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State Board · Class 11 · Physics

Lesson 7: Thermal Properties of Matter

7 concepts · 56 practice questions with answers · Free to read

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Temperature and its measurement

From daily life

We judge tea hot or cold by touch; a thermometer gives a number. Link to earlier classes : heat, temperature and states of matter of classes 9-10 are now measured with formulas.

Simple definition

Temperature measures how hot a body is; it decides the direction of heat flow.

Temperature scales°C°FK100212373032273C/5 = (F − 32)/9 = (K − 273)/5

In the figure : Freezing and boiling points on all three scales.

Rule 1: Scales and conversion — (F − 32) ÷ 9 = C ÷ 5 = (K − 273) ÷ 5. Freezing 0 °C = 32 °F = 273 K; boiling 100 °C = 212 °F = 373 K.

  • Easy : 37 °C = 37 × 9/5 + 32 = 98.6 °F.
  • Medium : 100 °F = (100 − 32) × 5/9 = 37.8 °C.
  • Exam (CET) : At what temperature are °C and °F equal? (1) −40 (2) 0 (3) 40 (4) 32 → answer 1) Trick : C = F ⇒ C = 1.8C + 32 ⇒ −40.

Rule 2: Change in temperature — ΔC : ΔF : ΔK = 5 : 9 : 5; so a change of 1 K = 1 °C = 1.8 °F.

  • Easy : A rise of 20 °C ⇒ 36 °F and 20 K.
  • Medium : 27 °C = 300 K; 300 K = 80.6 °F.
  • Exam (CET) : ΔT = 50 K ⇒ ΔC? (1) 50 (2) 90 (3) 122 (4) 223 → answer 1)

Rule 3: Gas thermometer — Constant-volume gas thermometer : T = 273.16 × P ÷ P_tp (tp = triple point). Absolute zero 0 K = −273 °C. Zeroth law : A~B and B~C ⇒ A~C.

  • Easy : P_tp = 80 kPa, P = 100 kPa ⇒ T = 341.5 K.
  • Medium : P halved ⇒ T halved (in K).
  • Exam (CET) : T ∝ ? (constant volume) (1) P (2) 1/P (3) P² (4) √P → answer 1)

Derivation : Conversion formula

  • Step 1 : A point lies at the same fraction on both scales : (F − 32) ÷ (212 − 32) = (C − 0) ÷ (100 − 0).
  • Step 2 : (F − 32) ÷ 180 = C ÷ 100.
  • Step 3 : Divide by 20 : (F − 32) ÷ 9 = C ÷ 5.
  • Step 4 : F = 9C/5 + 32; K = C + 273.
Remember

F = 9C/5 + 32; K = C + 273; ΔC = ΔK.

Common mistake / CET trap

Use K, not °C, in ratios; 273 cancels in a change (Δ).

Shortcut

Memory clue : "C-five, F-nine".

Spot the difference

ScaleFreezingBoilingDivisions
°C0100100
°F32212180
K273373100

Practice questions

Easy
  1. 1Convert 37 °C to °F.
  2. 20 K = ? °C
  3. 3State the zeroth law.
Medium
  1. 450 °C = ? °F
  2. 5300 K = ? °C
  3. 6When is F = 2C?
Challenging
  1. 7When is F = 3C?
  2. 8P_tp = 50 kPa, P = 60 kPa : T?
Show answers (try them yourself first)
  1. 198.6 °F
  2. 2−273 °C
  3. 3A~B, B~C ⇒ A~C
  4. 4122 °F
  5. 527 °C
  6. 6C = 160 °C
  7. 7C = 26.7 °C
  8. 8327.8 K

Thermal expansion

From daily life

Railway tracks keep small gaps — they lengthen in summer.

Simple definition

Thermal expansion is the increase in size of a body when temperature rises.

Thermal expansionLL + ΔLΔL = α L ΔTβ = 2αγ = 3αcold above, hot below

In the figure : The hot rod is longer.

Rule 1: Linear expansion — ΔL = α L ΔT; α = coefficient of linear expansion (K⁻¹); steel ≈ 1.2 × 10⁻⁵ K⁻¹.

  • Easy : L = 10 m, ΔT = 30 K ⇒ ΔL = 1.2 × 10⁻⁵ × 10 × 30 = 3.6 mm.
  • Medium : Aluminium α = 2.4 × 10⁻⁵, L = 2 m, 20→120 °C ⇒ ΔL = 4.8 mm.
  • Exam (CET) : ΔL ∝ ? (1) L ΔT (2) L/ΔT (3) ΔT/L (4) L² → answer 1)

Rule 2: Area and volume — β = 2α, γ = 3α; ΔA = β A ΔT, ΔV = γ V ΔT. A hole expands like the material.

  • Easy : α = 1.2 × 10⁻⁵ ⇒ γ = 3.6 × 10⁻⁵.
  • Medium : Plate A = 0.5 m², ΔT = 100, α = 1.2 × 10⁻⁵ ⇒ ΔA = 1.2 × 10⁻³ m².
  • Exam (CET) : α : β : γ = ? (1) 1 : 2 : 3 (2) 1 : 3 : 2 (3) 3 : 2 : 1 (4) 1 : 1 : 1 → answer 1)

Rule 3: Liquids and thermal stress — γ_real = γ_apparent + γ_vessel. A rod held rigid develops stress = Y α ΔT.

  • Easy : Hg γ = 1.8 × 10⁻⁴, glass γ = 2.7 × 10⁻⁵ ⇒ apparent = 1.53 × 10⁻⁴.
  • Medium : Y = 2 × 10¹¹, α = 1.2 × 10⁻⁵, ΔT = 50 ⇒ stress 1.2 × 10⁸ Pa.
  • Exam (CET) : Water is densest at? (1) 4 °C (2) 0 °C (3) 100 °C (4) −4 °C → answer 1) Trick : 'densest at four'.

Derivation : γ = 3α

  • Step 1 : Cube of side L : V = L³.
  • Step 2 : dV = 3 L² dL.
  • Step 3 : dL = α L ΔT ⇒ ΔV = 3 L³ α ΔT.
  • Step 4 : ΔV = γ V ΔT ⇒ γ = 3α.
Remember

ΔL = αLΔT; β = 2α; γ = 3α.

Common mistake / CET trap

Unit of α is K⁻¹; a hole grows, not shrinks.

Shortcut

Memory clue : "One-two-three : α, β, γ".

Spot the difference

CoefficientFormulaRelation
αΔL/(LΔT)α
βΔA/(AΔT)2α
γΔV/(VΔT)3α

Practice questions

Easy
  1. 1Formula for ΔL?
  2. 2Relation between γ and α?
  3. 3At what temperature is water densest?
Medium
  1. 4L = 5 m, α = 1.2 × 10⁻⁵, ΔT = 40 : ΔL?
  2. 5α = 2 × 10⁻⁵ : γ?
  3. 6What happens to a hole when the plate is heated?
Challenging
  1. 7Prove γ = 3α.
  2. 8Apparent γ of Hg in a glass tube?
Show answers (try them yourself first)
  1. 1αLΔT
  2. 2γ = 3α
  3. 34 °C
  4. 42.4 mm
  5. 56 × 10⁻⁵
  6. 6The hole becomes larger
  7. 7V = L³ ⇒ ΔV = 3L²ΔL
  8. 81.53 × 10⁻⁴

Specific heat and heat capacity

From daily life

At noon sand burns your feet, but the sea nearby is only warm.

Simple definition

Specific heat is the heat needed to raise 1 kg by 1 K.

Specific heatlarge csmall cTQQ = m c ΔT

In the figure : Small c ⇒ temperature rises faster.

Rule 1: Formula — Q = m c ΔT; c = specific heat (J kg⁻¹ K⁻¹). Water 4200, ice 2100, aluminium 900, copper 400.

  • Easy : 2 kg water 20→70 °C ⇒ Q = 2 × 4200 × 50 = 4.2 × 10⁵ J.
  • Medium : 0.5 kg copper ΔT = 10 ⇒ Q = 0.5 × 400 × 10 = 2000 J.
  • Exam (CET) : Same Q and m : larger c ⇒ ΔT? (1) smaller (2) larger (3) same (4) zero → answer 1)

Rule 2: Heat capacity — C = m c = Q ÷ ΔT (J/K); molar : C_m = Q ÷ (n ΔT).

  • Easy : m = 0.5 kg water ⇒ C = 2100 J/K.
  • Medium : A 1 kW heater warms 5 kg water by 20 K ⇒ t = 5 × 4200 × 20 ÷ 1000 = 420 s.
  • Exam (CET) : Unit of C? (1) J/K (2) J (3) K (4) J/kg → answer 1)

Rule 3: Why water matters — Water's c is large ⇒ it heats and cools slowly : sea breeze, coolants, hot-water bags.

  • Easy : Same Q : sand warms quickly, water does not.
  • Medium : Same m and Q : ΔT ∝ 1/c.
  • Exam (CET) : Water in a hot-water bag because? (1) large c (2) small c (3) light (4) cheap → answer 1)

Derivation : Q = m c ΔT

  • Step 1 : Experiment : Q ∝ m.
  • Step 2 : Q ∝ ΔT.
  • Step 3 : Together : Q ∝ m ΔT.
  • Step 4 : The constant is c ⇒ Q = m c ΔT; c = Q/(mΔT).
Remember

Q = mcΔT; C = mc.

Common mistake / CET trap

c (per kg) and C (whole body) differ.

Shortcut

Memory clue : "m-c-delta-T".

Spot the difference

Materialc (J/kg K)
Water4200
Ice2100
Aluminium900
Copper400

Practice questions

Easy
  1. 1Formula for Q?
  2. 2Unit of C?
  3. 3c of water?
Medium
  1. 43 kg water 10→40 °C : Q?
  2. 5m = 0.2 kg Al, ΔT = 50 : Q?
  3. 62 kW heater, 10 kg water, 10 K : t?
Challenging
  1. 7Same Q and m, c₁ = 2c₂ : ΔT₁/ΔT₂?
  2. 8Derive Q = mcΔT.
Show answers (try them yourself first)
  1. 1mcΔT
  2. 2J/K
  3. 34200 J/kg K
  4. 43.78 × 10⁵ J
  5. 59000 J
  6. 6210 s
  7. 7½
  8. 8From experiment Q ∝ mΔT

Calorimetry

From daily life

Pouring cold milk into hot tea gives a temperature in between.

Simple definition

Calorimetry is the measurement of heat exchange.

Calorimeterwater + bodylost = gainedm₁c₁(T₁−T) = m₂c₂(T−T₂)

In the figure : Mixing inside a calorimeter.

Rule 1: Principle of calorimetry — With no loss to surroundings, heat lost = heat gained. Mixture temperature T = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂).

  • Easy : 200 g water 80 °C + 300 g water 20 °C ⇒ T = 44 °C.
  • Medium : 100 g Al (900) at 100 °C + 200 g water at 20 °C ⇒ T = 27.74 °C.
  • Exam (CET) : Same substance, equal m, T₁ and T₂ ⇒ T? (1) (T₁+T₂)/2 (2) T₁ (3) T₂ (4) T₁T₂ → answer 1) Trick : average.

Rule 2: Water equivalent — w = m c ÷ c_water; the mass of water that absorbs as much heat as the body (kg).

  • Easy : 100 g Al ⇒ w = 0.1 × 900 ÷ 4200 = 0.0214 kg.
  • Medium : Add the calorimeter's w to the water mass in the equation.
  • Exam (CET) : Unit of w? (1) kg (2) J (3) K (4) W → answer 1)

Rule 3: Electric heater — V I t = m c ΔT (energy = heat); power P = Q ÷ t.

  • Easy : 12 V, 5 A, 100 s ⇒ 6000 J.
  • Medium : 6000 J ⇒ 0.5 kg water ΔT = 2.86 K.
  • Exam (CET) : P = ? (1) Q/t (2) Qt (3) t/Q (4) Q² → answer 1)

Derivation : Mixture temperature

  • Step 1 : The hot body cools from T₁ to T : Q₁ = m₁c₁(T₁ − T).
  • Step 2 : The cold body warms from T₂ to T : Q₂ = m₂c₂(T − T₂).
  • Step 3 : Q₁ = Q₂ : m₁c₁T₁ − m₁c₁T = m₂c₂T − m₂c₂T₂.
  • Step 4 : Solve for T : T = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂).
Remember

Lost = gained; T = Σ mcT ÷ Σ mc.

Common mistake / CET trap

Use (T₁ − T) and (T − T₂); do not forget the calorimeter's heat capacity.

Shortcut

Memory clue : "Hot loses, cold gains".

Spot the difference

QuantityFormulaUnit
Qm c ΔTJ
wm c / c_waterkg
PQ / tW

Practice questions

Easy
  1. 1State the principle of calorimetry.
  2. 2What is water equivalent?
  3. 3P = ?
Medium
  1. 41 kg water at 60 °C + 1 kg at 20 °C : T?
  2. 50.5 kg Cu (400) at 100 °C + 1 kg water at 20 °C : T?
  3. 60.2 kg Al : w?
Challenging
  1. 7Derive the formula for T.
  2. 8A 500 W heater, 2 kg water, 25 K : t?
Show answers (try them yourself first)
  1. 1Heat lost = heat gained
  2. 2Mass of water absorbing the same heat
  3. 3Q/t
  4. 440 °C
  5. 523.64 °C
  6. 60.043 kg
  7. 7Equate Q₁ = Q₂
  8. 8420 s

Change of state and latent heat

From daily life

A drink stays cold while ice melts; boiling water stops getting hotter.

Simple definition

Latent heat is the heat for a change of state (temperature constant).

Heating curve (ice → steam)icemeltingwaterboiling0°C100°CQTQ = m L

In the figure : The flat parts = change of state.

Live figure

Change of state and latent heat

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Rule 1: Latent heat — Q = m L; the state changes at constant temperature. Ice melting L_f = 3.35 × 10⁵ J/kg; boiling L_v = 2.26 × 10⁶ J/kg.

  • Easy : 2 kg ice 0 °C → water : Q = 6.70 × 10⁵ J.
  • Medium : 0.5 kg ice at −10 °C → steam at 100 °C : Q = 1.518 × 10⁶ J.
  • Exam (CET) : Temperature while melting? (1) constant (2) rises (3) falls (4) zero → answer 1) Trick : 'latent = temperature silent'.

Rule 2: Heating curve — Four parts : ice (mcΔT) → melting (mL) → water (mcΔT) → boiling (mL). Total Q is the sum.

  • Easy : 0.5 kg ice −10→0 : 10500 J.
  • Medium : Melting : 167500 J; water 0→100 : 210000 J; boiling 1130000 J.
  • Exam (CET) : Most heat? (1) boiling (2) melting (3) heating ice (4) heating water → answer 1)

Rule 3: Ice + water — First check : heat the water gives while cooling to 0 °C versus heat needed to melt all the ice.

  • Easy : 100 g ice at 0 °C + 400 g water at 50 °C : available 84000 J > melting 33500 J ⇒ all melts; final T = 24.0 °C.
  • Medium : Conversely : little warm water + lots of ice ⇒ T = 0 °C.
  • Exam (CET) : Temperature of a mixture of ice and water? (1) 0 °C (2) 4 °C (3) 100 °C (4) −10 °C → answer 1)

Derivation : Ice + water : final T

  • Step 1 : Heat from water cooling to 0 °C : m_w c_w (T_w − 0).
  • Step 2 : Heat to melt the ice : m_i L_f.
  • Step 3 : The balance warms all the water : m_w c_w T_w − m_i L_f = (m_w + m_i) c_w T.
  • Step 4 : T = (m_w c_w T_w − m_i L_f) ÷ ((m_w + m_i) c_w).
Remember

Q = mL; T constant while melting and boiling.

Common mistake / CET trap

Do not use mcΔT during a change of state; use mL.

Shortcut

Memory clue : "Heat ice → melt → heat → boil".

Spot the difference

ChangeFormulaTemperature
Melting/freezingm L_fconstant
Boiling/condensingm L_vconstant
Heatingm c ΔTchanges

Practice questions

Easy
  1. 1What is latent heat?
  2. 2L_f of ice?
  3. 3Temperature while boiling?
Medium
  1. 4Melt 1 kg ice at 0 °C : Q?
  2. 50.2 kg water at 100 °C → steam : Q?
  3. 60.5 kg ice −20→0 °C : Q (c = 2100)?
Challenging
  1. 7200 g ice at 0 °C + 1 kg water at 40 °C : T?
  2. 8Why does steam scald more than boiling water?
Show answers (try them yourself first)
  1. 1Heat that changes state at constant temperature
  2. 23.35 × 10⁵ J/kg
  3. 3Constant
  4. 43.35 × 10⁵ J
  5. 54.52 × 10⁵ J
  6. 621000 J
  7. 720.0 °C
  8. 8Steam also releases L_v on condensing

Conduction

From daily life

A steel spoon in hot tea warms quickly; a wooden one does not.

Simple definition

Conduction is heat transfer through a material without bulk motion.

ConductionT₁T₂H = Q/tLH = k A (T₁ − T₂) ÷ LA = cross-section

In the figure : Heat flows from the hot end to the cold end.

Rule 1: Law of conduction — In steady state H = Q ÷ t = k A (T₁ − T₂) ÷ L; k = thermal conductivity (W m⁻¹ K⁻¹). Copper 400, steel 50, glass ≈ 1.

  • Easy : Cu k = 400, A = 10⁻⁴, L = 0.5, ΔT = 100 ⇒ H = 8 W.
  • Medium : L doubled ⇒ H halves.
  • Exam (CET) : H ∝ ? (1) A ΔT / L (2) L / A (3) A L (4) ΔT L → answer 1)

Rule 2: Series — Equal L, A : rods in series ⇒ same H; junction T : k₁(T₁ − T) = k₂(T − T₂). R = L ÷ kA, R adds.

  • Easy : k₁ = 400, k₂ = 200, 100 °C and 0 °C ⇒ T = 66.7 °C.
  • Medium : k_eq = 2k₁k₂ ÷ (k₁ + k₂) = 266.7.
  • Exam (CET) : Series R? (1) adds (2) multiplies (3) same (4) zero → answer 1)

Rule 3: Parallel — Same ΔT : H adds; k_eq = (k₁A₁ + k₂A₂) ÷ (A₁ + A₂); equal A : k_eq = (k₁ + k₂) ÷ 2.

  • Easy : k = 400 and 200, equal A ⇒ k_eq = 300.
  • Medium : Add the H of both rods.
  • Exam (CET) : What is equal in parallel? (1) ΔT (2) H (3) R (4) L → answer 1)

Derivation : Law of conduction

  • Step 1 : For a thin slab dx : experiment gives H ∝ A.
  • Step 2 : H ∝ temperature gradient dT/dx.
  • Step 3 : H = −k A dT/dx (minus = toward the cold side).
  • Step 4 : For a linear gradient dT/dx = (T₁ − T₂)/L ⇒ H = kA(T₁ − T₂)/L.
Remember

H = kAΔT/L; R = L/kA; same H in series, same ΔT in parallel.

Common mistake / CET trap

In series k_eq is not the average (2k₁k₂/(k₁ + k₂)); equal L and A required.

Shortcut

Memory clue : "Series = same H; parallel = same ΔT".

Spot the difference

ConnectionSameTotal
SeriesHR = R₁ + R₂
ParallelΔTH = H₁ + H₂

Practice questions

Easy
  1. 1Formula for H?
  2. 2Unit of k?
  3. 3Name a good conductor.
Medium
  1. 4k = 50, A = 2 × 10⁻⁴, L = 0.2, ΔT = 80 : H?
  2. 5A doubled, L doubled : H?
  3. 6k = 400 and 100 in series : k_eq?
Challenging
  1. 7k₁ = 300, k₂ = 100 in series, 80 °C and 20 °C : T?
  2. 8Derive the law of conduction.
Show answers (try them yourself first)
  1. 1kA(T₁ − T₂)/L
  2. 2W m⁻¹ K⁻¹
  3. 3Copper
  4. 44 W
  5. 5Unchanged
  6. 6160
  7. 765 °C
  8. 8From H ∝ A dT/dx

Convection, radiation and Newton's law of cooling

From daily life

The Sun's heat crosses empty space; hot tea slowly cools.

Simple definition

Radiation is heat by electromagnetic waves; convection is heat by fluid motion.

Newton's law of coolingT₀TtT = T₀ + (T₁ − T₀) e^(−kt)−dT/dt = k (T − T₀)

In the figure : Cooling slows as the difference shrinks.

Live figure

Convection, radiation and Newton's law of cooling

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Rule 1: Convection — Heat is carried by the actual motion of a fluid. Natural (air, sea breeze) and forced (fan, pump).

  • Easy : Hot air rises and cold air sinks.
  • Medium : A heater is placed low, an AC high.
  • Exam (CET) : Convection occurs in? (1) fluids (2) solids (3) vacuum (4) everywhere → answer 1)

Rule 2: Radiation — Works even in vacuum; Stefan : P = σ e A T⁴ (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴); Wien : λ_m T = 2.9 × 10⁻³ m K.

  • Easy : A = 1 m², T = 500 K, e = 1 ⇒ P = 3544 W; T doubled ⇒ 16 times.
  • Medium : T = 5800 K ⇒ λ_m = 500 nm; net : σA(T⁴ − T₀⁴) = 6889 W/m² (600 K, 300 K).
  • Exam (CET) : T doubled ⇒ P? (1) 16 times (2) 2 times (3) 4 times (4) 8 times → answer 1) Trick : fourth power.

Rule 3: Newton's law of cooling — −dT/dt = k (T − T₀) (small difference); average form : (T₁ − T₂) ÷ t = k [(T₁ + T₂)/2 − T₀].

  • Easy : 80→60 °C in 10 min, T₀ = 20 ⇒ k = 0.04 min⁻¹.
  • Medium : 60→40 °C : t = 20 ÷ (0.04 × 30) = 16.7 min.
  • Exam (CET) : Rate of cooling depends on? (1) T − T₀ (2) T (3) T₀ (4) T² → answer 1)

Derivation : T = T₀ + (T₁ − T₀) e^(−kt)

  • Step 1 : Set up : dT/dt = −k (T − T₀).
  • Step 2 : Separate variables : dT ÷ (T − T₀) = −k dt.
  • Step 3 : Integrate : ln(T − T₀) = −kt + C.
  • Step 4 : At t = 0, T = T₁ ⇒ T − T₀ = (T₁ − T₀) e^(−kt).
Remember

P = σeAT⁴; λ_m T = 2.9 × 10⁻³; −dT/dt = k(T − T₀).

Common mistake / CET trap

T in kelvin in Stefan's law; the cooling law is for small differences.

Shortcut

Memory clue : "T-fourth, Wien = constant".

Spot the difference

ModeMediumLaw
ConductionsolidH = kAΔT/L
Convectionfluid—
Radiationworks in vacuumP = σeAT⁴

Practice questions

Easy
  1. 1Does radiation need a medium?
  2. 2State Stefan's law.
  3. 3Write Newton's law of cooling.
Medium
  1. 4T = 1000 K, A = 2 m², e = 1 : P?
  2. 5T = 3000 K : λ_m?
  3. 6T tripled : P?
Challenging
  1. 790→70 °C in 5 min, T₀ = 30 : k?
  2. 8Derive the cooling formula.
Show answers (try them yourself first)
  1. 1No
  2. 2P = σ e A T⁴
  3. 3−dT/dt = k (T − T₀)
  4. 41.13 × 10⁵ W
  5. 5967 nm
  6. 681 times
  7. 70.08 min⁻¹
  8. 8Separate variables and integrate

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
Conversion(F−32)/9 = C/5 = (K−273)/5—
ExpansionΔL = αLΔT; γ = 3αK⁻¹
HeatQ = mcΔTJ; M L² T⁻²
cQ/(mΔT)J kg⁻¹ K⁻¹; L² T⁻² K⁻¹
Latent heatQ = mLJ kg⁻¹; L² T⁻²
ConductionH = kAΔT/LW; k : W m⁻¹ K⁻¹
StefanP = σeAT⁴W
Cooling−dT/dt = k(T − T₀)K s⁻¹

🎯 Most important points for the exam

  • Change in C : F : K = 5 : 9 : 5.
  • γ = 3α, β = 2α; holes expand.
  • Q = mcΔT; water has a large c.
  • Heat lost = heat gained.
  • T constant in a change of state : Q = mL.
  • Series : same H; parallel : same ΔT.
  • P ∝ T⁴; cooling rate ∝ (T − T₀).
Study mantra

Temperature → expansion → mcΔT → mL → conduction, radiation, cooling.