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NCERT Solutions for Class 11 Physics Chapter 10 β 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 Newtonβs law of cooling, which are important for school exams and competitive exams like JEE and NEET.
NCERT Solutions for Class 11 Physics Chapter 10 β 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 β 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.
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?
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 0Β°C and 100Β°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 Β on the Celsius scale by
Β Β Β Β Β tc = T β 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.
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 Γ 105 Pa
0.200 Γ 105 Pa
Normal melting point of sulphur
1.797 Γ 105 Pa
0.287 Γ 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.
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 Γ 10-5Β oC-1.
Thermal Properties of Matter
Medium
Q.
A large steel wheel is to be lifted on a shaft of the same material. At 27Β°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 βdry iceβ. 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:
Ξ±steel = 1.20Γ10-5 Β K-1.
Thermal Properties of Matter
Medium
Q.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0Β°C. What is the change in the diameter of the hole when the sheet is heated to 227Β°C? Coefficient of linear expansion of copper = 1.70 Γ 10β5 Kβ1.
Thermal Properties of Matter
Medium
Q.
A brass wire 1.8 m long at 27 Β°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of β39 Β°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 Γ 10β5 Kβ1; Youngβs modulus of brass = 0.91 Γ 1011 Pa.
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 250Β°C, if the original lengths are at 40.0Β°C? Is there a βthermal stressβ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 Γ 10β5 Kβ1, steel = 1.2 Γ 10-5 K-1).
Thermal Properties of Matter
Medium
Q.
The coefficient of volume expansion of glycerin is 49 Γ 10β5 Kβ1. What is the fractional change in its density for a 30Β°C rise in temperature?
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 gβ1Kβ1.
Thermal Properties of Matter
Medium
Q.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 Β°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 Jgβ1Kβ1; heat of fusion of water = 335 J gβ1).
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 150Β°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27Β°C. The final temperature is 40Β°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.
Given below are observations on molar specific heats at room temperature of some common gases.
Gas
Molar specific heat (Cv)
(cal molβ1kβ1)
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.
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) β70 Β°C under 1 atm, (b) β60 Β°C under 10 atm, (c) 15 Β°C under 56 atm?
Thermal Properties of Matter
Medium
Q.
Answer the following questions based on the PβT phase diagram of CO2:
(a) CO2 at 1 atm pressure and temperature β60Β°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 β 65Β°C as it is heated up to room temperature at constant pressure.
(d) CO2 is heated to a temperature 70Β°C and compressed isothermally. What changes in its properties do you expect to observe?
Thermal Properties of Matter
Medium
Q.
A child running a temperature of 101Β°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 98Β°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 gβ1.
Thermal Properties of Matter
Medium
Q.
A βthermacoleβ 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 45Β°C, and co-efficient of thermal conductivity of thermacole is 0.01 J sβ1mβ1Kβ1. [Heat of fusion of water = 335 Γ 103 J kgβ1]
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 sβ1mβ1Kβ1; Heat of vaporisation of water = 2256 Γ 103 Jkgβ1.
NCERT Solutions for Class 11 Physics Chapter 10 β Thermal Properties of Matter
Q.1) Answer the following questions based on the PβT phase diagram of COβ:
(a) At what temperature and pressure can the solid, liquid, and vapour phases of COβ co-exist in equilibrium?
Answer: In the phase diagram of COβ, point C represents the triple point. At this point, the solid, liquid, and vapour phases of COβ coexist in equilibrium.
For COβ, the triple point occurs at a temperature of β56.6Β°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 COβ?
Answer: With a decrease in pressure, both the fusion point (melting point) and the boiling point of COβ decrease.
(c) What are the critical temperature and critical pressure of COβ? What is their significance?
Answer: The critical temperature of COβ is 31.1Β°C and the critical pressure is 73 atm.
Significance: Above the critical temperature, COβ cannot be liquefied into the liquid state, no matter how much pressure is applied.
(d) Identify the state of COβ under the following conditions:
Temperature
Pressure
State (Phase)
β70Β°C
1 atm
Vapour
β60Β°C
10 atm
Solid
15Β°C
56 atm
Liquid
Q.2) Advanced Analysis of COβ Phase Changes:
(a) COβ at 1 atm pressure and β60Β°C is compressed isothermally. Does it pass through the liquid phase?
Answer: No. At β60Β°C and 1 atm, the state of COβ lies to the left of the triple point (β56.6Β°C, 5.11 atm). Therefore, on isothermal compression, COβ changes directly from vapour to solid without passing through the liquid phase.
(b) What happens when COβ 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), COβ cannot exist in the liquid phase. On cooling at constant pressure, COβ changes directly from vapour to solid, skipping the liquid phase.
(c) Describe the changes in solid COβ at 10 atm and β65Β°C as it is heated to room temperature.
Answer: At a pressure of 10 atm, when solid COβ at β65Β°C is heated, it first melts into the liquid phase. On further heating, the liquid COβ changes into the vapour phase.
(d) COβ is heated to 70Β°C and compressed isothermally. What changes occur?
Answer: COβ does not liquefy under these conditions. Since 70Β°C is higher than the critical temperature (31.1Β°C), COβ remains in the vapour (supercritical fluid) state. On increasing pressure, its behaviour deviates increasingly from ideal gas behaviour.
FAQs: Class 11 Physics Chapter 10 β 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 Newtonβs 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.