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. 


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:

  1. Introduction
  2. Temperature and heat
  3. Measurement of temperature
  4. Ideal-gas equation and absolute temperature
  5. Thermal expansion
  6. Specific heat capacity
  7. Calorimetry
  8. Change of state
  9. Heat transfer
  10. Newton’s law of cooling
  11. Introduction: – The chapter begins with a summary where students get an overview of the concept and other topics they will learn.
  12. 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. 
  13. 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.
  14. 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.

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
  6. 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:

  1. 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.
  2. A detailed explanation will help students grasp concepts quickly and use them in their later stages of further studies as well. 
  3. 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. 
  4. 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.


RelationbetweenKelvin and Celsius scales isgiven as: T C = T K – 273.15 ( i ) RelationbetweenCelsius and Fahrenheit scales isgiven as: T F = 9 5 T C + 32 ( ii ) Incaseof neon, T K =24.57 K Fromequation(i),weobtain: T C =24.57273.15 =–248.58°C Fromequation(ii),weobtain: T F = 9 5 T C +32 T F = 9 5 ( –248.58 )+32 T F = 415 .44 o F Incaseof carbon dioxide: T K = 216.55 K Fromequation(i),weobtain: T C = 216.55273.15 = –56.60°C Fromequation(ii),weobtain: T F = 9 5 T C + 32 T F = 9 5 ( –56.60 )+32 = -69 .88 o F

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?


Here, triple point of water on scale A, T A =200 A Triple point of water on scale B,T B = 350 B Triple point of water on the Kelvin scale, T K = 273.15 K Accordingtothequestion: Temperature 273.15 K on Kelvin scale=200 A onabsolute scale A T 1 = T K 200A = 273.15 K A = 273.15 K 200 Accordingtothequestion, Temperature 273.15 K on Kelvin scale = 350 B onabsolute scale B T B = T K 350 B=273.15 B = 273.15 350 Relationbetween T A and T B isgivenas: 273.15 200 × T A = 273.15 350 × T B T A = 200 350 T B T A T B = 200 350 The ratio T A : T B = 4 : 7

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?


Given that, R = R 0 [1 + α(T – T 0 )] (i) Here, R 0 and T 0 representinitial resistanceandinitialtemperature respectively. R andTrepresentfinal resistanceandfinaltemperature respectively. α = a constant At triple point of water, T 0 = 273.15 K Resistance of thelead, R 0 = 101.6 Ω At thenormal melting point of lead,T = 600.5 K Resistance of thelead, R = 165.5 Ω Putting these values in equation (i), we obtain: R=R 0 [1 + α(T–T 0 )] 165.5 Ω = 101.6 Ω[1 + α(600.5 – 273.15)K] 1.629 Ω = 1 Ω + α(327.35)K α = 0.629 327.35K = 1 .92 × 10 -3 K -1 Incaseof resistance, R 1 =123.4 Ω R 1 = R 0 [1 + α(T–T 0 )] Here,Trepresentsthetemperaturewhenresistanceoflead=123.4 Ω 123.4 Ω = 101.6 [1 + 1 .92 ×10 -3 K -1 (T – 273.15)K] 1.214 Ω = 1Ω + 1 .92 ×10 -3 K -1 (T – 273.15)K 0.214 Ω 1 .92×10 -3 K -1 =T – 273.15 T = 384.61 K

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?


( a )The triple point of water has a unique value. At the certain values of volume and pressure, the triple point of water is always is equal to 273.16 K. The melting point of ice and the boiling point of water do not have specific values because they depend on the pressure and temperature. ( b )The absolute zero or 0 K is the other stable point on the Kelvin scale. ( c )On Kelvin scale, the temperature 273.16 K is the triple point of water. However, 273.16 K is not the melting point of ice. On Celsius scale, the temperature 0°C is the melting point of ice and its equivalent value on Kelvin scale is 273.15 K. Therefore, on Celsius scale, the absolute temperature ( Kelvin scale ) T, is related to temperature t c , as: t c = T – 273.15 d)Consider that T K is the temperature on absolute scale and T F is the temperature on Fahrenheit scale . The temperatures T F and T K can be related as: T F – 32 180 = T K – 273.15 100 (i) Consider T F1 be the temperature on Fahrenheit scale and T K1 be the temperature on absolute scale. The temperatures T F1 and T K1 canbe related as: T F1 – 32 180 = T K1 – 273.15 100 (ii) Given that: T K1 –T K = 1 K Subtractingequation(i)fromequation(ii),weobtain: T F1 – T F 180 = T K1 – T K 100 = 1 100 T F1 – T F = 1 × 180 100 = 9 5 Triplepointofwater = 273.16K Triplepointofwateronthenewscale= 273.16 × 9 5 = 491.69 K

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?


Here,triplepointofwater,T = 273.16K Atthetriplepointofwater,thepressureinthermometerA, P A = 1 .250 ×10 5 Pa Letthenormalmeltingpointofsulphur= T 1 Atthenormalmeltingpointofsulphur,thepressureinthermometerA, P 1 = 1 .797 × 10 5 Pa AsperCharles’law,wehave: P A T = P 1 T 1 T 1 = P 1 T P A T 1 = 1 .797×10 5 Pa × 273.16 K 1 .250 ×10 5 Pa = 392.69K Absolutetemperatureofnormalmeltingpointofsulphuras readbythermometerA=392.69K. Attriplepointofwater,pressureinthermometerB, P B =0.200× 10 5 Pa Attemperature T 1 ,pressureinthermometerB, P 2 =0.287× 10 5 Pa AsperCharles’law,wehave, P B T = P 2 T 1 T 1 = P 2 T P B T 1 = 0 .287×10 5 Pa × 273.16 K 0 .200×10 5 Pa T 1 = 391.98K Absolutetemperatureofnormalmeltingpointofsulphuras readbythermometerB=391.98K. (b)TheslightdifferenceinthereadingsofthermometersAandBisduetothefactthatoxygenandhydrogengasespresentinthermometers AandBrespectivelyarenotperfectlyideal.Toreducethe discrepancy betweenthetworeadings,thereadingsshouldbetakenatlowerpressure conditionsas inthatcase,thegasesapproachtotheidealgasbehaviour.

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.


Here,lengthofsteeltapeattemperatureT =27°C,l = 1m = 100cm Attemperature T 1 = 45 o C,lengthofthesteelrod, l 1 = 63cm Coefficientoflinearexpansionofthesteel,α = 1 .20×10 -5 K -1 Lettheactuallengthofthesteel rod = l 2 Letthelengthofthesteelrodat 45 o C = l’ l’ = l + αl( T 1 – T ) l’ = 100 cm + 1 .20×10 -5 K -1 × 100 cm ( 45 – 27 )°C l’ = 100.0216cm Actuallengthofthesteelrodmeasuredbythesteeltape at45°C, l 2 = 100.0216 cm 100 cm × 63 cm = 63.0136cm Actuallengthofthesteelrodat45°Cis63.0136cm anditslengthat27°Cis63.0cm.

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.


Here, temperature, T=27°C = 27°C + 273 =300 K Outer diameter of the shaft at T, d 1 = 8.70 cm Diameter of the central hole in the wheel at T, d 2 =8.69cm Coefficient of linear expansion of steel, α s =1.20× 10 –5 K –1 After cooling,lettemperatureoftheshaft= T 1 For thewheel to slip on the shaft,the change in diameter oftheshaft, Δd = 8.69 cm – 8.70 cm = –0.01 cm Temperature T 1 can be obtained from the relation: Δd = d 1 α s ( T 1 –T ) 0.01 cm = 8.70 cm × 1 .20 × 10 –5 K 1 ( T 1 –300 K ) T 1 –300K = 95.78 K T 1 = 204.21 K T 1 = 204.21 K – 273.16 K = – 68.95 K The wheel will slip on the shaft when the temperature of the shaft isapproximately–69 K.

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.


Here, initial temperature, T i = 27.0°C Diameter of hole at T i , d 1 = 4.24 cm Final temperature, T f = 227°C Let diameter of hole at T f = d 2 Co-efficient of linear expansion of copper, α Cu = 1 .70 × 10 –5 K –1 If theco-efficient of superficial expansionis βand changeintemperature isΔT,we have, Changeinarea( ΔA ) Originalarea( A ) =βΔT ( π d 2 2 4 π d 1 2 4 ) π( π d 1 2 4 ) = ΔA A ΔA A = d 2 2 – d 1 2 d 1 2 As,β = 2α d 2 2 – d 1 2 d 1 2 = 2αΔT d 2 2 d 1 2 -1 = 2α( T f – T i ) d 2 2 ( 4.24 m ) 2 = 2×1 .7×10 -5 K –1 ( 227 – 27 )°C +1 d 2 2 = 17.98 × 1 .0068 cm 2 = 18 .1 cm 2 d 2 = 4.2544cm Change in diameter = d 2 – d 1 = 4.2544 cm – 4.24 cm = 0.0144 cm Thediameter oftheholeincreases by 1 .44×10 –2 cm.

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.


Here, initial temperature,T i = 27°C Length of brass wire at T i ,l = 1.8 m Final temperature, T f = –39°C Diameter of wire,d = 2.0 mm = 2× 10 –3 m Coefficient of linear expansion of brass,α=2.0× 10 –5 K –1 Young’s modulus for brass,Y=0.91× 10 11 Pa Letcross-sectionalareaofbrasswire = A Lettension produced in the brasswire = F Young’s modulus is given as,Y= Stress Strain Y= F/A ΔL/L ΔL = F × L A × Y (i) Here, ΔL=Change inthe lengthofwire, Change inthe lengthofwireisgiven bytherelation, ΔL = αL( T 2 –T 1 ) ( ii ) Equating equations ( i ) and ( ii ), we obtain: αL( T 2 –T 1 ) = FL π ( d 2 ) 2 ×Y F = α( T 2 –T 1 )πY ( d 2 ) 2 F = 2 .0 × 10 –5 K –1 ×( -39 – 27 )°C × 3.14 × 0 .91×10 11 Pa × ( 2 .0 × 10 –3 m 2 ) 2 F = -3 .8 ×10 2 N (The negative sign shows that the tension is directed inward.) The tension produced in the wire is 3 .8 ×10 2 N.

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).


Here,initial temperature, T i =40°C Final temperature,T f =250°C Change in temperature, ΔT= T f T i =210°C Length of brass rod at T i , l i = 50 cm Diameter of brass rod at T i , d 1 = 3.0 mm Length of steel rod at T f , l 2 =50 cm Diameter of steel rod at T f , d 2 = 3.0 mm Coefficient of linear expansion of thebrass, α b = 2.0× 10 –5 K –1 Coefficient of linear expansion of thesteel, α s = 1 .2 × 10 –5 K –1 For expansion in the brass rod, we have: Changeinlength( Δl 1 ) Originallength( l 1 ) = α b ΔT Δl 1 =50 cm×( 2.1× 10 –5 K –1 )×210°C Δl 1 =0.2205cm For expansion in the steel rod, we have, Changeinlength( Δl 2 ) Originallength( l 2 ) = α s ΔT Δl 2 =50 cm×( 1 .2 × 10 –5 K –1 )×210°C Δl 2 = 0.126cm Total change in the lengths of brass and steelisgivenas: Δl = Δl 1 + Δl 2 Δl = 0.2205 + 0.126 = 0.346 cm Total change in the length of the joined rod=0.346 cm As the rod expands freely from both ends, no thermalstress is produced at the junction.

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?


Here,coefficient of volume expansion of glycerin, α V = 49 × 10 –5 K –1 Increase in thetemperature, ΔT = 30°C Fractional change in volume ofglycerin= ΔV V Thechange involumeis related to the change intemperature as: ΔV V = α V ΔT V T 2 – V T 1 = V T 1 α V ΔT m ρ T 2 m ρ T 1 = m ρ T 1 α V ΔT Here, m = Mass of glycerin ρ T 1 = Initial density at T 1 ρ T 2 = Final density at T 2 ρ T 1 – ρ T 2 ρ T 2 = α V ΔT Here, ρ T 1 – ρ T 2 ρ T 2 =Fractional change in thedensity Fractional change in density of glycerin = 49×10 –5 K –1 × 30°C =1 .47×10 –2

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.


Here, power of drilling machine, P = 10 kW = 10 × 10 3 W Mass of aluminum block, m = 8 .0 kg = 8 × 10 3 g Time for which the machine is used, t = 2.5 min t = 2.5 × 60s =150 s Specific heat of aluminium, c = 0 .91 J g –1 K –1 Letincrease in the temperature = ΔT Total energy of thedrilling machine = Pt Pt =10×10 3 W × 150 s = 1 .5 × 10 6 J As only 50% of energy is useful, Usefulenergy,ΔQ = 1 2 × 1 .5 × 10 6 J ΔQ = 7 .5 × 10 5 J AsΔQ = mcΔT ΔT = ΔQ mc ΔT = 7 .5 × 10 5 J 8×10 3 g × 0 .91 J g –1 K –1 ΔT = 103°C Increase in the temperature of the block = 103°C

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).


Here,mass of copper block, m = 2.5 kg = 2500 g Increase in temperature of the copper block, Δθ = 500°C Specific heat of copper, C = 0 .39 J g –1 °C –1 Heat of fusion of water, L = 335 J g –1 Heat lostbythe copper blockisgivenas: Q = mCΔθ Q = 2500 g × 0 .39 J g –1 °C –1 × 500°C = 487500 J Letmassoficemelted=m’ Heat attained by the melted ice, Q = m’L As,Heat attained by ice=Heat lost by copper m’L = 487500 J m’ = 487500 L = 487500 J 335 J g –1 = 1455.22g m’ =1.45kg Maximum amount of ice that can melt = 1.45 kg

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

Heat lost by the metal = Heat attained by the waterand colorimeter system mCΔT = (M+w)C w ΔT’ 200kg × C × 110°C = (150 + 25)kg × 4 .186 Jg -1 K × 13°C C = 175 kg × 4 .186 Jg -1 K × 13°C 110 °C × 200 kg C = 0.43 Jg -1 K -1 If some heat is lostto the surroundings, then the value of C will be less than the actual value of C.

Q.15 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?


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

Molar specific heat of a diatomic gas,C v = 5 2 R = 5 2 × 1.98 C v = 4.95 cal mol -1 K -1 Exceptchlorine, all the observations in the given table agreefairlywellwith( 5 2 R ).

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.


Here,Initial temperatureofthe body, T 1 =101°F Final temperature of the body,T 2 =98°F Change in temperatureofbody, ΔT = [ ( 101°F – 98 °F ) × 5 9 ]°C Time taken to decreasethe temperature,t=20 min Mass of child, m=30 kg = 30× 10 3 g Specific heat of human body=Specific heat ofwater=c c = 1000 cal kg -1 °C -1 Latent heat of evaporation of the water,L = 580 calg –1 Heat lost by the child is given as,Δθ=mcΔT Δθ =30 kg × 1000 cal kg -1 °C -1 × ( 101 – 98 ) °F × 5 9 Δθ= 50000cal Consider mass of the water evaporated fromthe body ofthechild in 20 min= m 1 Heat lossthrough water is given as,Δθ= m 1 L m 1 = Δθ L m 1 = 50000cal 580 calg –1 m 1 = 86.2g Average rate of extra evaporation dueto thedrug = m 1 t m 1 t = 86.2 g 20 min = 4.31 g min -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]


Here,side of the given cubical ice box, s = 30 cm = 0.3 m Thickness of ice box, l = 5.0 cm = 0.05 m Mass of theice placed in the ice box, m=4 kg Time difference, t=6 h = 6 × 60 × 60 s Outside temperature,T = 45°C Coefficient of thermal conductivity of thethermacole, K = 0 .01 J s –1 m –1 K –1 Latentheatoffusion,L=335× 10 3 Jkg –1 Let total amount of ice that melts in 6 h=m’ Heat lost by the foodisgivenas:θ = KA( T-0 )t l Here, A = Surface area of the box = 6s 2 A = 6× ( 0.3 m ) 2 = 0 .54 m 3 θ = 0.01 × 0 .54 m 3 × ( 45°C0°C ) × 6 × 60 × 60 s 0.05 m θ = 104976J Asθ = m’L m’ = θ L = 104976 J 335 × 10 3 Jkg –1 m’ = 0.313kg Mass of theice remained=(4 – 0.313) kg = 3.687 kg Theamount of ice left after 6 h is 3.687 kg.

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.


Here,base area of the brassboiler, A=0 .15 m 2 Thickness of thebrassboiler, l = 1.0 cm = 0.01 m Rate ofboiling ofwater, R = 6 .0 kg min -1 Mass, m = 6 kg Time, t = 1 min = 60 s Thermal conductivity of the brass, K = 109 J s –1 m –1 K –1 Heat of vaporisationofwater, L = 2256× 10 3 J kg –1 Heat flowing into water through the brass base of the boiler,θ = KA( T 1 -T 2 )t l (i) Here, T 1 and T 2 represent temperature of the flame in contact with boilerand boiling point of water( 100°C ) respectively. Heat required for boiling the waterisgivenas: θ = mL ( ii ) Equating equations ( i ) and ( ii ), we obtain: mL = KA( T 1 – T 2 )t l T 1 – T 2 = mLl KAt T 1 – T 2 = 6 kg × 2256 × 10 3 J kg –1 × 0.01 109 Js –1 m –1 K –1 × 0 .15 m 2 × 60 s T 1 – T 2 = 137.98°C Thetemperature of the part of the flame incontact withboiler = 237.98°C

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:

E=σ( T 4 T 0 4 )

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.


Here, temperature of the surroundings= T 0 =20°C Let temperature of the body=T According to Newton’s law of cooling, we have dT dt = -K( T – T 0 ) dT T – T 0 = -Kdt Here,Kisaconstant. Lettemperatureofthebodyfallsfrom T i to T f intimet, thenintegratingtheaboverelation,wehave T i T f dT T-T 0 = – 0 t Kdt [ log e ( T-T 0 ) ] T i T f = -Kt log e T f – T 0 T i – T 0 = -Kt 2 .3026 log 10 T f – T 0 T i – T 0 = -Kt 2 .3026 log 10 T i – T 0 T f – T 0 = Kt t = 2.3026 K log 10 T 1 – T 0 T 2 – T 0 (A) Accordingtothefirstconditionoftheproblem,wehave T i = 80°C, T f = 50°C, T 0 = 20°C,t = 5min = 5 × 60s Substitutingthesevaluesinequation(A),weobtain: 5 × 60s = 2.3026 K log 10 80°C – 20°C 50°C – 20°C 5 × 60s = 2.3026 K log 10 2 (i) Accordingtothesecondconditionoftheproblem, T i = 60°C, T f = 30°C, T 0 = 20°C,t = ? Substitutingthesevaluesinequation(A),weobtain: t = 2.3026 K log 10 60°C – 20°C 30°C – 20°C t = 2.3026 K log 10 4 (ii) Dividingequation(ii)byequation(i),weobtain: t 5 × 60 s = log 10 4 log 10 2 t 5 × 60 s = 0.6021 0.3010 = 2 t = 600s = 10min Timetakentocoolthebodyfrom60°Cto30°C = 10minutes

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