NCERT Solutions For Class 11 Physics Chapter 11
The unifying aspect of physical laws and the basic simplicity of nature form the underlying themes of Physics Chapter 11 Thermal Properties of Matter. In learning to apply these laws, students will come across some of the most important topics in Physics. More importantly, they will gain analytical abilities to help them apply these laws far beyond the scope of what can be included in a single book.
Chapter 11 physics class 11 helps you understand the basics and master the subject. This is why we bring you NCERT solutions class 11 Physics chapter 11 so that you can understand and learn the subject in a better way.
NCERT Solutions For Class 11 Physics Chapter 11 Thermal Properties Of Matter
Physics deals with the various properties of matter and nature. It forms the base for many scientific subjects that students take later in life. Chapter 11 deals with the concept of heat, its measurement process, and how it transfers from the body of one matter to another. During the study of this chapter, students will also come across real-life examples like why blacksmiths heat iron rings before fitting them on wooden wheels and the reason why beach wind changes direction after sunset, etc.
NCERT Solutions for Class 11 Physics will help students in understanding the concepts in an in-depth way.
NCERT SOLUTION FOR CLASS 11 PHYSICS CHAPTER 11 THERMAL PROPERTIES OF MATTER
The NCERT Solutions for Chapter 11 Class 11 Physics Thermal Properties of Matter are prepared by experts at Extramarks who have years of experience. By using the solutions, students can save a lot of time in finding the correct answer for the respective question.
In addition to NCERT Solutions, Extramarks also provides study materials, worksheets, question papers, and textbooks for the sake of students. Students are advised to take maximum help of the study materials to score good marks in examinations.
NCERT Solutions For Class 11 Thermal Properties Of Matter
Class 11 Physics Chapter 11 Thermal properties of matter talks about the effects of heat and temperature on various elements. In this chapter, students will get to know about the concept of heat and its measurement process, and how it transfers from the body of one matter to another.
NCERT Solutions Class 11 Physics Thermal Properties Of Matter – Topic Wise Discussion
The main topics covered in NCERT Solutions for Class 11 Physics Chapter 11 are:
- Temperature and heat
- Measurement of temperature
- Ideal-gas equation and absolute temperature
- Thermal expansion
- Specific heat capacity
- Change of state
- Heat transfer
- Newton’s law of cooling
- Introduction: – The chapter begins with a summary where students get an overview of the concept and other topics they will learn.
- Temperature And Heat: – NCERT Solutions Class 11 Physics Chapter 11 talks about the definition of heat and temperature. The heat of an object is the total energy of all the molecular motion inside that object. Temperature is the measure of the thermal energy or average heat of the molecules in a substance.
- Measurement Of Temperature: – Chapter 11 NCERT solutions explains the process of measuring temperature and the instruments used for that. This section also discusses how temperature is measured with the help of a thermometer. Mercury and Alcohol are commonly used liquids in liquid-in-glass thermometers. To construct a thermometer, two fixed points are to be chosen as a reference point, discussing in detail the thermometer, and its various scales.
- Ideal-Gas Equation And Absolute Temperature: – In this section of Chapter 11, a thermometer with a liquid-filled bulb at one end is taken, the most commonly used liquid is Mercury, Toluene, Alcohol, Pentane, .Creosote shows different readings for temperatures other than the fixed reading because of their different expansion properties and talks about various properties of a gas, and how thermometers that use gas can perform better than the liquid-filled ones.
It also explains the variables that describe the behaviour of a gas. The ideal gas equation, along with theories of different scientists is also a part of the discussion in this chapter.
- Thermal Expansion: – Thermal expansion is one of the most important sections of chapter 11 of Physics Class 11. Thermal expansion refers to the expansion or contraction of the dimensions of the solid, liquid or gas when their temperature is changed. There are three types of thermal expansion depending on the dimension that undergoes change and that are linear expansion, areal expansion and volumetric volume, it describes how various elements expand under the influence of heat. A real-life example mentioned here is that hot water is poured over metal lids in case it gets tightened. The reason here is that this hot water expands the metal and makes it easy to open.
- Specific Heat Capacity: – Every item has a particular boiling point, and they react differently when they absorb thermal energy. For instance, a bowl of water under heat starts to pop bubbles, until it becomes unstable when the water starts boiling. This concept of heat capacity is discussed in-depth, and students will also learn about its various aspects.
- Calorimetry: – Calorimetry is the process of the measurement of heat. This segment of Physics Class 11 Chapter 11 Thermal Properties Of Matter NCERT solutions discusses the concept of calorimetry in detail. With the help of real examples used here, students can easily and quickly grasp this topic.
- Change Of State: – Mainly there are three states – solid, liquid, gas. The transition of an element from one state to another is called change of state, and heat plays a pivotal role here. Students will also learn about the concept of triple point, latent heat, and conduction.
- Heat Transfer: – The transfer of heat is a common occurrence, and this ninth topic of Chapter 11 explains this regular event further. The concepts of radiation, convection, and others are a part of the discussion here.
- Newton’s Law Of Cooling: – Newton’s Law of Cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. This chapter on thermal properties of matter, Class 11 explains the concept in detail.
NCERT Ch 11 Physics Class 11 Distribution of Marks
Physics carries 100 marks in examinations. Out of these 100 marks, 30 marks are practicals and the remaining 70 marks will be for the theoretical examination. We recommend students solve all theory questions and numericals before appearing for the exams. This will prove fruitful in scoring high marks in physics and eventually improving their overall grades.
Why NCERT Solutions of Thermal Properties of Matter is a Must-Read?
Here are a few reasons why students should refer to NCERT Solutions:
- The language used is simple and easy to understand for students. This will help them in understanding the concepts in a better way and preparing for the exams.
- A detailed explanation will help students grasp concepts quickly and use them in their later stages of further studies as well.
- Real-life examples are used in the solutions for a better understanding of students and to help them clear all the concepts with a realistic perspective.
- The NCERT solutions Class 11 Physics follows the guidelines of CBSE. Hence, it will definitely not misguide the students in terms of solutions, steps, format, etc.
Q.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.
Q.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 TA and TB?
Q.3 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?
Q.4 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?
Q.5 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?
Q.6 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.
Q.7 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.
Q.8 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.
Q.9 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.
Q.10 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).
Q.11 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?
Q.12 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.
Q.13 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).
Q.14 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?
Here, mass of metal, m = 0.20 kg = 200 g
Initial temperature of metal, T1 = 150°C
Final temperature of metal, T2 = 40°C
Decrease in temperature of the metal: ΔT = T1 – T2 = 150°C – 40°C = 110°C
Water equivalent of calorimeter, w = 0.025 kg = 25 g
Volume of the water, V = 150 cm3
Mass (M) of the water at 27°C: 150 g × 1 = 150 g
Specific heat of the water, Cw = 4.186 Jg-1K
Let specific heat of metal = C
Q.15 Given below are observations on molar specific heats at room temperature of some common gases.
|Gas||Molar specific heat (Cv)
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?
The gases tabulated in the given table are diatomic in nature. In addition to the translational degree of freedom, they have other degrees of freedom (modes of motion).
Heat must be supplied to increase the temperature of the given gases. The average energy of all the modes of motion increases with it. Thus, the molar specific heat of diatomic gas is greater than that of monatomic gas. If we consider only the rotational mode of motion, then
This is because at room temperature, chlorine also has vibrational modes of motion in addition to rotational and translational modes of motion.
Q.16 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?
(a) The P-T phase diagram for CO2 is shown in the given figure. In the given phase diagram, C is the triple point of the CO2. It implies that the solid, liquid, and vapour phases of CO2 co-exist in equilibrium at the temperature and pressure corresponding to this point (i.e., at temperature = –56.6 °C and pressure = 5.11 atm).
(b) Both the fusion and boiling points of CO2 fall with the decrease in pressure.
(c) In case of CO2, the critical temperature and critical pressure are 31.1°C and 73 atm respectively. If the temperature of CO2 is greater than 31.1°C, it cannot be liquefied, howsoever large pressure we may apply.
(d) From the P-T phase diagram of CO2, it can be concluded that,
(a) At –70°C, under 1 atm pressure CO2 is a vapour.
(b) At –60°C, under 10 atm pressure CO2 is a solid.
(c) At 15°C, under 56 atm pressure CO2 is a liquid.
Q.17 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?
(a) The P-T phase diagram for CO2 is shown in the given figure. In the given phase diagram, at 1 atm pressure and at –60°C temperature, CO2 lies to the left of -56.6°C (triple point C), therefore, it lies in the region of vapour and solid phase. Therefore, CO2 condenses directly into the solid state, without becoming liquid.
(b) Since at the 4 atm pressure CO2 lies below 5.11 atm (triple point C), therefore, CO2 lies in the region of vapour and solid phase. Hence, CO2 condenses into the solid state directly, without becoming liquid.
(c) When solid CO2 (at 10 atm pressure and at–65°C) is heated, it changes to the liquid phase and then to the vapour phase. At pressure=10 atm if a horizontal line is drawn parallel to the temperature axis, then the fusion and boiling points of CO2 are given by the intersection point where this parallel line crosses the fusion and vaporisation curves.
(d) If CO2 is heated to 70°C and compressed isothermally, then it will not be converted to the liquid state. This is because; the critical temperature of CO2 is lower than 70°C. It will exist in the vapour state, but will depart more and more from its ideal behaviour with the increase in pressure.
Q.18 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.
Q.19 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]
Q.20 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.
Q.21 Explain why:
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water
(a) This is because a body with a large reflectivity is a poor absorber of light radiations and poor absorbers are poor emitters. Thus, a body with a large reflectivity is a poor emitter.
(b) Since brass is a good conductor of heat , when we touch a brass tumbler, heat is conducted from our body to the brass tumbler. Therefore, the temperature of our body decreases and we feel cooler.
On the contrary, wood is a poor conductor of heat.Thus, there is only a negligible fall in the temperature of our body and we feel comparatively less cool.
Hence, a brass tumbler feels much colder than a wooden tray on a chilly day.
(c) An optical pyrometer calibrated for an ideal black body radiation gives a very low value for temperature of a red hot iron piece placed in the open.
Black body radiation equation is given as:
Here, E represents the energy radiation
T and To represent temperature of the optical pyrometer and Temperature of the open space respectively.
σ is a constant
Thus, a rise in the temperature of open space lowers the radiation energy.
When the same piece of iron is kept in a furnace, its radiation energy is given as: E = σT4
(d) The atmospheric gases reflect infrared radiations from Earth back to its surface. Hence, the heat radiations collected by Earth from the Sun during day are kept trapped by the atmosphere.. All the heat would be radiated back from earth’s surface without atmospheric gases and the Earth would become too cold to live.
(e) The heating system based on the circulation of steam is more useful in warming a building than that based on the circulation of hot water. This is due to the fact that steam contains excessive heat in the form of latent heat which is 540 cal/g.
Q.22 A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.
FAQs (Frequently Asked Questions)
1. Why should I select NCERT solutions?
The sole reason for choosing NCERT solutions for referring is that it is well-structured, and made as per CBSE guidelines. The solutions are prepared in simple language making it easier for students to understand the reasoning behind every answer. Charts, graphs, and bullet points are further used for a more detailed and easy explanation for students.
2. What is the number of topics in NCERT Class 11 Physics Chapter 11?
In total, there are 10 topics in this particular chapter with all of them covering various properties of heat and temperature. With every passing chapter, the students will dive deeper into detailed explanations that will prove fruitful in further studies.
Apart from these, our focus is to provide all the study materials to students in the simplest of forms hence, we also have NCERT solutions, NCERT solutions class 1, NCERT solutions class 2, NCERT solutions class 3, NCERT solutions class 4, NCERT solutions class 5, NCERT solutions class 6, NCERT solutions class 7, NCERT solutions class 8, NCERT solutions class 9, NCERT solutions class 9, NCERT solutions class 10, NCERT solutions class 11 and NCERT solutions class 12.