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Thermometry, Thermal Expansion & Calorimetry

01
CONCEPT

Temperature, Scales & Thermometers

In one line
Temperature tells how hot or cold a body is; heat always flows from higher to lower temperature.
TEMPERATURE

12.1.1 What temperature means

  1. One of the seven fundamental quantities; dimension [θ], SI unit kelvin.
  2. Two bodies are in thermal equilibrium when no heat flows between them: same temperature.
  3. Kinetic theory: temperature measures the average translational KE of molecules.
  4. No upper limit; the lowest possible temperature is 0 K.
  5. Human body: 37 °C = 98.6 °F = 310.15 K. NTP: 0 °C = 32 °F = 273.15 K.
SCALES

12.1.2 Two fixed points

  1. Lower fixed point (LFP): ice point. Upper fixed point (UFP): steam point.
Fahrenheit212 °F32 °FCelsius100 °C0 °CKelvin373 K273 Ksteamice
Same ice and steam points on three scales
ScaleLFPUFPDivisions
Celsius0 °C100 °C100
Fahrenheit32 °F212 °F180
Kelvin273.15 K373.15 K100
Reaumur0 °R80 °R80
Rankine460 Ra672 Ra212
CONVERSION

12.1.3 From one scale to another

reading − LFPUFP − LFP = same on every scale
C5 = F − 329 = K − 2735 = R4
  1. For a difference: ΔC5 = ΔF9 = ΔK5.
  2. ΔC = ΔK: one degree Celsius = one kelvin.
  3. ΔF = 1.8 ΔC: one degree Fahrenheit is smaller.
  4. Same reading on C and F: −40°. Same reading on K and F: 574.25.
32−17.8°F°C
C against F: straight line, F-intercept 32
THERMOMETRY

12.1.4 Measuring with a thermometric property

  1. A property X that changes linearly with temperature: length of a liquid column, pressure of a gas at constant volume, resistance of a platinum wire.
  2. X0 and X100: values at 0 °C and 100 °C.
t = X − X0X100 − X0 × 100 °C
  1. Modern method: one fixed point, the triple point of water (273.16 K), and 0 K.
T = 273.16 × XXTr K
THERMOMETERS

12.1.5 Liquid, gas and resistance

  1. Mercury thermometer: large, uniform expansion, good conductor, low specific heat.
  2. Range about −50 °C to 350 °C (up to 550 °C with nitrogen above the mercury). Alcohol: −80 °C to 350 °C.
t = l − l0l100 − l0 × 100 °C
  1. Liquid from a bulb of volume V rising in a stem of area A:
l = VγΔθA
  1. Constant volume gas thermometer: P ∝ T.
t = P − P0P100 − P0 × 100 °C  ·  T = 273.16 PPTr K
  1. Platinum resistance thermometer: R = R0(1 + αt); range −200 °C to 1200 °C.
t = R − R0R100 − R0 × 100 °C
JEE · NEET

Corner

Other scales
  1. Reaumur: ice 0 °R, steam 80 °R. Rankine: ice 460 Ra, steam 672 Ra.
  2. In SI the kelvin has no degree sign: write 273 K, not 273 °K.
Constant pressure gas thermometer
  1. V ∝ T at constant P.
t = V − V0V100 − V0 × 100 °C  ·  T = 273.16 VVTr K
  1. Ranges: hydrogen −200 °C to 500 °C, nitrogen −200 °C to 1600 °C, helium −268 °C to 500 °C.
  2. Gas thermometers are the most sensitive and are used to standardise others.
Thermoelectric thermometer
  1. Two different metals form a thermocouple; a temperature difference between the junctions gives an emf (Seebeck effect).
e = at + bt²
ThermocoupleRange
copper–iron0 °C to 260 °C
iron–constantan0 °C to 800 °C
tungsten–molybdenum2000 °C to 3000 °C
  1. Low thermal capacity: reads quickly changing temperature.
Pyrometer and vapour pressure thermometer
  1. Pyrometer: uses radiation (E ∝ T⁴); measures 800 °C to 6000 °C, not below 800 °C.
  2. Vapour pressure thermometer: very low temperatures, 0.71 K to 120 K.
log P = a + bT + cT
  1. Germanium resistance thermometer: 4 K to 77 K. Magnetic thermometer: near 2 K.
Practise PYQs on this topicThermometry & Thermal Expansion · 74 questions with solutions
02
CONCEPT

Linear Expansion

In one line
A heated solid gets longer; the increase is proportional to its original length and the rise in temperature.
THERMAL EXPANSION

12.2.1 Why bodies expand

  1. Atoms vibrate with larger amplitude when heated.
  2. The potential energy curve is not symmetric, so the average spacing increases.
  3. Expansion: least in solids, most in gases.
  4. Solids: linear, area and volume expansion. Liquids and gases: volume only.
LINEAR EXPANSION

12.2.2 Change in length

L₀ (at T)ΔLT + ΔT
Rod of length L₀ heated by ΔT
ΔL = L0αΔT
L = L0(1 + αΔT)
α = ΔLL0ΔT
  1. Unit: °C−1 or K−1 (same value); dimension [θ−1].
  2. α depends on the material, not on the unit of length.
  3. Percentage change in length = αΔT × 100; invar and quartz have very small α.
Materialα (10−5 K−1)
steel1.2
copper1.7
brass2.0
aluminium2.4
CONSTANT DIFFERENCE

12.2.3 Two rods whose difference never changes

  1. Both rods must expand by the same amount.
L1α1 = L2α2
  1. The longer rod has the smaller α.
  2. If L1 − L2 = d (with α1 < α2):
L1 = dα2α2 − α1  ·  L2 = dα1α2 − α1
COMPOSITE ROD

12.2.4 Two rods joined end to end

  1. Total expansion = sum of the two expansions.
αeff = L1α1 + L2α2L1 + L2
  1. Equal lengths: αeff = α1 + α22.
  2. Each part expands by the same amount ⇒ again L1α1 = L2α2.
GAPS AND RINGS

12.2.5 Everyday problems

  1. Rail gap = expansion of one rail = L0αΔT.
  2. A ring or tyre slightly smaller than the wheel is heated until it fits:
ΔT = ΔDDα
  1. The hole of a ring grows exactly like the metal around it.
  1. Equal rise twice: ΔL2 = ΔL1(1 + αΔT), since the rod is already longer.
Practise PYQs on this topicThermometry & Thermal Expansion · 74 questions with solutions
03
CONCEPT

Area & Volume Expansion

In one line
Every length grows by the same fraction, so area grows twice as fast and volume three times as fast.
AREA EXPANSION

12.3.1 Superficial expansion

A₀L₀L₀ΔLΔL
Each side grows by ΔL
ΔA = A0βΔT
A = A0(1 + βΔT)  ·  β = ΔAA0ΔT
VOLUME EXPANSION

12.3.2 Cubical expansion

V₀L₀L₀L₀
Each edge grows by ΔL
ΔV = V0γΔT
V = V0(1 + γΔT)  ·  γ = ΔVV0ΔT
α, β AND γ

12.3.3 One material, three coefficients

α = β2 = γ3  ·  α : β : γ = 1 : 2 : 3
  1. % change in area = 2 × % change in length.
  2. % change in volume = 3 × % change in length.
  3. Sphere: ΔV = 4πR³αΔT. Cube of side a: ΔV = 3a³αΔT.
  4. Unequal (anisotropic) solid: γ = αx + αy + αz.
Materialαγ
steel1.23.6
copper1.75.1
brass2.06.0
aluminium2.47.2
  1. Values in 10−5 K−1.
HOLES AND CAVITIES

12.3.4 Photographic enlargement

Aplate C with hole ABsolid disc Bexpands like
Hole A grows like a solid disc B of the same metal
  1. A hole or a cavity expands as if it were filled with the same material.
  2. A hollow box or sphere expands like the solid one of the same outer size.
  3. Given the same heat, the hollow sphere has less mass, so it heats more and expands more.
DENSITY

12.3.5 Density falls on heating

  1. Mass stays the same while volume grows: ρ ∝ 1V.
ρ′ = ρ1 + γΔT ≈ ρ(1 − γΔT)
  1. % fall in density = γΔT × 100 = 3 × % rise in length.
HEAT AND EXPANSION

12.3.6 When heat Q is given instead of ΔT

  1. Rise in temperature from the heat given:
ΔT = ΔQms ⇒ ΔA = A(2α)ΔQms
JEE · NEET

Corner

Expansion of gases
  1. Gases show only volume expansion (the vessel’s expansion is negligible).
V′ = V(1 + αΔθ)  ·  P′ = P(1 + βΔθ)
ideal gas: α = β = 1273 °C−1 (at 0 °C)
  1. At constant P, V ∝ T ⇒ γ = 1T.
  2. For a process V ∝ Tⁿ: γ = nT.
Heating against compression
  1. Pressure P compresses a solid by ΔVV = PK (K = bulk modulus).
to undo it by heating: ΔT = P3αK
Practise PYQs on this topicThermometry & Thermal Expansion · 74 questions with solutions
04
CONCEPT

Thermal Stress, Pendulum Clock & Liquids

In one line
When expansion is blocked or unequal, heating gives force, bending or a clock error.
THERMAL STRESS

12.4.1 Rod fixed between two walls

force on walls F = YAαΔθheated, cannot expand
The rod presses on the supports
thermal strain = αΔθ
stress = YαΔθ  ·  F = YAαΔθ
  1. A rod free to expand has no thermal stress.
  2. F does not depend on the length of the rod.
  3. Heating: compressive stress. Cooling: tensile stress.
PENDULUM CLOCK

12.4.2 Time lost or gained

ll + Δl
Heating lengthens the pendulum
ΔTT = 12αΔθ
time lost or gained per day = 12αΔθ × 86400 s
  1. θ′ > θ (summer): clock runs slow, loses time.
  2. θ′ < θ (winter): clock runs fast, gains time.
  3. Independent of the period T; pendulums are made of invar.
BIMETALLIC STRIP

12.4.3 Two metals joined

steelbrasscold: straighthot: bendsbrass (α bigger)on the outer side
Metal with larger α on the outer side
  1. Heating: bends with the larger-α metal on the convex (outer) side.
  2. Cooling: bends towards the larger-α metal.
  3. Used in thermostats and fire alarms.
EXPANSION OF LIQUIDS

12.4.4 Real and apparent

QPRPQ: vessel expands (level falls)QR: real expansion of liquidPR: apparent expansion
Level first falls (P → Q), then rises (Q → R)
γreal = γapparent + γvessel
ΔVapp = V(γreal − 3αvessel)Δθ
  1. γreal > 3α: level rises. Equal: level stays. Smaller: level falls.
  2. Vacant space in a vessel stays constant when Vliquidγliquid = Vvesselγvessel.
JEE · NEET

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Anomalous expansion of water
4 °C0 °CVθV min, ρ max
Water has least volume (greatest density) at 4 °C
  1. From 0 °C to 4 °C water contracts on heating (γ negative).
  2. A lake freezes from the top; water at the bottom stays at 4 °C, so aquatic life survives.
Effect on upthrust
Th′Th = 1 + γsΔθ1 + γLΔθ
  1. γs < γL ⇒ upthrust falls on heating; apparent weight rises.
More applications
  1. Gaps are left between rails; transmission cables are not pulled tight.
  2. Test tubes and crucibles use pyrex (very small α).
  3. A stuck glass stopper is freed by warming the neck.
  4. Rubber-like materials contract on heating.
  5. Scale error: true value = scale reading [1 + α(θ′ − θ)].
Practise PYQs on this topicThermometry & Thermal Expansion · 74 questions with solutions
05
CONCEPT

Specific Heat, Thermal Capacity & Joule’s Law

In one line
Heat needed to warm a body is Q = mcΔθ; work and heat are linked by W = JQ.
HEAT

12.5.1 Energy in transit

  1. Heat is energy exchanged because of a temperature difference.
  2. Scalar; units joule, calorie, kcal. 1 kcal = 1000 cal.
  3. 1 cal: heat to warm 1 g of water from 14.5 °C to 15.5 °C.
  4. 1 cal = 4.18 J ≈ 4.2 J.
  5. Heat flows from hot to cold until the temperatures are equal.
SPECIFIC HEAT

12.5.2 Heat per unit mass per degree

Q = mcΔθ
c = QmΔθ
  1. Units: J kg−1 K−1 or cal g−1 °C−1; dimension [L²T−2θ−1].
  2. Depends on the material, not on its mass.
Substancecal g−1 °C−1J kg−1 K−1
water14200
ice0.52100
steam0.47≈ 1970
aluminium0.21900
copper0.09386
  1. Water has the largest specific heat of common solids and liquids: used in radiators and hot-water bottles.
MOLAR SPECIFIC HEAT

12.5.3 Per mole

C = Mc = QμΔθ
  1. Unit J mol−1 K−1; M = molar mass, μ = number of moles.
THERMAL CAPACITY

12.5.4 And water equivalent

thermal capacity = mc = QΔθ
  1. Heat needed to raise the whole body by 1 °C; unit J/K.
  2. Depends on both mass and material: Q ∝ m ∝ r³ for spheres of one metal.
water equivalent W = mc
  1. Mass of water that takes the same heat for the same rise; unit kg (or g).
  2. Numerically equal to the thermal capacity in cal°C.
HEATERS

12.5.5 Power and rise in temperature

Pt = mcΔθ
  1. With a heat loss: (P − Ploss)t = mcΔθ.
  2. Flowing water: P = mtcΔθ.
  3. Fuel needed per second = heat per second ÷ heat of combustion.
  4. Liquid heated with negligible expansion: ΔU = Q = mcΔθ.
JOULE’S LAW

12.5.6 Mechanical equivalent of heat

W = JQ  ·  J = WQ
  1. J = 4.2 J/cal = 4.2 × 107 erg/cal = 4.2 × 103 J/kcal.
  2. J is a conversion factor, not a physical quantity.
  1. Waterfall of height h:
mgh = mcΔθ ⇒ Δθ = ghc  (Δθ = ghJc with c in cal)
  1. Bullet stopping, fraction f of its KE becoming heat in it:
f × 12mv² = J msΔθ ⇒ Δθ = fv²2Js
  1. Bullet that also melts:
12mv² = J(msΔθ + mL)
  1. Ice falling from h and melting completely: h = JLg.
JEE · NEET

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Specific heat of a gas
  1. Compressed suddenly (ΔQ = 0) with a rise in T: c = 0.
  2. Heated while expanding at constant T: c = ∞.
  3. Can be negative if it cools while taking heat.
  4. Two useful values: CP and CV (see Kinetic Theory).
Specific heat of solids
  1. CV → 0 as T → 0; at low T, C ∝ T³ (Debye).
  2. Dulong–Petit: at room temperature molar specific heat of metals ≈ 3R ≈ 25 J mol−1 K−1 (6 cal mol−1 K−1).
  3. Specific heat of water is least near 37 °C.
  4. 1 BTU = 252 cal = 1055 J.
Practise PYQs on this topicCalorimetry, Latent Heat & Mechanical Equivalent of Heat · 56 questions with solutions
06
CONCEPT

Latent Heat, Change of State & Heating Curve

In one line
During a change of state heat is taken or given out, but the temperature stays constant.
CHANGE OF STATE

12.6.1 Solid, liquid, vapour

  1. At 1 atm, ice melts at 0 °C and water boils at 100 °C.
  2. Melting and boiling: heat is absorbed; molecules move apart, internal energy rises.
  3. Freezing and condensing: the same heat is given out.
LATENT HEAT

12.6.2 Heat for a change of state

Q = mL
  1. Unit J/kg or cal/g; dimension [L²T−2].
Changecal/gJ/kg
ice → water (Lf)80336 × 10³
water → steam (Lv)5402260 × 10³
  1. Lv > Lf: boiling needs a large increase in volume.
  2. Steam burns more than boiling water: each gram gives 540 cal extra on condensing.
  3. Melting point falls with pressure for ice (regelation); boiling point rises with pressure.
ΔU ON BOILING

12.6.3 Part of the heat does work

ΔU = mL − PΔV
  1. 1 g water → 1671 cm³ steam at 105 Pa: W = 167 J, ΔU = 2256 − 167 = 2089 J.
HEATING CURVE

12.6.4 Heat at constant rate P

ABCDT₁T₂solidmeltingliquidboilinggasheat (time)T
Temperature against time for a solid heated at a constant rate
  1. OA: solid warms. Slope ∝ 1csolid.
  2. AB: melting at T1; flat; length ∝ Lf.
  3. BC: liquid warms. Slope ∝ 1cliquid.
  4. CD: boiling at T2; flat and longer; length ∝ Lv.
  5. DE: vapour warms.
c = Pm × slope  ·  L = P(t2 − t1)m
  1. Same power, equal masses: cA : cB = slopeB : slopeA.
  2. Time ratio of two flat parts = ratio of latent heats.
ICE TO STEAM

12.6.5 Total heat in stages

  1. Add every stage separately:
Q = m[cice(0 − θ1) + Lf + cw(100) + Lv + csteam(θ2 − 100)]
  1. A body cooled to 0 °C on ice melts m = Q ÷ Lf of it.
  2. Heat Q given to ice below 0 °C: first warm it to 0 °C, then the rest melts Q ÷ Lf.
JEE · NEET

Corner

Phase diagram and triple point
triple pointcritical pointSLVPT
P–T diagram of water
  1. Three curves: sublimation (S–V), vaporisation (L–V), fusion (S–L).
  2. Triple point: solid, liquid and vapour together (water: 273.16 K).
  3. Water (and ice-like substances) contract on melting: melting point falls with pressure (regelation).
  4. Boiling point rises with pressure; latent heat of vaporisation falls as boiling point rises.
Vapour and humidity
  1. Saturated vapour pressure depends only on temperature.
  2. Boiling: SVP = outside pressure. Dew point: air just saturated.
RH = mM × 100% = SVP at dew pointSVP at given temperature × 100%
  1. Freezing mixture (salt + ice) goes below 0 °C.
  2. Evaporation causes cooling: if water is made to evaporate by pumping, mvLv = mfLf.
Practise PYQs on this topicCalorimetry, Latent Heat & Mechanical Equivalent of Heat · 56 questions with solutions
07
CONCEPT

Principle of Calorimetry: Mixtures

In one line
In an insulated mixture, heat lost by the hot bodies = heat gained by the cold bodies.
PRINCIPLE

12.7.1 Conservation of heat energy

icehot body givescold body takeswaterheat lost = heat gained
No heat lost to the surroundings
heat lost = heat gained
  1. The mixture temperature lies between the two: θL ≤ θmix ≤ θH.
  2. The calorimeter takes heat too: add its water equivalent to the water.
  3. Rise of one body need not equal the fall of the other; heats are equal.
NO CHANGE OF STATE

12.7.2 Two bodies mixed

m1c1(θ1 − θmix) = m2c2(θmix − θ2)
θmix = m1c1θ1 + m2c2θ2m1c1 + m2c2
Conditionθmix
same materialm1θ1 + m2θ2m1 + m2
same massc1θ1 + c2θ2c1 + c2
same mass and materialθ1 + θ22
  1. Hot metal in water: heat capacity of metal = heat gained by water ÷ its fall in temperature.
  2. Three liquids: first find the c-ratios from the two given mixtures, then mix the third pair.
ICE AND WATER

12.7.3 Mixing with a change of state

  1. Water mw at θw and ice mi at 0 °C, all ice melting:
mwcw(θw − θmix) = miL + micwθmix
θmix = mwθw − miL ÷ cwmw + mi
  1. If this comes out negative, θmix = 0 °C and only part of the ice melts:
mmelted = mwcwθwL
  1. Equal masses of ice and water: all ice melts only if θw ≥ 80 °C.
  2. Ice below 0 °C: add micice(0 − θi) to the heat gained.
STEAM

12.7.4 Steam with ice or water

  1. 1 g steam at 100 °C gives 540 cal on condensing and 100 cal more on cooling to 0 °C: 640 cal in all.
  2. 1 g ice at 0 °C needs 80 + 100 = 180 cal to become water at 100 °C.
  3. 1 g steam + 1 g ice ⇒ mixture stays at 100 °C (steam left over).
  4. Steam just enough to melt m g of ice: m ÷ 8 g.
METHOD

12.7.5 Solving any mixture question

  1. Step 1: find the heat the hot side can give down to 0 °C (or to the lower temperature).
  2. Step 2: compare with the heat the cold side needs to melt fully.
  3. Step 3: write heat lost = heat gained stage by stage, with water equivalents.
Practise PYQs on this topicCalorimetry, Latent Heat & Mechanical Equivalent of Heat · 56 questions with solutions
EXAM-DAY CARD

Just see this before the exam

Thermometry,Expansion & Heat
C5 = F − 329
scales
ΔL = L0αΔT
linear
α : β : γ = 1 : 2 : 3
area, volume
Q = mcΔθ
specific heat
Q = mL
latent heat
heat lost = heat gained
mixtures
FORMULA SHEET

Whole chapter on one page

Temperature

C5 = F − 329 = K − 2735

ΔF = 1.8ΔC, ΔK = ΔC

t = X − X0X100 − X0 × 100

T = 273.16 PPTr K

Linear

ΔL = L0αΔT

constant difference: L1α1 = L2α2

αeff = L1α1 + L2α2L1 + L2

ring: ΔT = ΔDDα

Area & volume

β = 2α, γ = 3α

ΔV = V0γΔT

ρ′ = ρ(1 − γΔT)

hole grows like solid

Stress & liquids

F = YAαΔθ

ΔTT = 12αΔθ

γr = γa + γv

l = VγΔθA (stem)

Heat

Q = mcΔθ, Q = mL

W = JQ, J = 4.2 J/cal

Δθ = ghc, Δθ = fv²2Js

ΔU = mL − PΔV

Mixtures

heat lost = heat gained

θmix = m1c1θ1 + m2c2θ2m1c1 + m2c2

Lf = 80, Lv = 540 cal/g

mmelt = mwcwθwL

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Temperature difference: ΔF = ΔC

γ = 2α or γ = α³

Pendulum: ΔTT = αΔθ

Ice + water: a negative θmix

CORRECT ✓

ΔF = 1.8 ΔC, but ΔK = ΔC

α : β : γ = 1 : 2 : 3

ΔTT = 12αΔθ

θmix = 0 °C, part of the ice melts