length of the steel rod on that day? What is the length of the same steel rod on the day when the temperature is 27.0 oC? Coefficient of linear expansion of steel = 1.20 × 10-5  oC-1.
Thermal Properties of Matter
Medium
Q.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:
Temperature
Pressure thermometer A
Pressure thermometer B
Triple-point of water
1.250 × 105 Pa
0.200 × 105 Pa
Normal melting point of sulphur
1.797 × 105 Pa
0.287 × 105 Pa
(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Thermal Properties of Matter
Medium
Q.
Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?
(c) The absolute temperature (Kelvin scale) T is related to the temperature tc  on the Celsius scale by
      tc = T – 273.15
Why do we have 273.15 in this relation, and not 273.16?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
Thermal Properties of Matter
Medium
Q.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = Ro [1 + α (T – To)]. The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?
Thermal Properties of Matter
Medium
Q.
A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg min-1 when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s–1m–1K–1; Heat of vaporisation of water = 2256 × 103 Jkg–1.

NCERT Solutions for Class 11 Physics Chapter 10 – Thermal Properties of Matter

Q.1) Answer the following questions based on the P–T phase diagram of CO₂:

(a) At what temperature and pressure can the solid, liquid, and vapour phases of CO₂ co-exist in equilibrium?

Answer:
In the phase diagram of CO₂, point C represents the triple point. At this point, the solid, liquid, and vapour phases of CO₂ coexist in equilibrium.

For CO₂, the triple point occurs at a temperature of –56.6°C and a pressure of 5.11 atm.

(b) What is the effect of decrease of pressure on the fusion point and boiling point of CO₂?

Answer:
With a decrease in pressure, both the fusion point (melting point) and the boiling point of CO₂ decrease.

(c) What are the critical temperature and critical pressure of CO₂? What is their significance?

Answer:
The critical temperature of CO₂ is 31.1°C and the critical pressure is 73 atm.

Significance:
Above the critical temperature, CO₂ cannot be liquefied into the liquid state, no matter how much pressure is applied.

(d) Identify the state of CO₂ under the following conditions:

Temperature Pressure State (Phase)
–70°C 1 atm Vapour
–60°C 10 atm Solid
15°C 56 atm Liquid

Q.2) Advanced Analysis of CO₂ Phase Changes:

(a) CO₂ at 1 atm pressure and –60°C is compressed isothermally. Does it pass through the liquid phase?

Answer:
No.
At –60°C and 1 atm, the state of CO₂ lies to the left of the triple point (–56.6°C, 5.11 atm). Therefore, on isothermal compression, CO₂ changes directly from vapour to solid without passing through the liquid phase.

(b) What happens when CO₂ at 4 atm pressure is cooled from room temperature at constant pressure?

Answer:
Since 4 atm is less than the triple point pressure (5.11 atm), CO₂ cannot exist in the liquid phase. On cooling at constant pressure, CO₂ changes directly from vapour to solid, skipping the liquid phase.

(c) Describe the changes in solid CO₂ at 10 atm and –65°C as it is heated to room temperature.

Answer:
At a pressure of 10 atm, when solid CO₂ at –65°C is heated, it first melts into the liquid phase. On further heating, the liquid CO₂ changes into the vapour phase.

(d) CO₂ is heated to 70°C and compressed isothermally. What changes occur?

Answer:
CO₂ does not liquefy under these conditions. Since 70°C is higher than the critical temperature (31.1°C), CO₂ remains in the vapour (supercritical fluid) state. On increasing pressure, its behaviour deviates increasingly from ideal gas behaviour.

FAQs: Class 11 Physics Chapter 10 – Thermal Properties of Matter

Q1. Is this chapter important for exams?
Yes, it is a high-weightage chapter in thermodynamics basics.

Q2. Which topics are most important here?
Thermal expansion, calorimetry, heat transfer, and Newton’s law of cooling.

Q3. Are numericals asked from this chapter?
Yes, calorimetry and expansion-based numericals are common.

Q4. Are graphs important in this chapter?
Yes, cooling curves and temperature-time graphs are frequently asked.

Q5. How do NCERT Solutions help?
They provide NCERT-aligned, exam-ready explanations with solved numericals.