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

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

Vishal kumarUpdated on 23 Aug 2025, 01:12 AM IST

Have you ever wondered why a metal chair, when cold, is much colder than a wooden one because even though they may be at the same temperature? And why are iron rims heated when placed on wooden cart wheels by blacksmiths? These common experiences are discussed in Class 11 Physics Chapter- Thermal Properties of Matter NCERT Solutions, where one can get to know about heat, temperature, thermal expansion, specific heat capacity and methods of heat transfer.

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  1. NCERT Solutions for Class 11 Physics Chapter 10: Download Solution PDF
  2. NCERT Solutions for Class 11 Physics Thermal Properties of Matter - Exercise Solution
  3. Class 11 Physics NCERT Chapter 10: Higher Order Thinking Skills (HOTS) Questions
  4. NCERT Chapter 10: Thermal Properties of Matter Topics
  5. NCERT Solutions for Class 11 Chapter 10: Important Formulae
  6. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  7. NCERT Solutions for Class 11 Physics Chapter-Wise
  8. NCERT solutions for class 11 subject-wise
NCERT Solutions for Class 11 Physics Chapter 10 Thermal Properties of Matter
NCERT SOlution for class 11 chapter 10 thermal properties of matter

These NCERT solutions are well planned by subject experts to act as step-by-step explanations to all the exercise questions, hence making even the difficult topics simple to understand. Students, through regular practice, not only develop a good conceptual base but also develop their problem-solving ability, which becomes essential during examinations. Be it the board exam, the CBSE, JEE, NEET, or the Olympiads, these NCERT solutions for Class 11 Physics can turn out to be a superb revision tool and help in relating physics in real life, such as boiling water, sea breeze, or other regular thermal effects.

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NCERT Solutions for Class 11 Physics Chapter 10: Download Solution PDF

The observations like expansion of gases on heating or the changing pressure of the gases in a container are simply and clearly explained by the NCERT Solutions of this Class 11 Physics Chapter 10 Thermal Properties of Gases. These solutions assist students in studying the behaviour of gases, temperature, pressure, and the thermodynamic principles behind them to make exam preparation more convenient and meaningful.

Download PDF

NCERT Solutions for Class 11 Physics Thermal Properties of Matter - Exercise Solution

The NCERT Solutions for Class 11 Physics Chapter 10 Thermal Properties of Matter addresses every question in the exercises with step-by-step explanations. Students develop an insight into topics such as heat transfer, heat expansion, and specific heat, and can solve numerical problems, and prepare better to take board exams, JEE, and NEET.

Q. 10.1 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:

The relation between Kelvin and Celcius scale is T K = T C + 273.15

Triple Point of Neon in Kelvin T K = 24.57 K

Triple Point of Neon in Celcius T C = T K -273.15 = 24.57 -273.15 =-248.58 o C

Triple Point of carbon dioxide in Kelvin T K = 216.55 K

Triple Point of carbon dioxide in Celcius T C = T K -273.15 = 216.55 - 273.15 = -56.60 o C

The relation between Celcius and Fahrenheit scale is $T_{F}=\frac{9}{5}T_{C} + 32$

Triple Point of Neon in Fahrenheit is

$T_{F}=\frac{9}{5}\times (-248.58) + 32$

T F = -415.44 o C

Triple Point of carbon dioxide in Fahrenheit is

$T_{F}=\frac{9}{5}\times (-56.60) + 32$

T F = -69.88 o C

Q. 10.2 Two absolute scales $A$ and $B$ have triple points of water defined to be $200A$ and $350B$ . What is the relation between $T_{A}$ and $T_{B}$ ?

Answer:

200 A = 273 K

$T_{K}=\frac{273}{200}T_{A}$

350 B = 273 K

$T_{K}=\frac{273}{350}T_{B}$

Equating T K from the above two equations, we have

$\frac{273}{200}T_{A}=\frac{273}{350}T_{B}$


$T_{A}=\frac{4}{7}T_{B}$

Q. 10.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:$R=R_{0}\left [ 1+\alpha (T-T_{0}) \right ]$ The resistance is $101.6\Omega$ at the triple-point of water $273.16\; K,$ and $165.5\; \Omega$ at the normal melting point of lead $(600.5\; K).$ What is the temperature when the resistance is $123.4\; \Omega$?

Answer:

$R=R_{0}\left [ 1+\alpha (T-T_{0}) \right ]$

R 0 = $101.6\Omega$

T 0 = $273.16 K$

R = $165.5\; \Omega$

T = $600.5 K$

Putting the above values in the given equation, we have

$\\\alpha =\frac{165.5-101.6}{600.5-27316}\\ $

$\alpha =1.92\times 10^{-2}\ K^{-1}$

For R = $123.4\; \Omega$

$\\T=T_{0}+\frac{1}{\alpha }\left ( \frac{R}{R_{0}}-1 \right )\\ $

$T=273.16+\frac{1}{1.92\times 10^{-2}}\left ( \frac{123.4}{101.6}-1 \right )\\ $

$T=384.75K$

Q. 10.4 (a) Answer the following:
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)?

Answer:

Unlike the melting point of ice and the boiling point of water, the triple point of water has a fixed value of 273.16 K. The melting point of ice and the boiling point of water vary with pressure.

Q. 10.4 (c) Answer the following :

The absolute temperature (Kelvin scale) $T$ is related to the temperature $t_{c}$ on the Celsius scale by $t_{c}=T-273.15$. Why do we have $273.15$ in this relation, and not $273.16?$

Answer:

This is because 0oC on the Celsius scale corresponds to the melting point at standard pressure is equal to 273.15 K, whereas 273.16 K is the triple point of water. The triple point of water is 0.01oC, not 0oC

Q. 10.4 (d) Answer the following :

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?

Answer:

Let at a certain temperature, the readings on the Fahrenheit and Kelvin Scales be T F and T K, respectively

$T_{F}-32=\frac{9}{5}(T_{K}-273)$ $(i)$

Let at another temperature, the reading on the Fahrenheit and Kelvin Scales be T' F and T' K respectively

$T'_{F}-32=\frac{9}{5}(T'_{K}-273)$ $(ii)$

Subtracting equation (ii) from (i)

$T_{F}-T'_{F}=\frac{9}{5}(T_{K}-T'_{K})$

For T K - T' K = 1 K,

T F - T' F = 9/5

Therefore, corresponding to 273.16 K, the absolute scale whose unit interval size is equal to that of the Fahrenheit scale

$\\T_{F}=\frac{9}{5}\times 273.16\\ $

$T_{F}=491.688$

Q. 10.5 (a) 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
Normal melting point of sulphur

What is the absolute temperature of the normal melting point of sulphur as read by thermometers A and B?

Answer:

$\\PV=nRT\\$

$\frac{P}{T}=\frac{nR}{V}$

As the moles of oxygen and hydrogen inside the thermometers and the volume occupied by the gases remain constant P/T would remain constant.

The triple point of water(T 1 ) = 273.16 K

Pressure in thermometer A at a temperature equal to the triple point of water (P 1 ) = $1.250\times 10^{5}Pa$

Pressure in thermometer A at a temperature equal to the Normal melting point of sulphur (P 2 ) = $1.797\times 10^{5}Pa$

The normal melting point of sulphur, as read by thermometer A, T2 would be given as

$T_{2}=\frac{P_{2}T_{1}}{P_{1}}$

$\\T_{2}=\frac{1.797\times 10^{5}\times 273.16}{1.250\times 10^{5}}\\ $

$T_{2}=392.69\ K$

Pressure in thermometer B at a temperature equal to the triple point of water (P1' ) = $0.200\times 10^{5}Pa$

Pressure in thermometer B at a temperature equal to the Normal melting point of sulphur (P2' ) = $0.287\times 10^{5}Pa$

The normal melting point of sulphur, as read by thermometer B, T2' would be given as

$\\T'_{2}=\frac{P'_{2}T_{1}}{P'_{1}}\\ $

$T'_{2}=\frac{0.287\times 10^{5}\times 273.16}{0.200\times 10^{5}}\\$

$ T'_{2}=391.98\ K$

Q. 10.5 (b) 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
Normal melting point of sulphur

What do you think is the reason behind the slight difference in the 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?

Answer:

The slight difference in the answers of thermometers A and B occurs because the gases used in the thermometers are not ideal gases. To reduce this discrepancy, the experiments should be carried out at low pressures where the behaviour of real gases tends close to that of ideal gases.

Q. 10.6 A steel tape 1m long is correctly calibrated for a temperature of $27.0^{\circ}C.$ 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 $45.0^{\circ}C.$ 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 $27.0^{\circ}C\; ?$ Coefficient of linear expansion of steel $=1.20\times 10^{-5}K^{-1}.$

Answer:

At 27 o C the 63 cm ($l_1$) mark on the steel tape would be measuring exactly 63 cm as the tape is calibrated at 27 o C

Coefficient of linear expansion of steel $=1.20\times 10^{-5}K^{-1}.$

Actual length when the scale is giving a reading of 63 cm at 45 oC is $l_2$

$\\l_{2}=l_{1}(1+\alpha \Delta T)\\ =63\times (1+1.20\times 10^{-5}\times (45-27))\\ =63.013608cm$

The actual length of the steel rod on a day when the temperature is 45oC is 63.013608 cm.

Length of the same steel rod on a day when the temperature is 63°C.

Q. 10.7 A large steel wheel is to be fitted on to a shaft of the same material. At $27^{\circ}C$ , 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 : $\alpha _{steel}=1.20\times 10^{-5}K^{-1}$ .

Answer:

Diameter of the steel shaft at 27 o C (T 1 ) d 1 = 8.70 cm

The diameter of the central hole in the wheel d2 = 8.69 cm

Coefficient of linear expansion of the steel $\alpha _{steel}=1.20\times 10^{-5}K^{-1}$ .

The wheel will slip on the shaft when the diameter of the steel shaft becomes equal to the diameter of the central hole in the wheel.

Let this happen at temperature T

$\\d_{2}=d_{1}(1+\alpha (T-T_{1}))\\ 8.69=8.7(1+1.2\times 10^{5}(T-27))\\ $

$T=-68.79\ ^{o}C$

Q. 10.8 A hole is drilled in a copper sheet. The diameter of the hole is $4.24\; cm$ at $27.0^{\circ}C.$ What is the change in the diameter of the hole when the sheet is heated to $227^{\circ}C\; ?$ Coefficient of linear expansion of copper $=1.70\times 10^{-5}K^{-1}.$

Answer:

Coefficient of linear expansion of copper $\alpha$ $=1.70\times 10^{-5}K^{-1}.$

Coefficient of superficial expansion of copper is $\beta$

$\beta =2\alpha \\ $

$\beta =2\times 1.7\times 10^{-5}\\ $

$\beta =3.4\times 10^{-5}K^{-1}$

Diameter of the hole at 27 o C (d 1 ) = 4.24 cm

Area of the hole at 227 o C is

$A_{2}=A_{1}(1+\beta \Delta T)\\ $

$A_{2}=\pi \left ( \frac{4.24}{2} \right )^{2}(1+3.4\times 10^{-5}(227-27))$
$A_{2}=14.215\ cm^{2}$

Let the diameter at 227 oC be d 2

$\\\pi \left ( \frac{d_{2}}{4} \right )^{2}=14.215\\$

$ d_{2}=4.254cm$

Change in diameter is d 2 -d 1 = 4.24 -4.254 = 0.014 cm.

Q. 10.9 A brass wire $1.8\; m$ long at $27^{\circ}C$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-39^{\circ}C,$ what is the tension developed in the wire if its diameter is $2.0\; mm\; ?$ Co-efficient of linear $expansion\; of \; brass=2.0\times 10^{-5}K^{-1};$ $Young's\; modulus \; of\; brass=0.91\times 10^{11}Pa.$

Answer:

Youngs Modulus of Brass, $Y=0.91\times 10^{11}$

Co-efficient of linear expansion of Brass, $\alpha =2.0\times 10^{-5}K^{-1}$

The diameter of the given brass wire, d = 2.0 mm

Length of the given brass wire, $l$ = 1.8 m

Initial Temperature T 1 = 27 o C

Final Temperature T 2 = -39 o C

$\\Y=\frac{Stress}{Strain}\\ $

$Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}\\ $

$F=\frac{\Delta lAY}{l}\\ $

$F=\frac{(l\alpha \Delta T)AY}{l}\\ $

$F=\alpha \Delta TAY\\ $

$F=\alpha \Delta TY\pi \frac{d^{2}}{4}\\ $

$F=\frac{2.0\times 10^{-5}\times (-39-27)\times 0.91\times 10^{11}\times \pi \times (2\times 10^{-3})^{2}}{4}$

$F=-378N$

The tension developed in the wire is 378 N. The negative sign signifies that this tension is inward.

Q. 10.10 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 $250^{\circ}C,$ if the original lengths are at $40.0^{\circ}C\;?$ Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Coefficient of linear $expansion\; of\; brass=2.0\times 10^{-5}K^{-1},\; steel=1.2\times 10^{-5}K^{-1}$ ).

Answer:

Length of the rods $l$ = 50 cm

Co-efficient of linear expansion of brass, $\\\alpha _{b}=2\times 10^{-5}K^{-1}$

Co-efficient of linear expansion of steel, $\\\alpha _{s}=1.2\times 10^{-5}K^{-1}$

Initial Temperature T 1 = 40.0 o C

Final Temperature T 2 = 250 o C

Change in the length of the brass rod is

$\\\Delta l_{b}=l_{t}\alpha_{b} \Delta T\\ $

$\Delta l_{b}=50\times 2.0\times 10^{-5}\times (250-40)\\ $

$\Delta l_{b}=0.21cm$

Change in the length of the steel rod is

$\\\Delta l_{s}=l_{s}\alpha_{s} \Delta T\\ $

$\Delta l_{s}=50\times 1.2\times 10^{-5}\times (250-40)\\ $

$\Delta l_{s}=0.126cm$

Change in the length of the combined rod is

$\\\Delta l=\Delta l_{s}+\Delta l_{b}\\ $

$\Delta l=0.126+0.21\\ $

$\Delta l=0.336cm$

Q. 10.11 The coefficient of volume expansion of glycerine is $49\times 10^{-5}K^{-1}.$ What is the fractional change in its density for a $30^{\circ}C$ rise in temperature?

Answer:

Coefficient of volume expansion of glycerine is $\gamma =49\times 10^{-5}K^{-1}$

Let the initial volume and mass of a certain amount of glycerine be V and m, respectively.

Initial density is

$\rho =\frac{m}{V}$

Change in volume for a 30 oC rise in temperature will be

$\\\Delta V=V(\gamma \Delta T)\\ $

$\Delta V=V(49\times 10^{-5}\times 30)\\ $

$\Delta V=0.0147V$

Final Density is

$\\\rho' =\frac{m}{V+\Delta V}\\ $

$\rho'=\frac{m}{1.0147V}\\ $

$\rho'=\frac{0.986m}{V}$

Fractional Change in density is

$\\\frac{\rho'-\rho}{\rho}\\ =\frac{\frac{0.986m}{v}-\frac{m}{v}}{\frac{m}{v}}\\ =-0.014$

The negative sign signifies that with an increase in temperature, density will decrease.

Q. 10.12 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 $50^{o}/_{o}$ of power is used up in heating the machine itself or lost to the surrounding. $Specific\; heat\; of\; aluminium=0.91\; J\; g^{-1}K^{-1}.$

Answer:

Power of the drilling machine, P = 10 kW

Time. t = 2.5 min

Total energy dissipated E, is

$\\E=Pt\\$

$ E=10\times 10^{3}\times 2.5\times 60\\ $

$E=1.5\times 10^{6}J$

The thermal energy absorbed by the aluminium block is

$\\\Delta Q=\frac{E}{2}\\$

$ \Delta Q=\frac{1.5\times 10^{6}}{2}\\ $

$\Delta Q=7.5\times 10^{5}J$

Mass of the aluminium block, m = 8.0 kg

Specific heat of aluminium, c = 0.91 J g -1 K -1

Let the rise in temperature be $\Delta T$

$\\mc\Delta T=\Delta Q\\ $

$\Delta T=\frac{\Delta Q}{mc}\\ $

$\Delta T=\frac{7.5\times 10^{5}}{0.91\times 8\times 10^{3}}\\ $

$\Delta T=103.02\ ^{o}C$

Q. 10.13 A copper block of mass $2.5 \; kg$ is heated in a furnace to a temperature of $500^{\circ}C$ and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper $=0.39\; J\; g^{-1}K^{-1};$ heat of fusion of water $=335\; J\; g^{-1}$ ).

Answer:

Mass of copper block m = 2.5 kg

Initial Temperature of the copper block, T 1 = 500 o C

Final Temperature of Copper block, T 2 = 0 o C

Specific heat of copper, c = 0.39 J g -1 K -1

Thermal Energy released by the copper block is $\Delta Q$

$\\\Delta Q=mc\Delta T\\ $

$\Delta Q=2.5\times 10^{3}\times 0.39\times 500\\ $

$\Delta Q=487500\ J$

Latent heat of fusion of water, L = 335 j g -1

The amount of ice that can melt is

$\\w=\frac{\Delta Q}{L}\\ $

$w=\frac{487500}{335}\\$

$ w=1455.22g$

1.455 kg of ice can melt using the heat released by the copper block.

Q. 10.14 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 $150\; cm^{3}$ of water at $27^{\circ}C.$ The final temperature is $40^{\circ}C.$ 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?

Answer:

Let the specific heat of the metal be c.

Mass of metal block m = 200 g

Initial Temperature of the metal block = 150 o C

Final Temperature of the metal block = 40 o C

The heat released by the block is

$\\\Delta Q=mc\Delta T\\ $

$\Delta Q=200\times c\times (150-40)\\ $

$\Delta Q=22000c$

Initial Temperature of the calorimeter and water = 27 °C

Final Temperature of the calorimeter and water = 40 °C

Amount of water = 150 cm

Mass of water = 150 g

Water equivalent of calorimeter = 25 g

Specific heat of water = 4.186 J g -1 K -1

Heat absorbed by the Calorimeter and water is $\\\Delta Q'$

$\\\Delta Q'=(150+25)\times 4.186\times (40-27)\\ $

$\Delta Q'=9523.15J$

The heat absorbed by the Calorimeter and water is equal to the heat released by the block

$\\\Delta Q=\Delta Q'\\ 22000c=9523.15\\ c=0.433\ J\ g^{-1}\ K^{-1}$

The above value would be less than the actual value since some heat must have been lost to the surroundings as well, which we haven't accounted for.

Q. 10.15 Given below are observations on molar specific heats at room temperature of some common gases.

Gas Molar specific heat (Cv)

$(cal\; mol^{-1}K^{-1})$

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, the 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?

Answer:

Monoatomic gases have only a translational degree of freedom, but diatomic gases have rotational degrees of freedom as well. The temperature increases with an increase in the spontaneity of motion in all degrees. Therefore, to increase the temperature of diatomic gases, more energy is required than that required to increase the temperature of monoatomic gases by the same value, owing to higher degrees of freedom in diatomic gases.

If we only consider rotational modes of freedom, the molar specific heat of the diatomic gases would be given as

$\\c=\frac{fR}{2}$


$ c=\frac{5}{2}\times 1.92$


$ c=4.95\ cal\ mol^{-1}\ K^{-1}$

The number of degrees of freedom = 5 (3 translational and 2 rotational)

The values given in the table are more or less in accordance with the above calculated ones. The larger deviation from the calculated value in the case of chlorine is because of the presence of vibrational motion as well.

Q. 10.16 A child running a temperature of $101^{\circ}F$ 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 $98^{\circ}F$ in 20 minutes, 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 $580\; cal\; g^{-1}$ .

Answer:

Initial Temperature of the boy = 101°F

Final Temperature of the boy = 98°F

Change in Temperature is

$\\\Delta T=3\ ^{o}F\\ $

$\Delta T=3\times \frac{5}{9}\\ $

$\Delta T=1.67\ ^{o}C$

Mass of the child is m = 30 kg

Specific heat of human body = 1000 cal kg -1 °C -1

Heat released is $\\\Delta Q$

$\\\Delta Q=mc\Delta T\\$

$\Delta Q=30\times 1000\times 1.67\\ $

$\Delta Q=50000\ cal$

Latent heat of evaporation of water = 580 cal g -1

The amount of heat lost by the boy has been absorbed by water.

Let the mass of water which has evaporated be m'

$\\\Delta Q=m'L\\ $

$ m'=\frac{Q}{L}\\ $

$m'=\frac{50000}{580}\\$

$ m'=86.2\ g$
Time in which the water has evaporated, t = 20 min.

The rate of evaporation is m'/t

$\\\frac{m'}{t}=\frac{86.2}{20}\\ $

$\frac{m'}{t}=4.31\ g\ min^{-1}$

Q. 10.17 A ‘thermacole’ icebox is a cheap and an 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 $45^{\circ}C.$ and co-efficient of thermal conductivity of thermacole is $0.01\; J\; s^{-1}m^{-1}K^{-1}.$ $[Heat\; of\; fusion\; of\; water=335\times 103\; J\; kg^{-1}]$

Answer:

Side of the box s = 30 cm

Area available for conduction A

A = 6s 2

A=6(30) 2

A=5400 cm 2 = 0.54 m 2

Temperature difference = 45 o C

Co-efficient of thermal conductivity of thermacole is k = 0.01 J s -1 m -1 K -1

Width of the box is d = 5 cm

Heat absorbed by the box in 6 hours is $\Delta Q$

$\\\Delta Q=\frac{kA\Delta T}{l}\\ $

$\Delta Q=\frac{0.01\times 0.54\times 45\times 6\times 60\times 60}{0.05}\\ $

$\Delta Q=104976\ J$

The heat of fusion of water is $L=335\times 10^{3}\ J\ kg^{-1}$

The amount of ice which has melted is m'

$\\m'=\frac{\Delta Q}{L}\\ $

$m'=\frac{104976}{335\times 10^{3}}\\ $

$m'=0.313\ kg$

Amount of ice left after 6 hours = 4 - 0.313 = 3.687 kg

Q. 10.18 A brass boiler has a base area of $0.15\; m^{2}$ and thickness $1.0\; cm.$ It boils water at the rate of $6.0 \; kg/min$ when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass $=109\; J\; s^{-1}m^{-1}K^{-1};$ Heat of vaporisation of water $=2256\times 103\; J\; kg^{-1}.$

Answer:

The rate at which water boils, R = 6.0 kg min -1

The heat of vaporisation of water, $L=2256\times 10^{3}\ J\ kg^{-1}$

The rate at which heat enters the boiler

$\\\frac{dQ}{dT}=RL\\ $

$\frac{dQ}{dT}=\frac{6\times 2256\times 10^{3}}{60}\\ $

$\frac{dQ}{dT}=2.256\times 10^{5}Js^{-1}$

The base area of the boiler, A = 0.15 m 2

Thickness, $l$ = 1.0 cm

Thermal conductivity of brass $=109\; J\; s^{-1}m^{-1}K^{-1};$

The temperature inside the boiler = Boiling point of water = 100 o C

Let the temperature of the flame in contact with the boiler be T

The amount of heat flowing into the boiler is

$\begin{aligned}
& \frac{d Q}{d t}=\frac{K A \Delta T}{l} \\
& 2.256 \times 10^5=\frac{109 \times 0.15 \times(T-100)}{1 \times 10^{-2}} \\
& T-100=137.98 \\
& T=237.98^{\circ} \mathrm{C}
\end{aligned}$

The temperature of the flame in contact with the boiler is 237.98 o C

Q. 10.19 (a) Explain whya body with large reflectivity is a poor emitter

Answer:

A body with a large reflectivity is a poor absorber. As we know a body which is a poor absorber will as well be a poor emitter. Therefore, a body with large reflectivity is a poor emitter.

Q .10.19 (b) Explain whya brass tumbler feels much colder than a wooden tray on a chilly day

Answer:

Brass is a good conductor of heat. Therefore once someone touches brass heat from their body flows into it and it feels cold, in case of a wooden tray, no such conduction of heat from the body takes place as wood is a very poor conductor of heat.

Q. 10.19 (c) Explain why 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

Answer:

An optical pyrometer relates the brightness of a glowing body with its temperature. In the open because of other sources of light the sensor in the optical pyrometer does not detect the true brightness of a red hot piece of iron and thus does not predict its temperature correctly whereas in the furnace the piece of iron is the only source of light and the sensor detects its brightness correctly thus giving the correct value of the temperature.

Q. 10.19 (d) Explain why the Earth without its atmosphere would be inhospitably cold

Answer:

The sun's rays contain infrared radiation. These are reflected back by the lower part of the atmosphere after being reflected by the surface of the Earth and are trapped inside the atmosphere, thus maintaining the Earth's temperature at a hospitable level. Without these rays being trapped, the temperature of the Earth would go down severely, and thus the Earth without its atmosphere would be inhospitably cold.

Q. 10.19 (e) Explain why heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water

Answer:

Heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water because the same amount of steam at 100 °C contains more energy available for heat dissipation than the same amount of water at 100 °C in the form of latent heat of vaporisation.

Q. 10.20 A body cools from $80^{\circ}C$ to $50^{\circ}C$ in 5 minutes. Calculate the time it takes to cool from $60^{\circ}C$ to $30^{\circ}C.$ The temperature of the surroundings is $20^{\circ}C.$

Answer:

Let a body initially be at temperature T 1

Let its final Temperature be T 2

Let the surrounding temperature be T 0

Let the temperature change in time t.

According to Newton's Law of cooling

$\\-\frac{dT}{dt}=K(T-T_{0})\\ $

$\frac{dT}{T-T_{0}}=-Kdt\\$

$ \int_{T_{1}}^{T_{2}}\frac{dT}{T-T_{0}}=-K\int_{0}^{t}dt\\$


$ \left [ ln(T-T_{0}) \right ]_{T_{1}}^{T_{2}}=-K[t]_{0}^{t}\\ $


$ln\left ( \frac{T_{2}-T_{0}}{T_{1}-T_{0}} \right )=-Kt$

where K is a constant.

We have been given that the body cools from 80 o C to 50 o C in 5 minutes when the surrounding temperature is 20 o C.

T 2 = 50 o C

T 1 = 80 o C

T 0 = 20 o C

t = 5 min = 300 s.

$\\ln\left ( \frac{50-20}{80-20} \right )=-K\times 300\\ K=\frac{ln(2)}{300}\\$

For T 1 = 60 o C and T 2 = 30 o C we have

$\\ln\left ( \frac{30-20}{60-20} \right )=-\frac{ln2}{300}t\\ $


$t=\frac{ln(4)\times 300}{ln(2)} \\$


$t=\frac{2ln(2)\times 300}{ln(2)}\\ $


$t=600\ s\\ $


$t= 10\ min$

The body will take 10 minutes to cool from 60 o C to 30 o C at the surrounding temperature of 20 o C.

Class 11 Physics NCERT Chapter 10: Higher Order Thinking Skills (HOTS) Questions

The Class 11 Physics NCERT Chapter 10 HOTS (Higher Order Thinking Skills) Questions are designed to challenge students’ understanding of thermal properties of matter beyond textbook exercises. These questions encourage critical thinking, application-based learning, and problem-solving skills, preparing students for competitive exams like JEE and NEET.

Q1: When 100 g of liquid A at 100oC is added to 50 g of liquid B at a temperature of 75oC, the temperature of the mixture becomes 90oC. The temperature of the mixture (in oC), if 100 g of liquid A at 100oC is added to 50 g of liquid B at 50oC, will be :

Answer:

Specific Heat -

$
\begin{aligned}
& S=\frac{Q}{m \cdot \Delta \theta} \\
& \text { - wherein } \\
& \mathrm{S}=\text { specific heat } \\
& \Delta \theta=\text { Change in temperature } \\
& \mathrm{m}=\text { Amount of mass } \\
& \text { For (I) } \\
& 100 \times S_A \times(100-90)=50 \times S_B \times(90-75) \\
& S_A=\frac{1.5}{2} S_B=\frac{3}{4} S_B \cdots(1) \\
& \text { now for }(\mathrm{II}) \\
& 100 \times S_A \times(100-T)=50 \times S_B \times(T-50) \\
& \Rightarrow 2 S_A(100-T)=S_B(T-50) \cdots(2)
\end{aligned}
$
From (1) & (2)

$
\begin{aligned}
& 2 \times \frac{3}{4} \times(100-T)=(T-50) \\
& 3(100-T)=2(T-50) \\
& 300-3 T=2 T-100 \\
& 400=5 T \\
& T=80^{\circ} \mathrm{C}
\end{aligned}
$


Q2: If the length of cylinder on heating increases by 2% then the area of base of cylinder will increase by

Answer:

Assuming the initial length of the cylinder is $\ell_1$, the linear expansion coefficient is $\alpha$, and the temperature rise is $\Delta \mathrm{t}^{\circ} \mathrm{C}$,

$
\begin{aligned}
& \Delta \ell=\ell \alpha \Delta \mathrm{t} \Rightarrow \frac{\Delta \ell}{\ell} \times 100=\alpha \Delta \mathrm{t} \times 100 \\
& \therefore 2=\alpha \Delta \mathrm{t} \times 100
\end{aligned}
$


Now, for the area

$
\begin{aligned}
& \Delta A=A \beta \Delta t \Rightarrow \frac{\Delta A}{A}=\beta \Delta t \\
& \Rightarrow \frac{\Delta A}{A} \times 100=\beta \Delta t \times 100 \\
& \% \text { change in area }=\beta \Delta t \times 100 \\
& =\beta \Delta t \times 100
\end{aligned}
$


We know that $2 \times \alpha \times \Delta \mathrm{t}=\beta \times \Delta \mathrm{t}$

$
\begin{aligned}
& =2 \alpha \Delta \mathrm{t} \times 100 \\
& =2(\alpha \Delta \mathrm{t} \times 100) \\
& =2 \times 2 \\
& =4 \%
\end{aligned}
$


Q3: A cylindrical metal rod of length L0 is shaped into a ring with a small gap as shown in Figure. On heating the system,

A. x will decrease, r will increase

B. Both x and r will decrease

C. Both x and r will increase

D. r will increase, x will remain same

Answer:

Linear Expansion - When a solid is heated and its length increases, then the Expansion is called linear Expansion.
On the heating, the system x and r both will increase since the expansion of isotropic solids is similar to true photographic enlargement.


Q4: The coefficient of thermal expansion of a rod varies with temperature as $\alpha=\alpha_0+\alpha_0 \frac{T}{T_0}$ where T is the temperature of the body and all others are positive constants. If the temperature of the rod of length $\ell_0$ is changed from $T_0 / 2$ to $T_0$ uniformly with time, then what is the ratio of the rate of fractional change of length initially and finally?

Answer:

We know the coefficient of thermal expansion is given as

$
\begin{gathered}
\quad \frac{\mathrm{d} L}{L \mathrm{~d} T}=\alpha \\
\Rightarrow \frac{\mathrm{d} L}{L \mathrm{dt}}=\alpha \frac{\mathrm{d} T}{\mathrm{~d} t} \\
\frac{\left(\frac{d L}{L d t}\right)_1}{\left(\frac{d L}{L d t}\right)_2}=\frac{\alpha_1 \frac{d T}{d t}}{\alpha_2 \frac{d T}{d t}} \\
\Rightarrow \frac{\left(\frac{d L}{L d}\right)_1}{\left(\frac{d L}{L d t}\right)_2}=\frac{\alpha_1}{\alpha_2} \\
\Rightarrow
\frac{\alpha_0+\alpha_0 \frac{T_0 / 2}{T_0}}{\alpha_0+\alpha_0 \frac{T_0}{T_0}}=\frac{3 / 2}{2}=\frac{3}{4}
\end{gathered}
$


Q5: A red-hot iron (mass 25 g) has a temperature of 100ºC. It is immersed in a mixture of ice and water at thermal equilibrium. The volume of the mixture is found to be reduced by $0.15 \mathrm{~cm}^3$with the temperature of the mixture remains constant. Find the specific heat (in cal gm ${ }^0 \mathrm{C}$) of the metal.

Given specific gravity of ice = 0.92, specific gravity of water at $0^{\circ} \mathrm{C}=1.0$, latent heat of fusion of ice $=80 \mathrm{calg}^{-1}$.

Answer:

As the ice melts, its density changes, which causes change in volume. Let m g of ice melt.

$\begin{aligned} & \frac{m}{\rho_{i c e}}-\frac{m}{\rho_{\mathrm{H}_2 \mathrm{O}}}=0.15 \\ & m=\frac{0.15 \times 0.92}{0.08} \\ & \therefore \quad H_{\text {lost }}=H_{\text {gaired }} \\ & \Rightarrow \quad 25 \times S \times 100=\frac{0.15 \times 0.92}{0.08} \times 80 \\ & S=0.06 \mathrm{cal} \mathrm{gm}^{\circ} \mathrm{C}\end{aligned}$


NCERT Chapter 10: Thermal Properties of Matter Topics

Chapter 10: Thermal Properties of Matter: Key topics include heat, temperature, thermal expansion and heat conduction. The chapter describes temperature-dependant behaviour of matter and the physics behind the phenomena we experience in our daily lives such as boiling of water, expansion of metals and sea breezes. These are the basic points of thermodynamics to understand and the applications in real life.

10.1 Introduction
10.2 Temperature And Heat
10.3 Measurement Of Temperature
10.4 Ideal-Gas Equation And Absolute Temperature
10.5 Thermal Expansion
10.6 Specific Heat Capacity
10.7 Calorimetry
10.8 Change Of State
10.8.1 Latent Heat
10.9 Heat Transfer
10.9.1 Conduction
10.9.2 Convection
10.9.3 Radiation
10.9.4 Blackbody Radiation
10.10 Newton’s Law of Cooling

NCERT Solutions for Class 11 Chapter 10: Important Formulae

Class 11 Physics NCERT Chapter 10: Important Formulae include all essential equations related to heat, temperature, thermal expansion, and heat transfer. These formulas help students solve numerical problems effectively and are crucial for board exams, JEE, NEET, and other competitive tests. Mastery of these formulas simplifies understanding and application of thermal concepts.

1. Heat and Temperature Relation

$
Q=m c \Delta T
$
Where:

$
\begin{aligned}
& Q=\text { heat added or removed } \\
& m=\text { mass } \\
& c=\text { specific heat capacity } \\
& \Delta T=\text { change in temperature }
\end{aligned}
$

2. Specific Heat Capacity

$c=\frac{Q}{m \Delta T}$

3. Heat Capacity

$\begin{aligned} & C=m c \\ & C=\frac{Q}{\Delta T}\end{aligned}$

4. Latent Heat

$
Q=m L
$
Where:
$L=$ latent heat of fusion or vaporisation
(Use for phase changes with no temperature change)

5. Linear, Areal and Volume Expansion

$
\Delta L=L \alpha \Delta T
$

$
\Delta A=A \beta \Delta T
$
Where $\beta=2 \alpha$


$
\Delta V=V \gamma \Delta T
$
Where $\gamma=3 \alpha$

8. Calorimetry Principle

Heat lost=Heat gained
(Applies to thermal equilibrium problems)

9. Thermal Conduction

$
Q=\frac{k A\left(T_1-T_2\right) t}{d}
$


Where:
$k=$ thermal conductivity
$A=$ area
$d=$ thickness
$t=$ time
$T_1-T_2=$ temperature difference

10. Newton’s Law of Cooling

$\frac{d T}{d t} \propto\left(T-T_{\text {env }}\right)$

Approach to Solve Questions of Thermal Properties of Matter

  • Have a Clear Conceptual Understanding. Revise the simple terms like heat, temperature, specific heat, thermal expansion, and types of heat transfer.
  • Categorise the Kind of Problem. Find out whether the question involves temperature change, thermal expansion, heat capacity, phase change or heat transfer.
  • These should be Calculated Using the Appropriate Formula. Choose the right formula according to the topic
  • Apply conduction, convection, and radiation formulae where necessary.
  • Convert Units, if required. Put everything in SI (e.g. mass as kg, temperature in Kelvin, time in seconds).
  • Use the principle of Calorimetry: Heat lost=Heat gained
  • Include Latent Heat in the case of Phase Change. Remember to add or delete latent heat in melting, boiling or condensation questions.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

This comparison table shows the divergence between textbook focus, depth and the kind of questions that are asked in a competitive examination like JEE and NEET on the topic of thermal properties of matter. NCERT gives a good conceptual knowledge along with a basic practice set, whereas JEE/NEET questions require application of techniques, application-based reasoning and advanced numerical practice. This table assists the students in areas they need to focus on in preparation of the exams.

Frequently Asked Questions (FAQs)

Q: How important is Thermal Properties of Matter Class 11 for JEE Main
A:

One question may be asked from Class 11 Physics chapter 10 Thermal Properties of Matter  for JEE Main exam. This is one important chapter for KVPY and NSEP exms. Students can get more problems on the chapter from the NCERT Exemplar.

Q: How important is the chapter Thermal Properties of Matter for NEET
A:

One question can be expected for NEET exam from the chapter Thermal Properties of Matter. Students can practice more questions from previous year NEET papers.

Q: What are the important topics of the chapter Thermal Properties of Matter
A:

Topics covered in NCERT book Class 11 chapter Thermal Properties of Matter are-

  • Temperature and heat   
  • Measurement of temperature   
  • Ideal-gas equation and absolute temperature   
  • Thermal expansion   
  • Specific heat capacity   
  • Calorimetry   
  • Change of state   
  • Heat transfer   
  • Newton’s law of cooling   
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