Class 11 · Physics · Chapter 10

Thermal Properties of Matter

This chapter looks at how matter responds to heat — temperature scales, thermal expansion of solids, liquids and gases, calorimetry, change of state, and the three modes of heat transfer (conduction, convection, radiation). There is no Class 11 board exam, but the chapter carries real weight in your school final and it is the foundation for thermodynamics in Class 12 and for JEE/NEET.

11 sections11 NCERT questions25 board questions15-question quiz

Before you startWhat the board asks from this chapter

No board exam at Class 11, but this chapter is a regular 4–6 mark contributor in the school final/pre-board paper, and its formulas (calorimetry, Newton's law of cooling, Stefan's law) are reused directly in Class 12 boards and in JEE/NEET.

Question typeMarksUsually asked
MCQ1Conceptual traps on which scale reads what, direction of anomalous expansion of water, which mode of heat transfer needs a medium.
Assertion–Reason1Linking a physical fact (e.g. bimetallic strip bending) to its correct thermal reasoning.
Very short answer2State a law, define a coefficient, or a one-line 'why' (why does a gap remain in a bimetallic strip design, why is a cavity in a solid also expands like solid material).
Short answer3Small numericals — one thermal expansion or one calorimetry calculation.
Long answer5Full derivation (Newton's law of cooling, or Wien's/Stefan's law statement with a multi-step calorimetry numerical), or a case study combining two ideas.
Case-based4A short passage (e.g. a thermometer, a calorimetry setup, or a cooling curve) followed by 4 sub-questions.

Hot topics, in order

  1. Calorimetry — principle of mixtures, specific heat, latent heat
  2. Thermal expansion of solids (linear, area, volume) and the anomalous expansion of water
  3. Newton's law of cooling — statement, graph, and derivation
  4. Conduction — thermal conductivity, steady state, and series/parallel rods
  5. Stefan–Boltzmann law and Wien's displacement law
  6. Temperature scales and their conversion (especially Celsius–Fahrenheit–Kelvin)
  7. Change of state and latent heat, with the temperature-vs-heat graph

The whole chapter, in orderNotes

01 10.1 Introduction

Heat is energy that flows between two bodies because they are at different temperatures. Temperature is the physical quantity that tells us which way heat will flow — heat always flows from a body at higher temperature to one at lower temperature, until they reach a common temperature called thermal equilibrium. This chapter studies how temperature is measured, how objects respond to heating (they expand, they store heat, they change state), and how heat actually travels from one place to another.

Keep the distinction clear for the exam: heat is energy in transit (measured in joule), while temperature is the degree of hotness (measured in kelvin or degree Celsius). A hot body does not 'contain' heat; it contains internal energy, and heat is what flows when there is a temperature difference.

Exam tip Examiners often frame a 1-mark question as 'heat and temperature are the same physical quantity — true or false'. Always answer false and be ready to say why in one line.

02 10.2 Temperature and Heat

Temperature measures the average kinetic energy of the molecules of a substance. The SI unit of temperature is the kelvin (K); Celsius (°C) is the commonly used practical scale. Heat is the energy transferred due to temperature difference and is measured in joule (older texts also use calorie; 1 cal = 4.186 J).

A thermometer works because some measurable property of a substance changes uniformly with temperature — this is called a thermometric property. Examples: the length of a mercury column, the pressure of a gas at constant volume, the resistance of a metal wire. The constant-volume gas thermometer is the most accurate and is used as the standard against which other thermometers are calibrated, because all gases (at low pressure) expand by nearly the same fraction for the same rise in temperature, independent of the gas used.

Exam tip If asked 'why is a gas thermometer preferred over a liquid thermometer for standard calibration', the answer is: all gases show nearly identical thermal expansion behaviour at low pressure, giving a temperature scale independent of the choice of gas.

03 10.3 Measurement of Temperature

A thermometer is calibrated by choosing two fixed, reproducible reference temperatures and dividing the interval between them into a fixed number of equal parts. On the Celsius scale the ice point (melting point of pure ice at 1 atm) is 0 °C and the steam point (boiling point of water at 1 atm) is 100 °C, with 100 equal divisions between them. On the Fahrenheit scale these are 32 °F and 212 °F, with 180 divisions.

For a linear thermometric property X, temperature is found from tX = [(XT − X0)/(X100 − X0)] × 100 °C, where X0 and X100 are the property's values at the ice point and steam point. Since the same interval is divided differently on different scales, a general relation connects any two scales: the fraction of the way up from the lower fixed point to the upper fixed point is the same number on every scale.

(C − 0)/100 = (F − 32)/180 = (K − 273.15)/100
Ctemperature in degree CelsiusFtemperature in degree FahrenheitKtemperature in kelvin
Exam tip The only temperature at which Celsius and Fahrenheit readings are numerically equal is −40° (−40 °C = −40 °F). This is a favourite MCQ trap.

04 10.4 Ideal-Gas Equation and Absolute Temperature

At low density, all gases obey the ideal gas equation PV = μRT, where P is pressure, V is volume, μ is the number of moles, R is the universal gas constant (8.31 J mol−1 K−1) and T is absolute temperature. A constant-volume gas thermometer plots pressure against temperature; extrapolating this straight-line graph backwards, all gases (in the low-pressure, ideal limit) reach zero pressure at the same temperature, −273.15 °C. This temperature is called absolute zero, and it defines the zero of the Kelvin scale. The Kelvin scale (SI unit of temperature) is therefore related to Celsius by T(K) = t(°C) + 273.15, and its size of one degree is identical to that of the Celsius scale.

The triple point of water — the unique temperature and pressure at which ice, water and water vapour coexist in equilibrium — is used as the single fixed reference point for the modern Kelvin scale and is assigned the value 273.16 K exactly.

T(K) = t(°C) + 273.15
Tabsolute (Kelvin) temperaturetCelsius temperature
PV = μRT
PpressureVvolumeμnumber of molesRuniversal gas constant, 8.31 J mol−1 K−1Tabsolute temperature (K)
Exam tip Triple point of water = 273.16 K, not 273 K and not 0 °C — examiners deduct marks if you write 273 K here.

05 10.5 Thermal Expansion PYQ CBSE 2019

Most substances expand when heated because the average distance between their molecules increases as thermal vibrations grow with temperature. Three types of expansion are defined for solids: linear expansion (change in length), area (superficial) expansion (change in area), and volume (cubical) expansion (change in volume).

For a small temperature change ΔT, the change is proportional to the original dimension and to ΔT:

Here α, β and γ are the coefficients of linear, area and volume expansion respectively, each with SI unit K−1 (or °C−1). For an isotropic solid, it can be shown that β = 2α and γ = 3α, because area scales as length² and volume as length³, and the cross terms in α·ΔT are negligible for small ΔT.

Liquids only have a volume expansion coefficient (they have no fixed shape). Because the containing vessel also expands, we distinguish apparent expansion of the liquid (relative to the vessel) from its real (absolute) expansion: γreal = γapparent + γvessel.

Anomalous expansion of water: water contracts on heating from 0 °C to 4 °C and expands beyond 4 °C — its density is maximum at 4 °C. This is why ice forms first at the top of a lake in winter and floats, insulating the water below and letting aquatic life survive; the water at the bottom stays near 4 °C, its densest state.

Gases expand far more than solids or liquids for the same rise in temperature, and at constant pressure obey V = V0(1 + γΔT) with γ nearly the same (1/273 per °C) for all gases at low pressure.

ΔL = α L ΔT
αcoefficient of linear expansion (K−1)Loriginal lengthΔTrise in temperature
ΔA = β A ΔT, β ≈ 2α
βcoefficient of area expansionAoriginal area
ΔV = γ V ΔT, γ ≈ 3α
γcoefficient of volume (cubical) expansionVoriginal volume
Exam tip A hole/cavity inside a solid expands exactly as if it were filled with the same solid material — this trips up many students who instinctively think the hole shrinks.

06 10.6 Specific Heat Capacity PYQ CBSE 2018

The heat capacity of a body is the amount of heat needed to raise its temperature by 1 K: S = ΔQ/ΔT. Specific heat capacity is heat capacity per unit mass, s = S/m = ΔQ/(mΔT), and molar specific heat capacity is per mole, C = ΔQ/(μΔT). Specific heat capacity is a property of the material, not of a particular body, and it can depend on the process (constant pressure vs constant volume for gases — hence CP and CV).

Water has an unusually high specific heat capacity (4186 J kg−1 K−1, i.e. 1 cal g−1 °C−1). This is why coastal regions have a milder climate than inland regions at the same latitude: large water bodies absorb or release a great deal of heat with only a small temperature change, moderating the surrounding air temperature; it is also why water is used as a coolant in car radiators.

s = ΔQ / (m ΔT)
sspecific heat capacity (J kg−1 K−1)mmassΔQheat absorbed/releasedΔTtemperature change
Exam tip Do not confuse heat capacity (property of a given object, depends on mass) with specific heat capacity (property of the material, independent of mass).

07 10.7 Calorimetry PYQ CBSE 2020 · CBSE 2017

Calorimetry is the measurement of heat exchanged. It rests on the principle of calorimetry (principle of mixtures): when two bodies at different temperatures are mixed in an isolated system (no heat lost to surroundings), heat lost by the hotter body equals heat gained by the colder body, because energy is conserved.

Heat lost by hot body = Heat gained by cold body.

A calorimeter is an insulated vessel (usually with a stirrer and a low specific heat, e.g. copper) used to measure specific heat or latent heat experimentally by mixing a known mass at a known temperature with the substance under test and recording the common final temperature.

Worked method: a hot substance of mass m1, specific heat s1, at temperature T1 is mixed with a colder substance of mass m2, specific heat s2, at temperature T2 (T1 > T2). If Tf is the common final temperature, m1s1(T1 − Tf) = m2s2(Tf − T2), which can be solved for any one unknown, typically s1 or Tf.

m1s1(T1 − Tf) = m2s2(Tf − T2)
m1, m2masses of the hot and cold substancess1, s2their specific heat capacitiesTfcommon final (equilibrium) temperature
Exam tip In numericals, always include the calorimeter's own heat capacity (water equivalent) if the question gives it — a very common mark loss is forgetting the calorimeter's contribution.

08 10.8 Change of State PYQ CBSE 2019

Matter exists as solid, liquid or gas, and can change from one state to another at fixed temperature and pressure conditions. Melting is solid → liquid at the melting point; freezing is liquid → solid at the freezing point (same temperature as melting point for a given pressure). Boiling/vaporisation is liquid → gas at the boiling point; condensation is the reverse. Sublimation is the direct change of solid to vapour without passing through the liquid state (e.g. camphor, dry ice).

During a change of state, temperature stays constant even though heat is continuously supplied — this heat goes into breaking intermolecular bonds (changing the state), not into raising kinetic energy/temperature. The heat required per unit mass for this, at constant temperature, is called latent heat: L = Q/m. Latent heat of fusion (Lf) is for melting/freezing (for ice, 3.33 × 105 J kg−1); latent heat of vaporisation (Lv) is for boiling/condensation (for water, 22.6 × 105 J kg−1).

A temperature-vs-heat-supplied graph for ice heated steadily from below 0 °C to steam above 100 °C shows: a rising slope (ice warming), a flat plateau at 0 °C (melting), a rising slope (water warming), a flat plateau at 100 °C (boiling), then a rising slope (steam warming). The plateaus are wider than the slopes because latent heat of vaporisation is much larger than that of fusion.

Melting point changes with pressure: it decreases with increasing pressure for substances that expand on melting is wrong for ice — ice contracts on melting, so increasing pressure lowers its melting point (this is why pressure from a skate blade can melt ice slightly). Boiling point always increases with increasing pressure (this is the principle of the pressure cooker).

Q = mL
Qheat supplied/released during change of statemmassLlatent heat (fusion or vaporisation), J kg−1
Exam tip A pressure cooker cooks faster because raising pressure raises the boiling point of water, so food cooks at a temperature above 100 °C.

09 10.9 Heat Transfer PYQ CBSE 2020

Heat travels from a hotter region to a colder one by three distinct mechanisms.

Conduction is the transfer of heat through a medium (usually solids) without bulk motion of the material — energy passes from molecule to molecule through collisions and, in metals, through free electrons. Consider a rod of length L, uniform cross-section A, with its ends kept at steady temperatures TC and TD (TD > TC), the sides insulated. In the steady state, the rate of heat flow H = ΔQ/Δt is the same at every cross-section and is given by H = KA(TD − TC)/L, where K is the thermal conductivity of the material — a measure of how readily it conducts heat. SI unit of K is J s−1 m−1 K−1 (or W m−1 K−1). Metals (especially silver and copper) have high K; most non-metals, and gases in particular, are poor conductors and are used as insulators (thermos flasks trap air/vacuum for exactly this reason).

Convection is heat transfer by actual movement of the heated material itself — it occurs in fluids (liquids and gases), where the heated portion becomes less dense, rises, and is replaced by cooler fluid, setting up convection currents. Natural convection (density-driven, e.g. land and sea breezes, boiling water) is different from forced convection (fluid driven by a pump/fan, e.g. blood circulation, forced-air heating). Land and sea breezes occur because land heats up and cools down faster than the sea; during the day, warm air over land rises and the cooler sea breeze blows in to replace it; at night the reverse (land breeze) occurs.

Radiation is heat transfer by electromagnetic waves and needs no medium — this is how heat from the Sun reaches Earth through empty space. Every body above absolute zero radiates thermal (infrared) energy; a body that is a good absorber of radiation is also a good emitter (this is the basis of a perfectly black body, which absorbs all radiation falling on it and, at a given temperature, is the best possible emitter).

H = KA(TD − TC)/L
Hrate of heat conduction (W)Kthermal conductivity (W m−1 K−1)Across-sectional areaLlength/thickness of the conductorTD − TCtemperature difference across the ends
Exam tip Conduction needs a medium and no bulk motion of matter; convection needs a medium and bulk motion; radiation needs no medium at all — this three-line contrast is a favourite VSA/MCQ.

10 10.10 Newton's Law of Cooling PYQ CBSE 2020 · CBSE 2016

Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings, provided this difference is small. If T is the body's temperature and Ts the (constant) surrounding temperature, −dQ/dt ∝ (T − Ts).

Derivation: Let a body of mass m and specific heat s cool from T to Ts. Rate of loss of heat: −dQ/dt = ms(dT/dt) in magnitude. By Newton's law, −dQ/dt = k(T − Ts), where k is a positive constant depending on the area and nature of the surface. So −ms(dT/dt) = k(T − Ts), i.e. dT/(T − Ts) = −(k/ms) dt = −K dt, where K = k/ms. Integrating between t = 0 (temperature T1) and time t (temperature T2): ln[(T2 − Ts)/(T1 − Ts)] = −Kt, giving T2 − Ts = (T1 − Ts)e−Kt. This shows temperature excess over the surroundings falls off exponentially with time.

For exam numericals, a simpler averaged form is commonly used: (T1 − T2)/t = K[(T1 + T2)/2 − Ts], which avoids needing calculus — it approximates the instantaneous rate by the average temperature over the interval. A graph of temperature versus time for a cooling body is a curve that falls steeply at first (large T − Ts) and flattens as T approaches Ts, asymptotically approaching the surrounding temperature; a graph of loge(T − Ts) versus t is a straight line with negative slope −K.

−dQ/dt = k(T − Ts)
kpositive constantTinstantaneous temperature of the bodyTstemperature of surroundings
(T1 − T2)/t = K[(T1 + T2)/2 − Ts]
T1, T2initial and final temperatures over interval tKcooling constant, k/ms
Exam tip The law only holds for a small temperature difference between body and surroundings — state this assumption explicitly in a 5-mark derivation, it is a marking-scheme point.

11 10.11 Heat Transfer: Blackbody Radiation Laws

The spectrum of radiation emitted by a hot body depends on its temperature. As temperature increases, the wavelength at which the emission is maximum shifts towards shorter wavelengths — this is Wien's displacement law: λmT = b, where λm is the wavelength of maximum intensity and b is Wien's constant (2.9 × 10−3 m K). This is why an iron rod glows dull red, then orange, then white as it is heated further — the peak wavelength shifts from infrared towards the visible and then towards blue.

The total energy radiated per unit area per unit time by a perfect black body is given by the Stefan–Boltzmann law: E = σT4, where σ is Stefan's constant (5.67 × 10−8 W m−2 K−4). For a body that is not a perfect black body, E = εσT4, where ε (emissivity, 0 < ε ≤ 1) measures how close the body is to an ideal black body. If such a body radiates into surroundings at temperature T0, the net rate of loss of energy per unit area is Enet = εσ(T4 − T04) — this is the more complete law that Newton's law of cooling approximates for small (T − T0).

λmT = b
λmwavelength of maximum emitted intensityTabsolute temperature of the bodybWien's constant, 2.9 × 10−3 m K
Enet = εσ(T4 − T04)
εemissivityσStefan's constant, 5.67 × 10−8 W m−2 K−4T, T0absolute temperature of body and surroundings
Exam tip Newton's law of cooling is only the small-ΔT approximation of the Stefan–Boltzmann T4 law — a good 'derive Newton's law from Stefan's law' link question uses binomial approximation on (T0 + ΔT)4.

VocabularyKey terms

Thermal equilibriumThe state in which two bodies in contact have no net heat flow between them because they are at the same temperature.
Triple point of waterThe unique temperature and pressure (273.16 K) at which ice, liquid water and water vapour coexist in equilibrium; used to fix the Kelvin scale.
Coefficient of linear expansion (α)The fractional increase in length of a solid per unit rise in temperature, ΔL/(LΔT).
Anomalous expansion of waterThe unusual behaviour of water contracting when heated from 0 °C to 4 °C and expanding beyond 4 °C, giving water its maximum density at 4 °C.
Specific heat capacityHeat required to raise the temperature of unit mass of a substance by 1 K; a property of the material, s = ΔQ/(mΔT).
Molar specific heat capacityHeat required to raise the temperature of one mole of a substance by 1 K.
Principle of calorimetryIn an isolated system, heat lost by the hotter body equals heat gained by the colder body — a statement of conservation of energy.
Latent heatThe heat absorbed or released per unit mass by a substance during a change of state at constant temperature, L = Q/m.
SublimationThe direct change of a substance from the solid state to the vapour state without passing through the liquid state.
ConductionTransfer of heat through a medium from particle to particle (or via free electrons in metals) without any bulk movement of the medium.
Thermal conductivity (K)A material property that measures how readily heat is conducted through it; defined by H = KA(ΔT)/L in the steady state.
ConvectionTransfer of heat by actual bulk movement of a heated fluid (liquid or gas) from one place to another.
RadiationTransfer of heat as electromagnetic waves; the only mode of heat transfer that needs no material medium.
Black bodyAn idealised object that absorbs all electromagnetic radiation incident on it and, for its temperature, is the best possible emitter of radiation.
Newton's law of coolingThe rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings, valid for small temperature differences.
Wien's displacement lawThe wavelength at which a black body's emission is maximum is inversely proportional to its absolute temperature, λmT = constant.

Keep this on one pageFormula sheet

NameExpressionNotes
Scale conversion(C)/100 = (F − 32)/180 = (K − 273.15)/100Convert between Celsius, Fahrenheit and Kelvin readings.
Ideal gas equationPV = μRTR = 8.31 J mol−1 K−1; low-density/ideal-gas limit.
Linear expansionΔL = αLΔTα in K−1; use for rods, wires, rails.
Area and volume expansionβ ≈ 2α, γ ≈ 3α; ΔA = βAΔT, ΔV = γVΔTHolds for isotropic solids with small ΔT.
Specific heat capacitys = ΔQ/(mΔT)SI unit J kg−1 K−1.
Calorimetry (principle of mixtures)m1s1(T1 − Tf) = m2s2(Tf − T2)Heat lost by hot body = heat gained by cold body, isolated system.
Latent heatQ = mLLfusion(ice) = 3.33 × 105 J kg−1; Lvap(water) = 22.6 × 105 J kg−1.
Steady-state conductionH = KA(TD − TC)/LH is rate of heat flow (W); K is thermal conductivity.
Newton's law of cooling (average form)(T1 − T2)/t = K[(T1+T2)/2 − Ts]Valid for small temperature excess over surroundings.
Wien's displacement lawλmT = bb = 2.9 × 10−3 m K.
Stefan–Boltzmann lawE = εσT4σ = 5.67 × 10−8 W m−2 K−4; ε = 1 for a perfect black body.

Practise with a pencilDiagrams that earn marks

Constant-volume gas thermometer and P–T extrapolation

Draw a bulb of gas connected by a tube to a manometer with a movable mercury reservoir, arranged so gas volume stays fixed. Alongside, plot pressure P on the y-axis against temperature t (°C) on the x-axis as a straight line; extend the line backward (dashed) to show it cutting the x-axis at −273.15 °C, labelling this as absolute zero.

Temperature vs heat-supplied graph for ice → water → steam

Draw heat supplied (Q) on the x-axis and temperature (°C) on the y-axis. Show a rising line from below 0 °C up to 0 °C (ice warming), a horizontal plateau at 0 °C (melting, label 'latent heat of fusion'), a rising line from 0 °C to 100 °C (water warming), a horizontal plateau at 100 °C (boiling, label 'latent heat of vaporisation', drawn wider than the fusion plateau), then a rising line above 100 °C (steam warming).

Conduction through a rod in steady state

Draw a horizontal rod of length L and cross-section A with insulated (hatched) sides, hot end at temperature TD on the left and cold end at TC on the right. Mark arrows showing heat flow H from hot to cold end, and label the formula H = KA(TD − TC)/L beside it.

Newton's law of cooling curve

Plot temperature T (y-axis) against time t (x-axis) as a curve starting steep and flattening out asymptotically to the surrounding temperature Ts, shown as a dashed horizontal line. On a second small graph, plot loge(T − Ts) against t as a straight line with negative slope, labelling the slope as −K.

TextbookNCERT questions the board keeps asking

11.1 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.3 m

Celsius: t = T − 273.15.

Neon: t = 24.57 − 273.15 = −248.58 °C.

CO2: t = 216.55 − 273.15 = −56.60 °C.

Fahrenheit: F = (9/5)C + 32.

Neon: F = (9/5)(−248.58) + 32 = −447.44 + 32 = −415.44 °F.

CO2: F = (9/5)(−56.60) + 32 = −101.88 + 32 = −69.88 °F.

11.3 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.)3 m

Heat released by copper cooling from 500 °C to 0 °C:

Q = ms ΔT = 2500 g × 0.39 J g−1 K−1 × 500 K = 487500 J.

Mass of ice melted, mice = Q/L = 487500/335 ≈ 1455 g ≈ 1.45 kg.

11.5 A steel tape 1 m long is correctly calibrated for a temperature of 27 °C. The length of a steel rod measured by this tape is found to be 63 cm on a hot day when the temperature is 45 °C. What is the actual length of the rod on that day? What is the length of the same rod on a day when the temperature is 27 °C? (αsteel = 1.20 × 10−5 K−1.)3 m

At 45 °C the tape itself has expanded, so each 'cm' marked on it is actually slightly longer than 1 cm.

Actual length shown by an expanded tape: L' = L(1 + αΔT) = 63 × [1 + 1.20×10−5 × 18] = 63 × 1.000216 ≈ 63.0136 cm.

This is the actual length of the rod at 45 °C. Since the rod's own length also depends on temperature, but the question asks the actual length at 45 °C — it is 63.0136 cm ≈ 63.01 cm.

Length of the same rod at 27 °C (the calibration temperature) is simply the reading, 63 cm, because the tape is correctly calibrated at 27 °C.

11.9 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? (αbrass = 2.0 × 10−5 K−1; Ybrass = 0.91 × 1011 Pa.)3 m

ΔT = 27 − (−39) = 66 K (fall in temperature).

Thermal strain = αΔT = 2.0×10−5 × 66 = 1.32×10−3.

Since the wire is held taut between rigid supports, this strain is resisted, producing stress = Y × strain = 0.91×1011 × 1.32×10−3 = 1.2012×108 Pa.

Area, A = πr2 = π(1.0×10−3)2 = 3.14×10−6 m2.

Tension, F = stress × A = 1.2012×108 × 3.14×10−6 ≈ 377 N.

11.11 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−1.3 m

Total energy supplied in 2.5 min = 150 s: E = Pt = 10000 × 150 = 1.5×106 J.

Energy actually going into the block (50%) = 0.75×106 J.

ΔT = Q/(ms) = 750000/(8000 × 0.91) = 750000/7280 ≈ 103 °C.

11.14 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.5 m

Using the average form of Newton's law of cooling: (T1 − T2)/t = K[(T1+T2)/2 − Ts].

First cooling: (80−50)/5 = K[(80+50)/2 − 20] ⇒ 30/5 = K[65 − 20] = 45K ⇒ K = 6/45 = 0.1333 min−1.

Second cooling: (60−30)/t = 0.1333 × [(60+30)/2 − 20] = 0.1333 × [45 − 20] = 0.1333 × 25 = 3.333.

t = 30/3.333 ≈ 9 minutes.

10.14 Calculate the temperature at which a black body radiates energy at a rate of 1.0 W cm−2. (σ = 5.67×10−8 W m−2 K−4.)3 m

1.0 W cm−2 = 1.0×104 W m−2.

E = σT4 ⇒ T4 = E/σ = 1.0×104/5.67×10−8 ≈ 1.764×1011.

T = (1.764×1011)1/4 ≈ 648 K.

11.7 Given below are observations on molar specific heats at room temperature of some common gases. Why does hydrogen have a value of CV different from that of monatomic gases like helium and argon?2 m

Monatomic gases (He, Ar) have only 3 translational degrees of freedom, so by the equipartition theorem CV = (3/2)R.

Hydrogen is diatomic. Besides 3 translational degrees of freedom, it also has 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond), so its CV = (5/2)R, which is higher than that of a monatomic gas.

C1 Why is a small gap left between two rails of a railway track, and why is a gap left at the ends of steel/concrete bridge girders?2 m

Rails and girders are metal, which expands on heating in summer. A gap is left so that thermal expansion of adjacent rail/girder sections does not push against each other and buckle or crack the track/bridge. In winter this gap widens slightly since the metal contracts.

C2 Explain why a body with large surface area and low specific heat cools faster than a body with small surface area and high specific heat, for the same initial excess temperature.3 m

By Newton's law of cooling, the rate of heat loss depends on the surface area exposed and the temperature excess over the surroundings — a larger area radiates/convects heat away faster. Also, for a body of a given mass, a lower specific heat means a smaller amount of heat needs to leave the body to produce a given fall in temperature (since ΔT = ΔQ/(ms)), so its temperature falls faster for the same rate of heat loss. Both effects combine to make such a body cool more quickly.

10.21 Derive an expression showing that pressure exerted by a liquid column is independent of the cross-sectional area of the vessel — related conceptually to why liquid thermometers use a fine capillary bore. Why is a capillary of very small (uniform) cross-section used in a mercury thermometer?3 m

A fine capillary bore is used so that even a small change in the volume of mercury (due to a small temperature change) produces a large, easily readable displacement of the mercury column, since the same change in volume ΔV corresponds to a length change Δl = ΔV/A which is larger when the bore's cross-sectional area A is small. This gives the thermometer higher sensitivity.

Previous years and frequently askedBoard questions with model answers

MCQ The temperature at which the Celsius and Fahrenheit scales give the same numerical reading is:
  1. 0°
  2. 100°
  3. −40°
  4. −273°
Frequently asked1 m
Answer: (c)
Setting C = F in C/5 = (F−32)/9 gives C = (C−32)/9 × 5 ⟹ 9C = 5C − 160 ⟹ 4C = −160 ⟹ C = −40. So −40 °C = −40 °F.
MCQ Two rods, one of copper and the other of steel, of the same length and cross-section, are joined end to end and their free ends are maintained at different temperatures. In the steady state, the temperature at the junction of the two rods is:
  1. Equal to the mean of the two end temperatures
  2. Closer to the temperature of the end with the more conductive rod
  3. Closer to the temperature of the end with the less conductive rod
  4. Independent of the thermal conductivities
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Answer: (b)
In steady state the heat current through both rods is equal; the more conductive rod needs a smaller temperature drop to carry the same heat current, so the junction temperature sits closer to the hot end connected through the better conductor.
MCQ Water is used as a coolant in car radiators mainly because it has:
  1. Low boiling point
  2. High thermal conductivity
  3. High specific heat capacity
  4. Low density
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Answer: (c)
Water's unusually high specific heat capacity lets it absorb a large amount of heat from the engine with only a modest rise in its own temperature.
MCQ A metal ball is heated from 20 °C to 60 °C. Its area increases by 0.16%. The coefficient of linear expansion of the metal is approximately:
  1. 2 × 10−5 K−1
  2. 1 × 10−5 K−1
  3. 4 × 10−5 K−1
  4. 8 × 10−5 K−1
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Answer: (a)
ΔA/A = βΔT = 2αΔT. So 0.0016 = 2α × 40 ⟹ α = 0.0016/80 = 2×10−5 K−1.
MCQ Which of the following modes of heat transfer does NOT require a material medium?
  1. Conduction
  2. Convection
  3. Radiation
  4. Both conduction and convection
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Answer: (c)
Radiation travels as electromagnetic waves and can pass through vacuum; conduction and convection both need matter to carry the energy.
MCQ The triple point of water on the Kelvin scale is:
  1. 273 K
  2. 273.16 K
  3. 373 K
  4. 0 K
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Answer: (b)
The triple point of water is fixed at exactly 273.16 K and defines the Kelvin scale; 273 K is only an approximation of the ice point.
MCQ A black body at 227 °C radiates heat at a rate of 5 W m−2. What is the rate of radiation when its temperature is raised to 727 °C? (Use T in kelvin.)
  1. 5 W m−2
  2. 20 W m−2
  3. 40 W m−2
  4. 80 W m−2
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Answer: (d)
T1 = 500 K, T2 = 1000 K. E ∝ T4, and T2/T1 = 2, so E2 = E1 × 24 = 5 × 16 = 80 W m−2.
Assertion–Reason Assertion (A): A bimetallic strip made of brass and iron bends when heated. Reason (R): Brass and iron have different coefficients of linear expansion.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
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Answer: (a)
Since brass expands more than iron for the same temperature rise, the composite strip bends towards the iron (less-expanding) side; R correctly explains A.
Assertion–Reason Assertion (A): Two bodies at the same temperature must have the same amount of heat. Reason (R): Heat and temperature are the same physical quantity.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is false but R is true
  4. Both A and R are false
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Answer: (d)
Neither statement is true: heat content depends on mass and specific heat capacity too, and heat (energy in transit) is a different physical quantity from temperature (a measure of hotness).
Assertion–Reason Assertion (A): Newton's law of cooling is a special case of the Stefan–Boltzmann law. Reason (R): Newton's law of cooling holds only when the temperature difference between the body and its surroundings is small.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
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Answer: (a)
Expanding (Ts+ΔT)4 by the binomial theorem for small ΔT and keeping only the first-order term reduces Stefan's T4 law to a linear (Newton's) law — valid precisely because ΔT is small.
Very short answer Why does a wooden handle feel less hot than a metal vessel taken out of boiling water, even though both are at the same temperature?Frequently asked2 m
Metal has a much higher thermal conductivity than wood. When the hand touches the metal, heat flows into the hand quickly, so it feels hot; wood conducts heat very slowly (it is a poor conductor / good insulator), so very little heat reaches the hand in the same time and it feels much less hot, even though both are at the same actual temperature.
Very short answer State the principle of calorimetry.Frequently asked2 m
When two bodies at different temperatures are mixed in a system isolated from the surroundings (no heat is lost to or gained from outside), the heat lost by the body at the higher temperature is exactly equal to the heat gained by the body at the lower temperature. This follows from the law of conservation of energy: heat lost = heat gained.
Very short answer Why does a body cool faster initially and then more slowly as it approaches the temperature of the surroundings?Frequently asked2 m
By Newton's law of cooling, the rate of loss of heat is proportional to the temperature difference between the body and the surroundings. Initially this difference is large, so heat is lost quickly and the body cools fast. As the body's temperature falls and approaches the surrounding temperature, the difference (and hence the rate of cooling) becomes smaller, so cooling slows down and the temperature approaches the surrounding value asymptotically.
Very short answer A gap is always left between two consecutive iron rails on a railway track. State the reason.Frequently asked2 m
Iron rails expand in length when they get heated in summer (linear thermal expansion). If the rails were laid end to end with no gap, this expansion would have nowhere to go, causing the rails to buckle or bend out of shape. The gap allows each rail to expand freely into the space without pushing against the next one.
Very short answer Define coefficient of linear expansion and give its SI unit.Frequently asked2 m
The coefficient of linear expansion (α) of a solid is the fractional increase in its length per unit rise in temperature: α = ΔL/(LΔT), where L is the original length and ΔT is the rise in temperature. Its SI unit is K−1 (or °C−1).
Very short answer Why is the density of water maximum at 4 °C?Frequently asked2 m
As water is cooled from higher temperatures, ordinary thermal contraction competes with the formation of an increasing number of hydrogen-bonded, open hexagonal (ice-like) molecular clusters that take up more volume. Below about 4 °C this clustering effect dominates and the volume starts increasing again (density decreasing) even though the temperature keeps falling, so density is maximum at 4 °C, where the two competing effects balance out.
Short answer Explain, with reference to molecular behaviour, why gases expand much more than solids and liquids for the same rise in temperature.Frequently asked3 m
In a solid, molecules are tightly bound in a fixed lattice by strong intermolecular forces and can only vibrate about their mean positions; heating increases the amplitude of vibration only slightly, so expansion is small. In a liquid, intermolecular forces are weaker than in a solid, allowing somewhat more freedom of movement, so liquids expand more than solids for the same temperature rise. In a gas, intermolecular forces are negligible and molecules move almost freely; heating substantially increases their kinetic energy and the average distance between them (or the pressure they exert, if volume is fixed), so gases show by far the largest thermal expansion of the three states.
Short answer 150 g of ice at 0 °C is mixed with 300 g of water at 50 °C in a calorimeter of negligible heat capacity. Find the final temperature of the mixture. (Lice = 336 J g−1, swater = 4.2 J g−1 K−1.)Frequently asked3 m
Heat needed to melt all the ice: Qmelt = mL = 150 × 336 = 50400 J. Maximum heat available from 300 g water cooling from 50 °C to 0 °C: Qavail = ms ΔT = 300 × 4.2 × 50 = 63000 J. Since Qavail (63000 J) > Qmelt (50400 J), all the ice melts, and the remaining heat (63000 − 50400 = 12600 J) raises the temperature of the resulting 450 g of water (150 g melted ice + 300 g original water) above 0 °C. ΔT = 12600/(450 × 4.2) = 12600/1890 = 6.67 °C. Final temperature = 0 + 6.67 °C ≈ 6.7 °C.
Short answer Two metal rods A and B of the same length and area of cross-section have thermal conductivities 3 W m−1K−1 and 5 W m−1K−1 respectively. They are joined end to end. If the free end of A is at 100 °C and the free end of B is at 0 °C, find the temperature of the junction in the steady state.CBSE 20183 m
In steady state, the heat current through both rods is the same (they are in series): H = KAA(100 − T)/L = KBA(T − 0)/L, so KA(100 − T) = KBT. 3(100 − T) = 5T ⟹ 300 − 3T = 5T ⟹ 300 = 8T ⟹ T = 37.5 °C.
Short answer A block of ice at 0 °C and mass 200 g is dropped into 500 g of water at 25 °C in an insulated container. What is the final temperature of the mixture, given that Lice = 336 J g−1 and swater = 4.2 J g−1 K−1? Assume the calorimeter absorbs no heat.Frequently asked3 m
Heat needed to melt 200 g of ice: Qmelt = 200 × 336 = 67200 J. Maximum heat available from 500 g water cooling from 25 °C to 0 °C: Qavail = 500 × 4.2 × 25 = 52500 J. Since Qavail (52500 J) is less than Qmelt (67200 J), not all the ice can melt; the mixture settles at 0 °C with some ice remaining (mass melted = 52500/336 ≈ 156.25 g, leaving about 43.75 g of ice unmelted). Final temperature of the mixture = 0 °C.
Long answer State Newton's law of cooling. Derive the relation T2 − Ts = (T1 − Ts)e−Kt for a body cooling from temperature T1 to T2 in surroundings at temperature Ts, and sketch the temperature–time graph.CBSE 20205 m
Newton's law of cooling: the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided this difference is small. Derivation: Let a body of mass m and specific heat s be at temperature T, in surroundings at constant temperature Ts. Rate of loss of heat: −dQ/dt = ms(dT/dt) in magnitude. By Newton's law: −dQ/dt = k(T − Ts), k a positive constant. So −ms(dT/dt) = k(T − Ts) ⟹ dT/(T − Ts) = −(k/ms)dt = −K dt, where K = k/ms. Integrating from t = 0 (T = T1) to time t (T = T2): ∫dT/(T − Ts) = −K∫dt ⟹ ln[(T2−Ts)/(T1−Ts)] = −Kt. Taking exponentials: T2 − Ts = (T1 − Ts)e−Kt. Graph: temperature T on the y-axis falls exponentially with time t, starting steep and flattening out, approaching the dashed horizontal line T = Ts asymptotically but never quite reaching it.
Long answer Define thermal conductivity and derive the expression for the rate of heat flow through a rod in the steady state. A composite slab is made of two materials of the same area but different thickness and conductivity, placed in series. Write the expression for the effective thermal conductivity of the combination.Frequently asked5 m
Thermal conductivity K of a material measures its ability to conduct heat; it is defined through H = KA ΔT/L for a rod of length L, area A, with a temperature difference ΔT between its ends in the steady state. Derivation (qualitative, as required at this level): Consider a rod with insulated sides so heat flows only along its length, with the hot end at TD and cold end at TC (TD > TC). In the steady state, the temperature at each point along the rod is constant in time (though it varies with position), and the same quantity of heat ΔQ crosses every cross-section in the same time Δt (no heat is stored anywhere along the rod). Experimentally, the rate of heat flow H = ΔQ/Δt is found to be directly proportional to the area A and the temperature difference (TD − TC), and inversely proportional to the length L: H = KA(TD − TC)/L, K being the constant of proportionality (thermal conductivity). For two slabs of the same area A, thicknesses L1, L2 and conductivities K1, K2 joined in series, the same heat current H passes through both, so the effective conductivity Keff for the combined thickness (L1+L2) satisfies: (L1+L2)/Keff = L1/K1 + L2/K2.
Long answer What are the three types of thermal expansion in solids? Derive the relations β = 2α and γ = 3α for an isotropic solid, and explain why liquids and gases show only volume expansion.Frequently asked5 m
The three types are linear expansion (change in length, coefficient α), area/superficial expansion (change in area, coefficient β) and volume/cubical expansion (change in volume, coefficient γ), each defined by ΔL = αLΔT, ΔA = βAΔT, ΔV = γVΔT. Derivation of β = 2α: consider a square lamina of side L. Area A = L2. On heating by ΔT, new side L' = L(1+αΔT), so new area A' = L2(1+αΔT)2 = L2(1 + 2αΔT + α2ΔT2). Since α is very small, α2ΔT2 is negligible, so A' ≈ A(1 + 2αΔT). Comparing with A' = A(1+βΔT) gives β = 2α. Derivation of γ = 3α: similarly for a cube of side L, volume V = L3. V' = L3(1+αΔT)3 ≈ L3(1 + 3αΔT), neglecting higher powers of αΔT. Comparing with V' = V(1+γΔT) gives γ = 3α. Liquids and gases have no fixed shape, so a linear or area expansion cannot be meaningfully defined for them — only their volume is a well-defined quantity, so only the coefficient of volume expansion γ is used for fluids.
Case-based A metal sphere of mass 200 g is heated to 150 °C and then dropped into a copper calorimeter of mass 100 g containing 250 g of water at 25 °C. The specific heat of copper is 0.39 J g−1K−1 and that of water is 4.2 J g−1K−1. The final steady temperature recorded is 32 °C. Answer the following:
(i) State the principle used to find the specific heat of the metal in this experiment.
(ii) Write the heat-balance equation for this setup.
(iii) Calculate the specific heat capacity of the metal.
(iv) Name the device used for this experiment and one precaution to reduce error.
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(i) The principle of calorimetry (principle of mixtures): heat lost by the hot metal sphere equals heat gained by the calorimeter and the water, since the system is thermally isolated.
(ii) mmetalsmetal(150 − 32) = mcalscu(32 − 25) + mwaterswater(32 − 25).
(iii) 200 × smetal × 118 = (100 × 0.39 × 7) + (250 × 4.2 × 7). RHS = 273 + 7350 = 7623 J. So smetal = 7623/(200 × 118) = 7623/23600 ≈ 0.323 J g−1K−1.
(iv) A calorimeter (with stirrer, insulating jacket and lid) is used; a key precaution is to insulate the calorimeter well (e.g. keep it in a wooden box with a lid) and stir the water gently so that heat loss to the surroundings is minimised and the temperature is uniform.
Case-based A hot cup of tea at 90 °C is left on a table in a room where the surrounding temperature is 30 °C. It is observed that the tea cools to 70 °C in 5 minutes. Using this information, answer:
(i) State the law that governs this cooling process and its condition of validity.
(ii) Using the average form of the law, find the cooling constant K.
(iii) Estimate how long the tea takes to cool further from 70 °C to 55 °C.
(iv) Would the tea cool faster or slower if it were kept in a room at 10 °C instead of 30 °C, and why?
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(i) Newton's law of cooling: the rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings, valid only when this difference is small.
(ii) (T1−T2)/t = K[(T1+T2)/2 − Ts] ⟹ (90−70)/5 = K[(90+70)/2 − 30] ⟹ 4 = K[80−30] = 50K ⟹ K = 0.08 min−1.
(iii) (70−55)/t = 0.08 × [(70+55)/2 − 30] = 0.08 × [62.5−30] = 0.08 × 32.5 = 2.6. t = 15/2.6 ≈ 5.77 minutes.
(iv) It would cool faster in the 10 °C room, because the temperature difference between the tea and its surroundings would be larger throughout the process, and by Newton's law the rate of heat loss (and hence the rate of cooling) is directly proportional to this difference.

Test yourselfPractice quiz

Marks lost every yearCommon mistakes

  • Writing the triple point of water as 273 K or 0 °C instead of the correct 273.16 K.
  • Forgetting that a hole/cavity in a solid expands like the solid material itself, not like empty space that stays fixed or shrinks.
  • Confusing heat capacity (depends on the mass of a specific object) with specific heat capacity (a fixed property of the material only).
  • In calorimetry numericals, forgetting to include the heat absorbed or released by the calorimeter vessel itself when its heat capacity/water equivalent is given.
  • Applying Newton's law of cooling even when the temperature difference between the body and surroundings is large, where the law is not valid.
  • Mixing up β = 2α and γ = 3α, or applying them to non-isotropic materials where they do not simply hold.
  • Treating latent heat problems as if temperature changes during melting or boiling — temperature stays constant throughout a change of state at fixed pressure.
  • Using Celsius temperature directly in PV = μRT or in σT4 instead of first converting to kelvin.

How to prepareStrategy for this chapter

This chapter has two kinds of marks: definitions/laws that must be stated word-perfect (Newton's law of cooling, principle of calorimetry, Stefan's and Wien's laws), and numericals that follow a fixed method. Start by memorising the three scale-conversion formula and the three expansion coefficients with their α–β–γ relation — these feed almost every other numerical in the chapter.

For calorimetry and change-of-state problems, always write the heat-balance equation first (heat lost = heat gained), check whether a change of state happens partway through (compare the heat needed to melt/boil against the heat available before assuming a single final temperature), and only then substitute numbers. For conduction, learn the steady-state derivation properly — it is asked as a 5-marker on its own or combined with a series/parallel-rods numerical. For Newton's law of cooling, be ready for both the exponential derivation (5 marks) and the average-rate numerical shortcut (3 marks) — know when each is expected.

Do not skip the qualitative parts: why gases have CP and CV but solids do not, why water shows anomalous expansion, why a pressure cooker cooks faster, why land and sea breezes occur. These recur as VSA and Assertion–Reason questions almost every year and are easy marks if you have the reasoning ready in one or two clean sentences.

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