Class 11 Physics Chapter 10: Thermal Properties of Matter | NCERT Notes & Solutions

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Learn the chapter

Study the ideas and derivations, with worked examples beside the topics they use.

10.1–10.2

Heat, temperature and equilibrium

Temperature describes thermal state; heat describes energy in transit.

Temperature is an objective measure of hotness. Touch is subjective and limited; a thermometer uses a measurable thermometric property that varies with temperature.
Heat is energy transferred between systems solely because of a temperature difference. It flows spontaneously from higher to lower temperature; its SI unit is joule (J). Temperature is measured in kelvin (K).
When system and surroundings reach the same temperature, net heat transfer ceases. Heat is not a substance stored inside a body; internal energy is a state property.

10.3

Measuring temperature and temperature scales

A scale assigns values using reproducible fixed points and equal intervals.

Liquid-in-glass thermometers use volume expansion; gas thermometers use pressure at fixed volume. Different thermometric properties may disagree between calibration points.
At standard pressure, ice and steam points are 0 °C and 100 °C, or 32 °F and 212 °F. Linear interpolation gives (t_F−32)/180 = t_C/100.
Thus t_F=(9/5)t_C+32 and t_C=(5/9)(t_F−32). Celsius and Fahrenheit differ in both zero and interval size.

For the same temperature, (t_F−32)/180 = t_C/100; hence t_F = (9/5)t_C + 32.

10.4

Ideal-gas equation and absolute temperature

Low-density gases provide a common route to an absolute scale.

For a fixed amount of gas, Boyle’s law gives PV=constant at fixed T; Charles’ law gives V/T=constant at fixed P. Combining them gives PV/T=constant.
For μ moles, PV=μRT, where R=8.31 J mol⁻¹ K⁻¹. At constant volume P∝T; gas-thermometer lines extrapolate toward zero pressure at absolute zero.
T=t_C+273.15 K. Kelvin and Celsius have equal interval sizes, but different origins. Real gases depart from ideal behaviour, especially at low temperature; use dilute gas and extrapolation.

10.5

Linear, area and volume expansion

For small temperature intervals, expansion is proportional to original size and ΔT.

Linear expansivity α is defined by ΔL/L=αΔT, so L=L₀(1+αΔT). Area expansivity β=ΔA/(AΔT); for an isotropic solid β≈2α.
Proof: ΔA=aΔb+bΔa+ΔaΔb; put Δa=αaΔT and Δb=αbΔT. The product term is second order and negligible, leaving ΔA/A≈2αΔT.
Volume expansivity γ=ΔV/(VΔT). For a cube, ΔV=(L+ΔL)³−L³≈3L²ΔL, hence γ≈3α. These relations assume small expansion and isotropic solids.

For small expansions, second-order products such as ΔaΔb are negligible; area expansivity is 2α and volume expansivity is 3α for an isotropic solid.

NCERT Example 10.1 · Proof

Area expansivity of a rectangular sheet

Show that the area-expansion coefficient of a solid sheet is twice its linear expansivity.

Step 1. Let its sides be a and b. A rise ΔT gives Δa=αaΔT and Δb=αbΔT.

Step 2. The area change is ΔA=aΔb+bΔa+ΔaΔb=αabΔT(2+αΔT).

Step 3. Since αΔT≪1, neglect the second-order term: ΔA/A≈2αΔT.

Result: β=2α for small, isotropic expansion.

NCERT Example 10.2 · Solved

Heating an iron ring to fit a wheel

At 27 °C, an iron ring is 5.231 m across and a wheel rim is 5.243 m. Find the ring temperature required; use α=1.20×10⁻⁵ K⁻¹.

Step 1. Use D₂=D₁[1+α(T₂−T₁)] because every length of the free ring expands by the same fractional amount.

Step 2. T₂=27+[5.243/5.231−1]/(1.20×10⁻⁵)=218.4 °C.

Step 3. A small expansion is enough because the diameter deficit is only 0.012 m.

Result: About 218 °C, as in the NCERT rounded result.

10.5

Liquids, gases and constrained expansion

Expansion depends on state, temperature and mechanical constraints.

Liquids have volume expansivity γ. A liquid in a vessel appears to expand by γ_app=γ_liquid−γ_vessel; the vessel expands too.
Water is anomalous between 0 °C and 4 °C: it contracts when warmed, has maximum density at 4 °C, and expands on cooling below 4 °C. Lakes freeze at the surface, helping aquatic life survive below.
For an ideal gas at constant pressure, V∝T, so γ=1/T. If a rod is fully constrained, total strain is zero; mechanical strain cancels αΔT and |σ|=Yα|ΔT|. Heating gives compression and cooling gives tension; expansion joints prevent damaging stress.

Worked check · Thermal stress

Fully restrained thermal strain

A steel rail cannot expand. Find the compressive thermal stress for α=1.2×10⁻⁵ K⁻¹, Y=2×10¹¹ Pa and a 10 K rise.

Step 1. The free thermal strain would be αΔT=1.2×10⁻⁴.

Step 2. Rigid supports impose an equal-and-opposite elastic strain.

Step 3. The stress magnitude is YαΔT=(2×10¹¹)(1.2×10⁻⁴).

Result: 2.4×10⁷ Pa compression. (This is the chapter’s rail illustration, solved explicitly.)

10.6

Heat capacity and specific heat

Temperature change depends on mass, material and the amount of heat transferred.

Heat capacity S=ΔQ/ΔT is for the whole body (J K⁻¹). Specific heat s=ΔQ/(mΔT) is per unit mass (J kg⁻¹ K⁻¹); hence ΔQ=msΔT when no phase change occurs.
Molar heat capacity C=ΔQ/(μΔT), in J mol⁻¹ K⁻¹. For a gas, specify constant pressure C_p or constant volume C_v; generally C_p>C_v.
Specific heats depend on material and temperature. Water’s large s≈4186 J kg⁻¹ K⁻¹ makes it useful as a coolant and moderator of day–night and land–sea temperature changes.

10.7

Calorimetry: heat balance

An isolated mixture conserves energy: heat lost equals heat gained.

A calorimeter limits heat exchange with surroundings. In the ideal isolated model, ΣQ=0; hot components lose energy and colder components gain it until one common final temperature is reached.
For sensible heating/cooling use Q=ms(T_f−T_i) with sign. Include the calorimeter’s own heat capacity or water equivalent.
For a phase change include latent heat too. Check that the predicted final state is physically possible before solving; account for environmental losses if the apparatus is not well insulated.

NCERT Example 10.3 · Solved

Specific heat of aluminium by calorimetry

A 0.047 kg aluminium sphere at 100 °C is placed in 0.25 kg water and a 0.14 kg copper calorimeter, both at 20 °C. Final temperature: 23 °C. Find sAl.

Step 1. Heat lost by Al = 0.047sAl(100−23).

Step 2. Heat gained = [0.25(4180)+0.14(386)](23−20)=3,297 J.

Step 3. Set lost equal to gained: sAl=3297/[0.047×77]=911 J kg⁻¹ K⁻¹.

Result: sAl≈0.911 kJ kg⁻¹ K⁻¹, consistent with the NCERT worked result.

10.8

Change of state and latent heat

During a phase transition, supplied energy changes arrangement rather than temperature.

Fusion is solid→liquid and freezing is liquid→solid. Vaporisation is liquid→gas and condensation is gas→liquid. During an equilibrium transition, both phases coexist and temperature remains constant at fixed pressure.
Latent heat per unit mass is L=Q/m (J kg⁻¹); Q=mL. Use Lf for fusion and Lv for vaporisation. Heating a sample across several stages requires summing sensible-heat and latent-heat intervals.
Evaporation occurs at the surface below boiling and cools the remaining liquid; boiling is bulk vapour formation at the pressure-dependent boiling point. The triple point is where solid, liquid and vapour coexist; a phase diagram shows the coexistence curves.

At each plateau, energy changes the phase (latent heat); the temperature stays fixed while the two phases coexist at the specified pressure.

NCERT Example 10.4 · Solved

Finding latent heat of fusion

Mix 0.15 kg ice at 0 °C with 0.30 kg water at 50 °C; final temperature is 6.7 °C. Find Lf; sw=4186 J kg⁻¹ K⁻¹.

Step 1. Hot-water heat loss: 0.30×4186×(50−6.7)=54,376 J.

Step 2. Ice first melts, then the melt warms: 0.15Lf+0.15×4186×6.7.

Step 3. Equate gain and loss: Lf=(54,376−4,207)/0.15.

Result: Lf≈3.34×10⁵ J kg⁻¹.

NCERT Example 10.5 · Solved

Ice at −12 °C to steam at 100 °C

Find the heat to turn 3 kg ice at −12 °C into steam at 100 °C. Use sice=2100, sw=4186 J kg⁻¹ K⁻¹, Lf=3.35×10⁵ and Lv=2.256×10⁶ J kg⁻¹.

Step 1. Warm ice to 0 °C: Q₁=3×2100×12=75,600 J.

Step 2. Melt it: Q₂=3×3.35×10⁵=1,005,000 J; warm water to 100 °C: Q₃=3×4186×100=1,255,800 J.

Step 3. Vaporise at 100 °C: Q₄=3×2.256×10⁶=6,768,000 J. Add all four stages.

Result: Q=9.1044×10⁶ J≈9.1 MJ.

10.8

Pressure, regelation and phase diagrams

Phase boundaries shift with pressure; water is a notable exception.

Melting and boiling temperatures depend on pressure. Increased pressure lowers ice’s melting point because liquid water is denser than ice; a weighted wire can melt ice beneath it and water refreezes above it (regelation).
A P–T phase diagram separates solid, liquid and vapour regions. Sublimation, fusion and vaporisation curves mark coexistence of two phases; their intersection is the triple point.
Water’s triple point is 273.16 K at about 611 Pa. Its fusion line has negative slope; the fusion line of CO₂ and most substances slopes positively. At ordinary atmospheric pressure CO₂ sublimes instead of forming a liquid.

The fusion curve for water has negative slope because ice is less dense than liquid water; most substances and solid CO₂ have a positive slope.

10.9

Heat transfer: conduction and convection

Heat can move by neighbouring interactions, bulk flow or electromagnetic radiation.

Conduction transfers energy between neighbouring parts without bulk matter flow. For a uniform insulated bar at steady state, H=K A(T_hot−T_cold)/L; K is thermal conductivity (W m⁻¹ K⁻¹).
For layers in series the heat current is common and thermal resistances add: R_th=L/(KA), H=ΔT/ΣR_th. Metals conduct well; trapped air and foams insulate.
Convection requires moving fluid. Natural convection uses buoyancy (warm, less-dense fluid rises); forced convection uses a pump or fan. Land/sea breezes reverse from day to night; Earth’s rotation shapes large-scale convection into trade-wind circulation.

NCERT Example 10.6 · Solved

Steel–copper junction temperature

Steel (L=0.15 m, K=50.2) and copper (L=0.10 m, K=385 W m⁻¹ K⁻¹) are joined in series. The steel area is 2A and copper area A; ends are at 300 °C and 0 °C. Find junction temperature.

Step 1. At steady state, heat currents are equal: 2(50.2)(300−T)/0.15 = 385T/0.10.

Step 2. Cancel the common area and solve 669.3(300−T)=3850T.

Step 3. The same current passes every cross-section; side losses are neglected.

Result: T≈44.4 °C, consistent with the NCERT worked result.

NCERT Example 10.7 · Solved

Two rods in series: junction and equivalent K

Equal 0.10 m rods of iron (K₁=79) and brass (K₂=109 W m⁻¹ K⁻¹), each area 0.02 m², have ends at 373 K and 273 K. Find junction temperature, equivalent conductivity and heat current.

Step 1. Equal steady heat currents give K₁(T₁−T₀)=K₂(T₀−T₂), so T₀=(K₁T₁+K₂T₂)/(K₁+K₂).

Step 2. T₀=(79×373+109×273)/188=315 K; K′=2K₁K₂/(K₁+K₂)=91.6 W m⁻¹ K⁻¹.

Step 3. H=K′A(T₁−T₂)/(2L)=91.6×0.02×100/0.20.

Result: T₀=315 K; K′=91.6 W m⁻¹ K⁻¹; H≈916 W.

10.9

Radiation and blackbody laws

Thermal radiation crosses a vacuum; its spectrum and intensity reveal temperature.

Thermal radiation is electromagnetic energy emitted because of temperature; unlike conduction or convection it needs no material medium. Dark, matte surfaces absorb and emit more effectively than shiny reflective ones.
A blackbody absorbs all incident radiation and is an ideal emitter. Its continuous spectrum is universal at a given temperature. Wien’s displacement law gives λ_maxT=2.9×10⁻³ m K: hotter blackbodies peak at shorter wavelength.
Stefan–Boltzmann law: P=eσAT⁴ (σ=5.67×10⁻⁸ W m⁻² K⁻⁴; e=1 for a blackbody). In surroundings at T_s, net radiative loss is eσA(T⁴−T_s⁴). A thermos uses silvered walls and a vacuum gap to reduce radiation and conduction/convection.

As T rises, the peak shifts to shorter wavelength (λ_maxT = 2.9×10⁻³ m K) and total emitted power rises sharply (∝T⁴).

10.10

Newton’s law of cooling

For small excess temperatures, heat-loss rate is proportional to the body’s excess temperature.

For body temperature T and surroundings T_s, −dQ/dt=k(T−T_s) for a small temperature difference. k depends on exposed area and surface; multiple transfer modes may contribute.
If the body has mass m and specific heat s, dQ=ms dT. Thus d(T−T_s)/dt=−K(T−T_s), where K=k/(ms).
Integrating gives ln(T−T_s)=−Kt+C, or T=T_s+(T₀−T_s)e⁻ᴷᵗ. It is an approximation for small excess temperature; a plot of ln(T−T_s) against time is straight.

The straight-line test is ln(T−Tₛ) versus time; its negative slope is −k/(ms) in the heat-loss formulation.

NCERT Example 10.8 · Solved

Cooling time from Newton’s law

A pan cools 94→86 °C in 2 min in a 20 °C room. Estimate the time for 71→69 °C.

Step 1. For the first interval, mean excess temperature is 90−20=70 °C and cooling rate is 8/2=4 °C min⁻¹.

Step 2. For the second, mean excess is 70−20=50 °C; rate scales to 4×50/70=2.86 °C min⁻¹.

Step 3. Time for a 2 °C drop is 2/2.86=0.70 min.

Result: About 0.70 min = 42 s (small-excess, mean-temperature approximation).

MODEL

Interactive thermal models

Use the controls to see how temperature, pressure and heat transfer change in each model.

1 · A constrained metal bar expands

Raise ΔT and compare free expansion with a fixed end.

Temperature riseΔT = 0 °C

Use ΔL/L=αΔT; the displayed displacement is exaggerated.

2 · Heating curve and latent heat

Move through warming, melting, liquid heating and boiling.

Heat supplied along the stages0%

During each flat phase-change stage, temperature stays constant.

3 · Three routes for heat transfer

Choose a mode; play to see the idealised energy path.

Energy passes between neighbours without bulk material flow.

4 · Exponential cooling

Adjust elapsed time and room temperature in the Newton model.

Elapsed timeT ≈ 100 °CRoom temperature20 °C

Temperature difference decays exponentially.

The models illustrate the relationships described by the equations.

CHECK

Points to keep in mind

Kelvin’s zero and its unit interval are defined by the scale; 273.16 K is water’s triple point, while 0 °C at 1 atm is 273.15 K.
Coexisting phases have the same equilibrium temperature and pressure, but different molar volumes and densities.
Heat transfer requires a temperature difference; energy transfer without that cause is not heat.
A hot bar under running water loses heat to the water by conduction at the interface; convection is the subsequent bulk motion within the fluid.

DATA

Useful material data

Reference values at ordinary temperatures; exact values vary with temperature, purity and composition.

Open NCERT property tables

Linear expansion coefficients α (10⁻⁵ K⁻¹)
Material	α	Material	α
Aluminium	2.5	Brass	1.8
Iron	1.2	Copper	1.7
Silver	1.9	Gold	1.4
Pyrex glass	0.32	Lead	2.9

Specific heat s (J kg⁻¹ K⁻¹)
Material	s	Material	s
Aluminium	900	Water	4186
Ice	2060	Copper	386
Iron	450	Glass	840
Lead	128	Mercury	140
Silver	236	Carbon	507
Kerosene	2118	Edible oil	1965

Thermal conductivity K (W m⁻¹ K⁻¹)
Material	K	Material	K
Silver	406	Copper	385
Aluminium	205	Brass	109
Steel	50.2	Mercury	8.3
Glass / concrete	0.8	Water	0.8
Wood	0.12	Glass wool / felt	0.04
Air	0.024	Hydrogen	0.14

Water’s phase-change data at 1 atm
Transition	Temperature	Latent heat
Fusion / freezing	0 °C	Lf≈3.33×10⁵ J kg⁻¹
Vaporisation / condensation	100 °C	Lv≈2.26×10⁶ J kg⁻¹

QUIZ

Concept quiz

Choose an answer to see the explanation and compare the options.

0 / 36 AnsweredCorrect: 0 (0%)

10.1–10.2Heat, temperature and equilibrium

1. Which statement defines heat?

Heat is energy in transit driven by a temperature difference.

Recall: Heat is transfer; internal energy is stored in the system.

2. At thermal equilibrium between two bodies, the net heat flow is:

Equal temperatures remove the temperature difference that drives net heat transfer.

Recall: Thermal equilibrium means equal temperature.

3. The SI units of heat and temperature are respectively:

Heat is measured in joules and thermodynamic temperature in kelvin.

Recall: Use J for energy and K for temperature.

10.3Measuring temperature and temperature scales

1. A temperature of 20 °C equals:

tF=(9/5)20+32=68 °F.

Recall: Convert Celsius by multiplying by 9/5, then add 32.

2. Which thermometric property is commonly used in a liquid-in-glass thermometer?

The liquid volume changes with temperature and moves the column.

Recall: Thermometers infer temperature from a changing physical property.

3. Why is the triple point useful as a standard fixed point?

The specified three-phase equilibrium state is reproducible and unique.

Recall: Fixed points must be reproducible.

10.4Ideal-gas equation and absolute temperature

1. At constant volume, an ideal gas has:

From PV=μRT at fixed V, P/T=μR/V is constant.

Recall: At fixed volume, gas pressure is proportional to kelvin temperature.

2. The ideal-gas equation is:

For μ moles, PV=μRT.

Recall: Temperature in this equation must be absolute.

3. The Kelvin and Celsius scales have:

Their intervals are equal; T=tC+273.15.

Recall: Kelvin shifts the origin, not the interval size.

10.5Linear, area and volume expansion

1. A 2 m steel rod with α=1.2×10⁻⁵ K⁻¹ warms by 10 K. Its extension is:

ΔL=αLΔT=2.4×10⁻⁴ m=0.24 mm.

Recall: Check units after using ΔL=αLΔT.

2. For a small temperature rise, area expansivity β of an isotropic solid is approximately:

Two in-plane dimensions each expand by αΔT, giving β≈2α.

Recall: Area expansion has two linear directions.

3. A circular hole in a heated sheet:

Every linear dimension of the sheet, including the boundary of the hole, scales by the same factor.

Recall: A hole expands when the surrounding solid expands.

10.5Liquids, gases and constrained expansion

1. At what temperature is water densest near ordinary pressure?

Water contracts on warming from 0 to 4 °C and has maximum density at 4 °C.

Recall: Water’s expansion is anomalous below 4 °C.

2. For an ideal gas at constant pressure, the volume expansivity is:

V∝T, so (1/V)(dV/dT)=1/T.

Recall: The gas expansivity depends on absolute temperature.

3. The magnitude of stress in a fully restrained rod heated by ΔT is:

The imposed elastic strain cancels free thermal strain, so stress magnitude is YαΔT.

Recall: Constraint converts expansion strain into stress.

10.6Heat capacity and specific heat

1. Specific heat capacity s is:

s=ΔQ/(mΔT).

Recall: Distinguish specific heat from whole-body heat capacity.

2. Why do we specify Cp or Cv for a gas?

The work and resulting temperature rise depend on the constraint.

Recall: Gas molar heat capacity needs a process condition.

3. For heating with no phase change, the required heat is:

The definition of specific heat gives Q=msΔT.

Recall: Use sensible heat within a single phase.

10.7Calorimetry: heat balance

1. In an ideal isolated calorimetry experiment:

Energy conservation gives ΣQ=0 when the system is isolated.

Recall: Include the calorimeter in the energy balance.

2. A calorimeter’s water equivalent represents:

It is the mass of water with the same heat capacity.

Recall: Add the equivalent water mass to the sample water mass.

3. If the final mixture temperature is below the melting point of ice, ice originally at that melting point will:

There is no energy available to melt all the ice and reach a lower final temperature.

Recall: Check phase assumptions against the energy balance.

10.8Change of state and latent heat

1. During melting at fixed pressure, the temperature of a pure substance:

Added energy supplies latent heat while solid and liquid coexist.

Recall: A phase plateau stores energy without raising temperature.

2. Latent heat of fusion per unit mass has SI unit:

Lf=Q/m.

Recall: Latent heat is energy per mass.

3. Evaporation differs from boiling because evaporation:

Surface molecules can escape below the boiling temperature, cooling the liquid.

Recall: Evaporation can happen at many temperatures.

10.8Pressure, regelation and phase diagrams

1. At the triple point, which phases coexist?

All three phase-boundary curves meet at the triple point.

Recall: Triple point means three phases in equilibrium.

2. Why does water’s fusion curve have negative slope?

Pressure favours the denser liquid phase, lowering the melting temperature.

Recall: Ice expands on melting because it is less dense.

3. At ordinary atmospheric pressure, dry ice (solid CO₂) usually:

The 1 atm path does not cross the liquid region of the CO₂ phase diagram.

Recall: Sublimation is solid-to-vapour transition.

10.9Heat transfer: conduction and convection

1. Which heat-transfer mode can occur through a vacuum?

Electromagnetic radiation needs no material medium.

Recall: Sunlight warms Earth through space by radiation.

2. For an insulated bar at steady state, the heat current is proportional to:

Fourier’s steady one-dimensional relation is H=KAΔT/L.

Recall: Longer bars resist conduction; larger area/conductivity increase it.

3. Natural convection is mainly driven by:

Heating changes fluid density; buoyant warm fluid rises and cooler fluid sinks.

Recall: Convection transports heat with moving fluid.

10.9Radiation and blackbody laws

1. Wien’s displacement law predicts that when temperature rises, λmax:

λmaxT is constant, so a higher T means a shorter peak wavelength.

Recall: Hotter stars look bluer because the peak shifts shorter.

2. A perfect blackbody has emissivity:

A blackbody is the ideal emitter and absorber, e=1.

Recall: Emissivity is a fraction from 0 to 1.

3. The net radiation from a body in surroundings at Ts is proportional to:

Net power is eσA(T⁴−Ts⁴).

Recall: Temperatures in radiation laws are in kelvin.

10.10Newton’s law of cooling

1. Newton’s law of cooling is a good approximation when:

The linear rate law assumes a small temperature excess.

Recall: Do not extend the linear cooling law to arbitrary large differences.

2. For Newton cooling, ln(T−Ts) plotted against time is:

T−Ts=(T₀−Ts)e⁻ᴷᵗ; taking logs gives slope −K.

Recall: Exponential excess temperature becomes linear on a log plot.

3. A body’s temperature approaches ambient under Newton cooling:

The excess decays exponentially and tends to zero asymptotically.

Recall: The cooling rate slows as the excess falls.

NCERT · EXERCISE 10

Chapter exercises

Try each question before opening its step-by-step solution.

Ex 10.1NCERT

10.1The 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.

Related concept: Review the matching notes

Step 1. Use tC=T−273.15 and tF=(9/5)tC+32.

Step 2. Neon: −248.58 °C, then −415.44 °F.

Step 3. CO₂: −56.60 °C, then −69.88 °F.

Answer: Ne: −248.58 °C (−415.44 °F); CO₂: −56.60 °C (−69.88 °F).

Ex 10.2NCERT

10.2Two 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?

Related concept: Review the matching notes

Step 1. Absolute scales have the same zero; their readings are proportional.

Step 2. At water’s triple point, TA/TB=200/350=4/7.

Answer: TA=(4/7)TB (equivalently TB=(7/4)TA).

Ex 10.3NCERT

10.3The electrical resistance in ohms of a certain thermometer varies with temperature according to 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?

Related concept: Review the matching notes

Step 1. Linear interpolation in resistance gives T=273.16+(123.4−101.6)(600.5−273.16)/(165.5−101.6).

Step 2. Substitute the resistance differences and evaluate.

Answer: T≈384.8 K.

Ex 10.4NCERT

10.4Answer 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?

Related concept: Review the matching notes

Step 1. (a) The triple point is a unique state; ice/steam points shift with pressure and impurities.

Step 2. (b) The other defining reference is absolute zero, 0 K.

Step 3. (c) 273.16 K is the triple point, 0.01 K above water’s melting point at 1 atm, which defines 0 °C; hence the Celsius offset is 273.15.

Step 4. (d) A Fahrenheit-sized interval is 5/9 K, so one such absolute degree is 9/5 K.

Answer: (d) 273.16×9/5=491.688 degrees on that absolute scale.

Ex 10.5NCERT

10.5Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made: at the triple-point of water the pressures are 1.250×10⁵ Pa (A) and 0.200×10⁵ Pa (B); at the normal melting point of sulphur they are 1.797×10⁵ Pa (A) and 0.287×10⁵ Pa (B). (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 (the thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy?

Related concept: Review the matching notes

Step 1. At fixed volume, T/Ttp=P/Ptp. For A, T=273.16×1.797/1.250.

Step 2. For B, T=273.16×0.287/0.200.

Step 3. Real gases deviate slightly at finite pressure; repeat at lower gas densities and extrapolate to zero pressure.

Answer: A: about 392.6 K; B: about 392.4 K. Zero-pressure extrapolation reduces the difference.

Ex 10.6NCERT

10.6A steel tape 1 m 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.

Related concept: Review the matching notes

Step 1. At 45 °C one tape metre is 1[1+α(18)]=1.000216 m, so the indicated 0.630 m represents 0.630136 m.

Step 2. The tape and rod are the same material, so both expand by the same factor from 27 to 45 °C.

Step 3. Divide the hot-day physical length by 1.000216 to recover its 27 °C length.

Answer: Hot-day length ≈0.63014 m; length at 27 °C =0.63000 m.

Ex 10.7NCERT

10.7A 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 steel to be constant over the required temperature range: αsteel=1.20×10⁻⁵ K⁻¹.

Related concept: Review the matching notes

Step 1. The shaft must contract to 8.69 cm: 8.69=8.70[1+α(T−27)].

Step 2. Solve T=27+(8.69/8.70−1)/(1.20×10⁻⁵).

Answer: T≈−68.8 °C; cooling to dry-ice temperature is sufficient.

Ex 10.8NCERT

10.8A 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.

Related concept: Review the matching notes

Step 1. A hole expands as if it were filled with the same material: Δd=αdΔT.

Step 2. Δd=1.70×10⁻⁵×4.24 cm×200.

Answer: Δd=0.0144 cm=0.144 mm (increase).

Ex 10.9NCERT

10.9A 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? Coefficient of linear expansion of brass =2.0×10⁻⁵ K⁻¹; Young’s modulus of brass =0.91×10¹¹ Pa.

Related concept: Review the matching notes

Step 1. Cooling by 66 K would cause tensile strain α|ΔT| if free; the supports prevent it.

Step 2. σ=Yα|ΔT|=0.91×10¹¹×2×10⁻⁵×66=1.201×10⁸ Pa.

Step 3. A=π(1.0×10⁻³)² m²; tension F=σA.

Answer: F≈3.77×10² N (about 377 N).

Ex 10.10NCERT

10.10A 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).

Related concept: Review the matching notes

Step 1. Each rod warms by 210 K. ΔLtotal=0.50(2.0×10⁻⁵)(210)+0.50(1.2×10⁻⁵)(210).

Step 2. Because the outer ends are free, each section expands by its own free amount; no constraint requires elastic strain.

Answer: Total increase =3.36×10⁻³ m=3.36 mm; no thermal stress is required at the junction.

Ex 10.11NCERT

10.11The 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?

Related concept: Review the matching notes

Step 1. At fixed mass, ρ= m/V. With V₂/V₁=1+γΔT, Δρ/ρ=1/(1+γΔT)−1.

Step 2. Here γΔT=49×10⁻⁵×30=0.0147.

Answer: Δρ/ρ≈−0.0145 (about a 1.45% decrease; the small-change estimate is −1.47%).

Ex 10.12NCERT

10.12A 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.

Related concept: Review the matching notes

Step 1. Useful heat=0.50×10,000 W×150 s=750,000 J.

Step 2. The block heat capacity is 8000 g×0.91 J g⁻¹ K⁻¹=7280 J K⁻¹.

Answer: ΔT≈103 K.

Ex 10.13NCERT

10.13A 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.)

Related concept: Review the matching notes

Step 1. The block cools to 0 °C and releases Q=2.5×390×500=487,500 J.

Step 2. That heat melts m=Q/Lf=487,500/335,000 kg.

Answer: About 1.46 kg of ice.

Ex 10.14NCERT

10.14In 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?

Related concept: Review the matching notes

Step 1. Water plus calorimeter water-equivalent mass is 0.150+0.025=0.175 kg. Heat gained=0.175×4186×(40−27)=9.52×10³ J.

Step 2. Metal loses 0.20s(150−40)=22s joules. Equate to heat gained.

Step 3. If heat escaped outside, the metal gave more heat than the water/calorimeter received, so ignoring that loss underestimates s.

Answer: s≈433 J kg⁻¹ K⁻¹; with unaccounted heat loss the calculated value is smaller than the actual value.

Ex 10.15NCERT

10.15Given below are observations on molar specific heats at room temperature of some common gases: H₂ 4.87, N₂ 4.97, O₂ 5.02, NO 4.99, CO 5.01, Cl₂ 6.17 cal mol⁻¹ K⁻¹. 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⁻¹ K⁻¹. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Related concept: Review the matching notes

Step 1. Diatomic molecules store thermal energy in rotational as well as translational degrees of freedom; additional vibrational modes can contribute.

Step 2. The larger Cl₂ value indicates that more internal vibrational energy is accessible at room temperature than for the other listed gases.

Answer: Molecular internal degrees of freedom raise molar heat capacity above the monatomic value; Cl₂’s higher value signals greater vibrational contribution.

Ex 10.16NCERT

10.16A 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.

Related concept: Review the matching notes

Step 1. A 3 °F fall is ΔT=(5/9)×3=1.667 K. Heat removed≈30×4186×1.667=2.09×10⁵ J.

Step 2. Latent heat 580 cal g⁻¹≈2.43×10⁶ J kg⁻¹; evaporated mass≈0.086 kg.

Step 3. Divide by 20 min.

Answer: About 4.3 g min⁻¹ of extra evaporation.

Ex 10.17NCERT

10.17A “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]

Related concept: Review the matching notes

Step 1. Approximate area of six outside faces A=6(0.30)²=0.54 m²; inside stays near 0 °C.

Step 2. Conducted heat Q=K A ΔT t/L=0.01×0.54×45×21600/0.05≈1.05×10⁵ J.

Step 3. Melted ice≈Q/Lf≈0.313 kg; subtract from 4.0 kg.

Answer: About 3.69 kg remains (rough estimate; edge losses and temperature changes are neglected).

Ex 10.18NCERT

10.18A 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.

Related concept: Review the matching notes

Step 1. Heat current into boiling water=ṁLv=(6/60)×2.256×10⁶=225,600 W.

Step 2. Conduction through the base gives H=KA(Tflame−100 °C)/L.

Step 3. Solve ΔT=HL/(KA)=225600×0.01/(109×0.15).

Answer: Flame-contact temperature≈238 °C.

Ex 10.19NCERT

10.19Explain 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.

Related concept: Review the matching notes

Step 1. (a) At thermal equilibrium, strong reflection means weak absorption and hence weak emission.

Step 2. (b) Brass conducts heat away from skin much faster than wood, so the skin cools faster.

Step 3. (c) Open iron has emissivity below 1, so a blackbody-calibrated pyrometer under-reads; inside a hot furnace reflected radiation adds and the cavity approaches blackbody conditions.

Step 4. (d) The atmosphere absorbs and re-emits outgoing infrared radiation, warming the surface.

Step 5. (e) Steam releases latent heat when it condenses, in addition to sensible heat.

Answer: Each effect follows from emissivity/absorptivity, conductivity, cavity radiation, atmospheric infrared trapping or latent heat.

Ex 10.20NCERT

10.20A 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.

Related concept: Review the matching notes

Step 1. For the first interval the mean body temperature is 65 °C, so mean excess is 45 °C; rate≈30/5=6 °C min⁻¹.

Step 2. For the second interval mean body temperature is 45 °C, so mean excess is 25 °C. Newton’s small-interval approximation gives rate≈6×25/45=3.33 °C min⁻¹.

Step 3. Time≈30/3.33.

Answer: About 9 min by the textbook mean-excess approximation. Exact exponential integration gives 10 min; Newton’s law is approximate over these finite intervals.

REVISION

Revision

Review the key ideas, formulas and conditions.

10.1–10.2 · Heat, temperature and equilibrium

Review this concept

Temperature measures thermal state; heat is energy crossing a boundary because temperatures differ.
Heat flows hot → cold until thermal equilibrium; SI units are J and K.
A thermometer reads a thermometric property; touch alone is unreliable.

10.3 · Measuring temperature and temperature scales

Review this concept

Use tF=(9/5)tC+32 and tC=(5/9)(tF−32).
A scale needs a thermometric property plus fixed reference points.
The Celsius and Fahrenheit ice/steam points are (0,100) °C and (32,212) °F.

10.4 · Ideal-gas equation and absolute temperature

Review this concept

PV=μRT; Boyle: PV const at fixed T; Charles: V/T const at fixed P.
Kelvin temperature is T=tC+273.15; never put Celsius into the ideal-gas law.
Absolute zero is inferred by dilute-gas extrapolation; real gases do not remain ideal there.

10.5 · Linear, area and volume expansion

Review this concept

ΔL/L=αΔT; ΔA/A≈2αΔT; ΔV/V≈3αΔT for small isotropic solid expansion.
Area proof drops the second-order ΔaΔb term; cube proof drops (ΔL)² and (ΔL)³.
Always state the initial dimension and temperature interval used.

10.5 · Liquids, gases and constrained expansion

Review this concept

Liquid apparent expansion: γapp=γliquid−γcontainer.
Water has maximum density at 4 °C; below 4 °C it expands as it cools.
At fixed pressure ideal-gas γ=1/T; fully restrained rod stress magnitude is Yα|ΔT|.

10.6 · Heat capacity and specific heat

Review this concept

S=ΔQ/ΔT; s=ΔQ/(mΔT); C=ΔQ/(μΔT).
Use Q=msΔT only while the substance remains in one phase.
For gases name the condition: Cp or Cv; water has unusually high specific heat.

10.7 · Calorimetry: heat balance

Review this concept

Isolated system: heat lost + heat gained = 0.
Include water, calorimeter and every object; a water equivalent is its heat capacity divided by sw.
Add latent-heat terms at phase transitions and check the assumed final phase.

10.8 · Change of state and latent heat

Review this concept

At a phase plateau, Q=mL while T is constant and both phases coexist.
Evaporation is surface-only below boiling; boiling occurs throughout at a pressure-dependent temperature.
Triple point: solid + liquid + vapour coexist. Regelation is pressure melting followed by refreezing.

10.8 · Pressure, regelation and phase diagrams

Review this concept

Phase boundaries depend on pressure; ice melts under increased pressure because liquid water is denser.
The three phase-boundary curves meet at the triple point (water: 273.16 K, about 611 Pa).
Solid CO₂ sublimes at 1 atm; water’s fusion curve slopes negatively.

10.9 · Heat transfer: conduction and convection

Review this concept

Conduction: H=KAΔT/L in steady state; no bulk material flow.
Series layers: thermal resistances L/(KA) add.
Convection transports heat with fluid motion; natural flow uses buoyancy, forced flow uses a pump/fan.

10.9 · Radiation and blackbody laws

Review this concept

Radiation is electromagnetic energy and needs no medium; only this mode works across vacuum.
Wien: λmaxT=2.9×10⁻³ m K. Stefan–Boltzmann: P=eσAT⁴; net P=eσA(T⁴−Ts⁴).
Black/dull surfaces are strong absorbers and emitters; shiny surfaces reflect radiation.

10.10 · Newton’s law of cooling

Review this concept

Cooling rate ∝ (T−Ts) only for a small excess temperature.
Integrate to T−Ts=(T0−Ts)e⁻ᴷᵗ; the ln(excess)–time slope is −K.
Compare mean excess temperatures for textbook finite-interval estimates.

TESTS

Practice tests

Choose a level and check your understanding.

1. A 25 °C temperature on Fahrenheit scale is:

2. An ideal gas is at fixed volume. If absolute temperature doubles, pressure:

3. A 1 m rod with α=2×10⁻⁵ K⁻¹ is heated 50 K. ΔL is:

4. At a melting plateau, added energy primarily increases:

5. A material has K=200 W m⁻¹ K⁻¹, A=0.01 m², L=0.5 m, ΔT=20 K. H is:

1. The triple point of water is approximately:

2. A body absorbs 4.2 kJ and warms 10 K. Its heat capacity is:

3. A 0.2 kg object with s=500 J kg⁻¹ K⁻¹ cools 20 K. Heat lost magnitude:

4. When equal heat currents flow through rods in series at steady state, what is equal?

5. Hot air rises mainly because it:

1. A blackbody has peak wavelength 500 nm at 5800 K. Wien constant from these data is:

2. If absolute temperature of a blackbody doubles, emitted power per area changes by factor:

3. A silvered thermos wall mainly reduces heat transfer by:

4. A 30 °C body cools in surroundings at 20 °C. Its initial excess temperature is:

5. The slope of ln(T−Ts) against time is: