Home › NCERT Solutions › Class 11 Physics › Chapter 10

NCERT Solutions · Class 11 Physics · NCERT Physics Part I and II Part II · Chapter 10

Chapter 10: Thermal Properties of Matter (Properties of Bulk Matter)

Step-by-step solutions to all 20 exercises of Chapter 10, Thermal Properties of Matter (NCERT Class 11 Physics Part II, 2026-27): temperature scales and gas thermometers, a resistance thermometer, thermal expansion of tapes, shafts, holes and wires, thermal stress, calorimetry, specific heat of gases, evaporation, conduction through an icebox and a boiler, radiation and Newton's law of cooling. All 20 questions are answered, with the key answer highlighted.

NCERT textbook Chapter 10 (PDF)

Exam revision? Chapter 10 notes with key points, diagrams and practice questions: see the free sample · ₹199 for all 14 chapters

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-11-physics/chapter-10-thermal-properties-of-matter

$t_{F} = \frac{9}{5} t_{C} + 32$ , $T = t_{C} + 273.15$ . Linear expansion $Δ l = l α Δ T$ ; volume $Δ V = V γ Δ T$ . Thermal stress $= Y α Δ T$ . Heat $Q = m c Δ T$ , $Q = m L$ . Conduction: $H = \frac{K A Δ T}{L}$ . Newton's law of cooling: $\frac{Δ T}{Δ t} = k (\overset{ˉ}{T} - T_{s})$ (average-temperature form).

Exercises

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.

Solution

$t_{C} = T - 273.15$ , $t_{F} = \frac{9}{5} t_{C} + 32$ .

Neon: $t_{C} = 24.57 - 273.15 = - 248.58^{\circ}$ C; $t_{F} = \frac{9}{5} (- 248.58) + 32 = - 415.44^{\circ}$ F.

CO₂: $t_{C} = 216.55 - 273.15 = - 56.60^{\circ}$ C; $t_{F} = \frac{9}{5} (- 56.60) + 32 = - 69.88^{\circ}$ F.

Neon: −248.58 °C, −415.44 °F; CO₂: −56.60 °C, −69.88 °F

10.2

Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B?

Solution

On any absolute scale, temperature is proportional to the true absolute temperature, so $\frac{T _{A}}{T _{B}} = \frac{200}{350} = \frac{4}{7}$ .

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

10.3

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R₀ [1 + α (T – T₀)]. 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?

Solution

$R_{0} = 101.6 Ω$ at $T_{0} = 273.16$ K. From lead: $165.5 = 101.6 [1 + α (327.34)] \Rightarrow α = \frac{0.6289}{327.34} = 1.921 \times 1 0^{- 3} K^{- 1}$ .

For $R = 123.4$ : $T - T_{0} = \frac{123.4/101.6 - 1}{α} = \frac{0.2146}{1.921 \times 1 0 ^{- 3}} = 111.7$ K.

T = 384.8 K

10.4

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 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? (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?

Solution

(a) The triple point (ice, water and vapour together) occurs at one unique temperature and pressure. The melting and boiling points change with pressure, so they are not reproducible standards.

(b) Absolute zero (0 K).

(c) On the Celsius scale the melting point of ice (at 1 atm) is 0 °C, and it corresponds to 273.15 K. The triple point of water is 0.01 °C higher, i.e. 273.16 K.

(d) A Fahrenheit degree is $\frac{5}{9}$ of a kelvin, so the number is $273.16 \times \frac{9}{5} = 491.69$ .

(a) The triple point is unique; melting/boiling points depend on pressure (b) absolute zero (c) triple point is 0.01 °C (d) 491.69

10.5

Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made: at the triple-point of water, pressure in A = 1.250 × 10⁵ Pa and in B = 0.200 × 10⁵ Pa; at the normal melting point of sulphur, A = 1.797 × 10⁵ Pa and B = 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?

Solution

(a) For a constant-volume gas thermometer, $T = 273.16 \times \frac{p}{p _{t r}}$ .

$T_{A} = 273.16 \times \frac{1.797}{1.250} = 392.69$ K; $T_{B} = 273.16 \times \frac{0.287}{0.200} = 391.98$ K.

(b) Oxygen and hydrogen are not perfectly ideal gases. Take readings at lower and lower pressures and extrapolate the measured temperature to zero pressure, where all gases behave ideally; the two thermometers will then agree.

(a) 392.69 K (A), 391.98 K (B) (b) the gases are not ideal; extrapolate readings to zero pressure.

10.6

A steel tape 1m long is correctly calibrated for a temperature of 27.0 °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 °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 °C? Coefficient of linear expansion of steel = 1.20 × 10⁻⁵ K⁻¹.

Solution

At 45 °C each "cm" mark of the tape has expanded. The true length of 63.0 tape-cm is $63.0 (1 + α Δ T) = 63.0 (1 + 1.20 \times 1 0^{- 5} \times 18) = 63.0 + 0.0136 = 63.0136$ cm.

At 27.0 °C the tape is correct and the rod (same material) shrinks back in the same ratio, so it measures 63.0 cm.

63.0136 cm on the hot day (63.0 cm to 3 significant figures); 63.0 cm at 27.0 °C

10.7

A large steel wheel is to be fitted on to 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: α_steel = 1.20 × 10⁻⁵ K⁻¹.

Solution

The shaft's diameter must shrink to 8.69 cm: $8.69 = 8.70 [1 + 1.20 \times 1 0^{- 5} Δ T]$ .

$Δ T = \frac{- 0.01}{8.70 \times 1.20 \times 1 0 ^{- 5}} = - 95.8$ K, so $T = 27 - 95.8 = - 68.8^{\circ}$ C.

About −69 °C

10.8

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.

Solution

A hole expands just as if it were filled with the same material:

$Δ d = d α Δ T = 4.24 \times 1.70 \times 1 0^{- 5} \times 200 = 1.44 \times 1 0^{- 2}$ cm.

The diameter increases by 1.44 × 10⁻² cm

10.9

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.

Solution

The wire tries to contract by $Δ l = l α Δ T$ but is held fixed, so the strain is $α Δ T$ and the tension is $F = Y A α Δ T$ (independent of the length).

$F = 0.91 \times 1 0^{11} \times π (1.0 \times 1 0^{- 3})^{2} \times 2.0 \times 1 0^{- 5} \times 66 = 0.91 \times 1 0^{11} \times 3.14 \times 1 0^{- 6} \times 1.32 \times 1 0^{- 3} = 377$ N.

About 3.8 × 10² N

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 °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).

Solution

$Δ T = 210$ K. Brass: $50 \times 2.0 \times 1 0^{- 5} \times 210 = 0.21$ cm; steel: $50 \times 1.2 \times 1 0^{- 5} \times 210 = 0.126$ cm.

Total $= 0.336 \approx 0.34$ cm. The ends are free, so each rod expands freely: no thermal stress at the junction.

0.34 cm; no thermal stress (free ends)

10.11

The coefficient of volume expansion of glycerine is 49 × 10⁻⁵ K⁻¹. What is the fractional change in its density for a 30 °C rise in temperature?

Solution

Mass is fixed, so $ρ V$ is constant: $\frac{Δ ρ}{ρ} = - \frac{Δ V}{V} = - γ Δ T = - 49 \times 1 0^{- 5} \times 30$ .

−0.0147 (density falls by about 1.5 × 10⁻²)

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% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g⁻¹ K⁻¹.

Solution

Heat to the block $= 0.5 \times 1 0^{4} \times 150 = 7.5 \times 1 0^{5}$ J.

$Δ T = \frac{Q}{m c} = \frac{7.5 \times 1 0 ^{5}}{8000 \times 0.91}$

About 103 °C

10.13

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 J g⁻¹ K⁻¹; heat of fusion of water = 335 J g⁻¹).

Solution

Heat given by copper cooling to 0 °C: $Q = 2500 \times 0.39 \times 500 = 4.875 \times 1 0^{5}$ J.

Ice melted $= \frac{4.875 \times 1 0 ^{5}}{335} = 1455$ g.

About 1.5 kg

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 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?

Solution

Heat lost by metal $= 200 \times s \times (150 - 40) = 22000 s$ J.

Heat gained by water + calorimeter $= (150 + 25) \times 4.186 \times (40 - 27) = 9523$ J.

$s = \frac{9523}{22000} = 0.43 J g^{- 1} K^{- 1}$ .

If some heat is lost to the surroundings, the water gains less than the metal gives, so the calculated value is smaller than the actual value.

0.43 J g⁻¹ K⁻¹; smaller than the actual value

10.15

Given below are observations on molar specific heats at room temperature of some common gases: 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, 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?

Solution

A monatomic gas has only translational motion: $C_{v} = \frac{3}{2} R \approx 2.98$ cal mol⁻¹ K⁻¹. These gases are diatomic; their molecules can also rotate, so heat must also raise rotational energy: $C_{v} = \frac{5}{2} R \approx 4.97$ cal mol⁻¹ K⁻¹, which matches all the gases except chlorine.

Chlorine's larger value shows that, besides rotation, vibrational modes are also excited in chlorine at room temperature (its heavy atoms and weaker bond make vibration easier).

Diatomic gases have rotational modes too (≈ 5R/2); chlorine also vibrates at room temperature.

10.16

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 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.

Solution

Fall of 3 °F $= 3 \times \frac{5}{9} = 1.67^{\circ}$ C. Heat lost $= 30000 \times 1 \times 1.67 = 5.0 \times 1 0^{4}$ cal.

Sweat evaporated $= \frac{5.0 \times 1 0 ^{4}}{580} = 86.2$ g in 20 min.

About 4.3 g min⁻¹

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 °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 × 10³ J kg⁻¹]

Solution

Area of 6 faces $= 6 \times (0.30)^{2} = 0.54 m^{2}$ . Heat flow $H = \frac{K A Δ T}{L} = \frac{0.01 \times 0.54 \times 45}{0.05} = 4.86$ W.

In 6 h: $Q = 4.86 \times 21600 = 1.05 \times 1 0^{5}$ J; ice melted $= \frac{1.05 \times 1 0 ^{5}}{335 \times 1 0 ^{3}} = 0.31$ kg.

About 3.7 kg of ice remains

10.18

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 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 × 10³ J kg⁻¹.

Solution

Heat needed per second $= \frac{6.0}{60} \times 2256 \times 1 0^{3} = 2.256 \times 1 0^{5}$ W.

$H = \frac{K A Δ T}{L} \Rightarrow Δ T = \frac{2.256 \times 1 0 ^{5} \times 0.01}{109 \times 0.15} = 138$ K.

Flame temperature $= 100 + 138$ .

About 238 °C

10.19

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

Solution

(a) A good reflector absorbs little radiation, and by Kirchhoff's law a poor absorber is a poor emitter.

(b) Brass is a good conductor: it draws heat quickly from the hand, so the skin cools fast and it feels cold. Wood is a poor conductor.

(c) In the open, red-hot iron emits less than a black body at the same temperature ( $E = e σ T^{4}$ , e < 1), so the pyrometer reads low. Inside the furnace the iron is in equilibrium with the furnace walls and the radiation is that of a black body, so the reading is correct.

(d) The atmosphere traps infrared radiation from the earth (greenhouse effect). Without it, the heat would escape at night and the temperature would fall very low.

(e) Steam gives out its large latent heat (about 540 cal g⁻¹) when it condenses, much more than hot water gives when it cools by a few degrees.

(a) Poor absorber ⇒ poor emitter (b) brass conducts heat away fast (c) iron is not a black body in the open (d) greenhouse effect (e) latent heat of steam

10.20

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.

Solution

Newton's law (average temperature form): $\frac{Δ T}{t} = k (\overset{ˉ}{T} - T_{s})$ .

First: $\frac{30}{5} = k (65 - 20) \Rightarrow k = \frac{6}{45} = \frac{2}{15} min^{- 1}$ .

Second: $\frac{30}{t} = \frac{2}{15} (45 - 20) = \frac{10}{3} \Rightarrow t = 9$ min.

9 minutes

Cover of Complete Physics Notes for Class 11 (NCERT 2026-27)

Done with the answers?Now revise the whole chapter: Class 11 Physics notes

For NCERT Physics Class 11 Part I and Part II (2026-27 edition) - all 14 chapters
Typed concept notes with every derivation written step by step and labelled diagrams
Every NCERT exercise question solved

₹199₹24920% off · all 14 chapters · instant PDF

Read free sample

What is inside the Chapter 10 notes →

← Chapter 9: Mechanical Properties of Fluids Chapter 11: Thermodynamics →

More free NCERT solutions for Class 11Class 11 Maths · 14 chapters Class 11 Chemistry · 9 chapters Class 11 Biology · 19 chapters All classes →

Found a mistake or need help with a question? Message us on WhatsApp. Free study tips and new solutions: join our WhatsApp Channel.