Class 11 Physics NCERT Solutions

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Thermal Properties of Matter Class 11 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 10, Thermal Properties of Matter — 20 questions from 10.1 to 10.20, each worked through step by step in the CBSE marking pattern. Temperature scales, thermal expansion, specific heat, calorimetry and heat transfer.

Class:11Subject:PhysicsChapter:10

4 Key Formulas25 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 11 Physics Chapter 10?

Chapter 10 carries 20 exercise questions, numbered 10.1 to 10.20. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents21 sections

Revision & Practice

Chapter Overview

Thermal properties of matter brings together temperature scales, thermometers, thermal expansion, calorimetry and heat transfer. The barometer equivalents here are the gas thermometer and the resistance thermometer; the calorimetry questions — metal blocks in calorimeters, drills heating aluminium, copper on ice — are the most repeated board problems. Every question below is from the NCERT Class 11 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Keep the three scale links ready: t_C = T_K − 273.15 and t_F = (9/5)t_C + 32. Gas thermometers read T = 273.16 × P/P_tp. For expansion use Δl = lαΔT and treat a hole as though it were made of the sheet material. Always write heat gained = heat lost before substituting calorimetry numbers, and quote Newton's law of cooling with the log form for rate questions.

$Ex 10.1 — Neon and carbon dioxide on the Celsius and Fahrenheit scales$
$Ex 10.2 — Relation between two absolute scales A and B$
$Ex 10.3 — Resistance thermometer: temperature at R = 123.4 \Omega$
$Ex 10.4 — Triplet point, fixed points and the 273.15 connection$
$Ex 10.5 — Ideal gas thermometers with oxygen and hydrogen$
$Ex 10.6 — Actual length of a rod measured by an expanded steel tape$
$Ex 10.7 — Cooling a shaft so that the wheel slips on$
$Ex 10.8 — Change in the diameter of a heated hole$
$Ex 10.9 — Tension in a brass wire on cooling$
$Ex 10.10 — Expansion of a brass-steel composite rod$
$Ex 10.11 — Fractional change in density of glycerine$
$Ex 10.12 — Temperature rise of an aluminium block during drilling$
$Ex 10.13 — Ice melted by a hot copper block$
$Ex 10.14 — Specific heat of a metal by the method of mixtures$
$Ex 10.15 — Why molar specific heats of diatomic gases differ$
$Ex 10.16 — Rate of extra evaporation that brings down a fever$
$Ex 10.17 — Ice remaining in a thermacole icebox after 6 h$
$Ex 10.18 — Flame temperature from a brass boiler$
$Ex 10.19 — Reflection, conduction, pyrometers and steam heating$
$Ex 10.20 — Cooling time by Newton’s law of cooling$

NCERT Exercise 10.1 — Neon and Carbon Dioxide on the Celsius and Fahrenheit Scales

1Exercise question

Step-by-step solution

1Celsius: t_C = T_K − 273.15.
2Neon: t_C = 24.57 − 273.15 = −248.58 °C; t_F = (9/5)(−248.58) + 32 = −415.44 °F.
3Carbon dioxide: t_C = 216.55 − 273.15 = −56.60 °C; t_F = (9/5)(−56.60) + 32 = −69.88 °F.

Final answer

Neon: −248.58 °C, −415.44 °F; carbon dioxide: −56.60 °C, −69.88 °F.

NCERT Exercise 10.2 — Relation Between Two Absolute Scales A and B

1Exercise question

Step-by-step solution

1The triple point is 273.16 K: one unit of A = 273.16/200 K, one unit of B = 273.16/350 K.
2The same temperature reads T_A × (273.16/200) = T_B × (273.16/350).
3T_A = (200/350) T_B = (4/7) T_B.

Final answer

T_A = (4/7) T_B, i.e. T_A : T_B = 4 : 7.

NCERT Exercise 10.3 — Resistance Thermometer: Temperature at R = 123.4 Ω

1Exercise question

Step-by-step solution

1From the two calibration points, 165.5 = 101.6[1 + α(600.5 − 273.16)].
21.629 = 1 + α × 327.34, so α = 0.629/327.34 = 1.92 × 10⁻³ K⁻¹.
3For R = 123.4 Ω: 123.4 = 101.6[1 + 1.92 × 10⁻³(T − 273.16)].
4T − 273.16 = 0.214/1.92 × 10⁻³ = 111.5 K.
5T = 384.7 K ≈ 385 K.

Final answer

The temperature is about 385 K.

NCERT Exercise 10.4 — Triple Point, Fixed Points and the 273.15 Connection

1Exercise question

Step-by-step solution

1(a) The triple point (273.16 K) occurs at one unique combination of temperature and pressure, so it is reproducible every time. The ice and steam points vary with atmospheric pressure and dissolved impurities, so they are not unique fixed points.
2(b) The other fixed point of the Kelvin scale is absolute zero, 0 K.
3(c) 273.16 K is the triple point of water, while 0 °C is the melting point of ice, which lies at 273.15 K on the Kelvin scale; that is why t = T − 273.15.
4(d) A Fahrenheit-sized degree is 5/9 of a kelvin, so the triple point reads 273.16 × (9/5) = 491.7 on such a scale.

Final answer

(a) the triple point is unique, ice/steam points vary with pressure (b) absolute zero, 0 K (c) 0 °C is at 273.15 K, the melting point, not the triple point (d) 491.7.

NCERT Exercise 10.5 — Ideal Gas Thermometers With Oxygen and Hydrogen

1Exercise question

Step-by-step solution

1Charles' law (constant volume): T = 273.16 × P/P_tp.
2Thermometer A: T = 273.16 × 1.797 × 10⁵/1.250 × 10⁵ = 392.69 K.
3Thermometer B: T = 273.16 × 0.287 × 10⁵/0.200 × 10⁵ = 391.98 K.
4(b) The gases are not perfect ideal gases, so their different molecular interactions give slightly different readings.
5To reduce the discrepancy, take readings at progressively lower pressures and extrapolate to zero pressure, where every gas behaves ideally.

Final answer

(a) A reads 392.69 K, B reads 391.98 K (b) non-ideal gas behaviour; use low pressures and extrapolate to zero pressure.

NCERT Exercise 10.6 — Actual Length of a Rod Measured by an Expanded Steel Tape

1Exercise question

Step-by-step solution

1At 45 °C the tape has stretched: l' = 100 cm × (1 + 1.20 × 10⁻⁵ × 18) = 100.0216 cm.
2Each apparent 100 cm of reading now equals 100.0216 real cm, so the 63.0 cm reading corresponds to 63.0 × 1.000216 = 63.0136 cm actual.
3At 27 °C both tape and rod are at the calibration temperature, so the true length is the recorded 63.0 cm (neglecting the tiny contraction of the rod itself).

Final answer

Actual length at 45 °C ≈ 63.014 cm; length at 27 °C = 63.0 cm.

NCERT Exercise 10.7 — Cooling a Shaft So That the Wheel Slips On

1Exercise question

Step-by-step solution

1The shaft must shrink by Δd = 8.69 − 8.70 = −0.01 cm = −1 × 10⁻⁴ m.
2Δd = d₁α(T₁ − T): −1 × 10⁻⁴ = 0.087 × 1.20 × 10⁻⁵ × (T₁ − 300).
3T₁ − 300 = −95.8 K, so T₁ = 300 − 95.8 = 204.2 K = −69 °C.

Final answer

The wheel slips on when the shaft cools to about −69 °C.

NCERT Exercise 10.8 — Change in the Diameter of a Heated Hole

1Exercise question

Step-by-step solution

1The hole expands exactly as a disc of the same material would, so Δd = dαΔT.
2Δd = 4.24 cm × 1.70 × 10⁻⁵ × 200.
3Δd = 4.24 × 3.4 × 10⁻³ = 0.0144 cm ≈ 1.44 × 10⁻² cm.
4The hole diameter increases by 0.0144 cm.

Final answer

The hole diameter increases by about 0.0144 cm (1.44 × 10⁻² cm).

NCERT Exercise 10.9 — Tension in a Brass Wire on Cooling

1Exercise question

Step-by-step solution

1Free contraction sought: ΔL/L = α(T₂ − T₁) = 2.0 × 10⁻⁵ × (−39 − 27) = −1.32 × 10⁻³.
2Y = stress/strain: F/A = Y |ΔL/L|.
3A = π(1.0 × 10⁻³)² = 3.14 × 10⁻⁶ m².
4F = 0.91 × 10¹¹ × 1.32 × 10⁻³ × 3.14 × 10⁻⁶.
5F = 3.8 × 10² N (directed inward as a pull on the supports).

Final answer

A tension of 3.8 × 10² N develops in the wire.

NCERT Exercise 10.10 — Expansion of a Brass-Steel Composite Rod

1Exercise question

Step-by-step solution

1ΔT = 250 − 40 = 210 K.
2Brass: Δl₁ = 50 × 2.0 × 10⁻⁵ × 210 = 0.21 cm.
3Steel: Δl₂ = 50 × 1.2 × 10⁻⁵ × 210 = 0.126 cm.
4Total Δl = 0.21 + 0.126 = 0.346 cm ≈ 3.5 × 10⁻³ m.
5Both ends are free to expand, so no stress develops at the junction.

Final answer

Combined change in length ≈ 0.35 cm; no thermal stress since the ends are free.

NCERT Exercise 10.11 — Fractional Change in Density of Glycerine

1Exercise question

Step-by-step solution

1For a fixed mass, density falls as volume grows: Δρ/ρ ≈ −γΔT.
2Fractional change = 49 × 10⁻⁵ × 30 = 1.47 × 10⁻².
3Density decreases by a fraction 1.47 × 10⁻² = 1.47%.

Final answer

Density decreases by a fraction 1.47 × 10⁻² (1.47%).

NCERT Exercise 10.12 — Temperature Rise of an Aluminium Block During Drilling

1Exercise question

Step-by-step solution

1Total energy = Pt = 10 × 10³ × 150 = 1.5 × 10⁶ J.
2Heat reaching the block = 50% = 7.5 × 10⁵ J.
3ΔT = Q/mc = 7.5 × 10⁵/(8.0 × 10³ × 0.91).
4ΔT = 103 °C.

Final answer

The block temperature rises by 103 °C.

NCERT Exercise 10.13 — Ice Melted by a Hot Copper Block

1Exercise question

Step-by-step solution

1Heat released by copper cooling to 0 °C: Q = mcΔT = 2500 × 0.39 × 500 = 4.875 × 10⁵ J.
2Ice melted: m_ice = Q/L = 4.875 × 10⁵/335.
3m_ice = 1455 g = 1.455 kg ≈ 1.45 kg.

Final answer

About 1.45 kg of ice can melt.

NCERT Exercise 10.14 — Specific Heat of a Metal by the Method of Mixtures

1Exercise question

Step-by-step solution

1Heat lost by metal: Q₁ = 0.20 × c × (150 − 40) = 22c J (c in J kg⁻¹ K⁻¹).
2Heat gained by calorimeter + water (water equivalent 0.150 + 0.025 = 0.175 kg): Q₂ = 0.175 × 4186 × (40 − 27).
3Q₂ = 0.175 × 4186 × 13 = 9523 J.
4Q₁ = Q₂ gives c = 9523/22 = 433 J kg⁻¹ K⁻¹ ≈ 0.43 × 10³ J kg⁻¹ K⁻¹.
5If heat is lost to the surroundings, the gain used above is less than the metal's true loss, so the computed c is smaller than the actual value.

Final answer

c ≈ 0.43 × 10³ J kg⁻¹ K⁻¹; with heat losses the computed value is smaller than the actual one.

NCERT Exercise 10.15 — Why Molar Specific Heats of Diatomic Gases Differ

1Exercise question

Step-by-step solution

1These gases are diatomic: besides translation, energy must feed rotational (and eventually vibrational) modes.
2With rotation active: C_v = 5/2 R = 5/2 × 1.98 ≈ 4.95 cal mol⁻¹ K⁻¹, matching the observations.
3Chlorine has the largest value (6.17) because its low vibrational frequency lets vibrational modes excite at room temperature as well.

Final answer

Diatomic gases store extra energy in rotational (≈5/2 R) and for chlorine also vibrational modes; chlorine's large value shows its vibrations are active at room temperature.

NCERT Exercise 10.16 — Rate of Extra Evaporation That Brings Down a Fever

1Exercise question

Step-by-step solution

1Temperature drop = 3 °F = 3 × 5/9 = 1.67 °C.
2Heat lost = mcΔT = 30 × 10³ × 1 × (5/3) = 5 × 10⁴ cal (c = 1 cal g⁻¹ °C⁻¹).
3Extra sweat evaporated = 5 × 10⁴/580 = 86.2 g in 20 min.
4Average rate = 86.2/20 = 4.3 g min⁻¹.

Final answer

The average extra evaporation rate is 4.3 g min⁻¹.

NCERT Exercise 10.17 — Ice Remaining in a Thermacole Icebox After 6 h

1Exercise question

Step-by-step solution

1Surface area of the cube: A = 6 × (0.3)² = 0.54 m².
2Heat leaking in: H = kA(45 − 0)/d = 0.01 × 0.54 × 45/0.05 = 4.86 W.
3Heat in 6 h: Q = 4.86 × 6 × 3600 = 1.05 × 10⁵ J.
4Ice melted: m = Q/L = 1.05 × 10⁵/335 × 10³ = 0.313 kg.
5Ice remaining = 4.0 − 0.313 = 3.69 kg ≈ 3.7 kg.

Final answer

About 3.7 kg of ice remains after 6 hours.

NCERT Exercise 10.18 — Flame Temperature From a Brass Boiler

1Exercise question

Step-by-step solution

1Heat needed to boil 6 kg in 60 s: Q = mL = 6 × 2256 × 10³ = 1.354 × 10⁷ J.
2Rate of heat flow through the base: H = KA(T₁ − T₂)/d = 109 × 0.15 × (T₁ − 100)/0.01.
3H = Q/t = 1.354 × 10⁷/60 = 2.256 × 10⁵ W.
4T₁ − 100 = 2.256 × 10⁵ × 0.01/(109 × 0.15) = 138 K.
5T₁ = 238 °C.

Final answer

The flame part in contact with the boiler is at about 238 °C.

NCERT Exercise 10.19 — Reflection, Conduction, Pyrometers and Steam Heating

1Exercise question

Step-by-step solution

1(a) A good reflector absorbs little radiation; by Kirchhoff's law a poor absorber is also a poor emitter.
2(b) Brass conducts heat well, so it draws heat rapidly from the hand and feels cold; wood conducts poorly, taking hardly any heat.
3(c) In the open the red-hot iron radiates far less than a black body at the same temperature (emissivity < 1), so the black-body-calibrated pyrometer reads low; in a furnace the walls surround the piece, and the iron approaches black-body radiation, giving the right value.
4(d) The atmosphere traps outgoing infrared (greenhouse effect); without it nearly all heat would radiate back to space, and the surface would become very cold.
5(e) Steam carries surplus heat as latent heat of vaporisation (~540 cal g⁻¹), releasing far more energy per unit mass on condensing than hot water gives by cooling.

Final answer

(a) reflectors absorb (and so emit) little (b) brass conducts heat from the hand, wood does not (c) open-air iron is a poor radiator, furnace iron approaches a black body (d) the atmosphere traps heat (e) steam releases latent heat on condensing.

NCERT Exercise 10.20 — Cooling Time by Newton's Law of Cooling

1Exercise question

Step-by-step solution

1Newton's law of cooling: ln[(T₁ − T₀)/(T₂ − T₀)] = k t.
2First interval: ln[(80 − 20)/(50 − 20)] = ln 2 = k × 5.
3Second interval: ln[(60 − 20)/(30 − 20)] = ln 4 = k × t.
4t = 5 × (ln 4/ln 2) = 5 × 2 = 10 min.

Final answer

It takes 10 minutes to cool from 60 °C to 30 °C.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Linear expansion

Δ L = α L Δ T

Specific heat

Q = m c Δ T

Calorimetry (no loss)

m_{1} c_{1} (T_{1} - T) = m_{2} c_{2} (T - T_{2})

Newton's law of cooling

\frac{d Q}{d t} \propto (T_{amb} - T)

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

In calorimetry, heat lost by the hotter body equals heat gained by the cooler one.
Apparent expansion of a liquid is an artefact of the container expanding too — always subtract the container's own expansion.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Physics Chapter 10 (Thermal Properties of Matter)?

There are 20 exercise questions in this chapter, numbered 10.1 to 10.20. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Thermal Properties of Matter Class 11 Physics?

The formulas this chapter's questions actually turn on are: Linear expansion, Specific heat, Calorimetry (no loss), Newton's law of cooling. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Thermal Properties of Matter important for JEE Main and NEET?

Steady rather than spectacular — calorimetry and expansion numericals are frequent one- and two-mark board questions and appear in NEET almost every year.

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