Heat Transfer: Conduction, Radiation & Cooling: theory

CONCEPT

Conduction, Thermal Conductivity & Thermal Resistance

In one line
Heat flows from hot to cold; in conduction it passes from particle to particle while the particles stay in place.

MODES

13.1.1 Three ways heat travels

Conduction: particle to particle, no flow of matter. Needs a medium; slow.
Convection: hot fluid itself moves and carries the heat.
Radiation: electromagnetic waves; no medium needed (sun to earth).

Mode	Medium	Where
conduction	needed	mainly solids
convection	needed	liquids, gases
radiation	not needed	even vacuum

CONDUCTION

13.1.2 How solids pass on heat

Metals: free electrons carry heat, so they are good conductors.
Non-metals and fluids: only vibrations of molecules, so poor conductors.
Liquids and gases heated from the top transfer heat downward by conduction only.

STEADY STATE

13.1.3 When temperatures stop changing

Rod of length l and area A between θ₁ and θ₂

Variable state: every section is still getting hotter.
Steady state: temperature of each section stays constant; the rod absorbs no more heat.
In steady state, for a uniform rod with no side loss, θ falls linearly along the rod.

Steady state: θ falls linearly with x

GRADIENT

13.1.4 Temperature gradient

temperature gradient = −dθdx = θ₁ − θ₂l

Unit: K m⁻¹ or °C m⁻¹; dimension [L⁻¹θ].
Minus sign: θ falls along the direction of heat flow.

FOURIER’S LAW

13.1.5 Heat through a rod

Q = KA(θ₁ − θ₂)tl

H = Qt = KA(θ₁ − θ₂)l

H is the heat current (rate of flow of heat), in J s⁻¹ = W.
For a cylindrical rod A = πr², so H ∝ r²l: double every length ⇒ H doubles.

CONDUCTIVITY K

13.1.6 Thermal conductivity

K depends only on the material, not on length, area or mass.
Units: W m⁻¹ K⁻¹ (SI), cal cm⁻¹ s⁻¹ °C⁻¹; dimension [MLT⁻³θ⁻¹].
Perfect conductor K = ∞; perfect insulator K = 0.
K_Ag > K_Cu > K_Al; K_solid > K_liquid > K_gas.
Air pockets (fur, two blankets, foam, double walls) make poor conductors.

Substance	K (W m⁻¹ K⁻¹)
silver	420
copper	400
aluminium	240
iron	80
glass	0.9
water	0.6
air	0.024

THERMAL RESISTANCE

13.1.7 Heat flow like electric current

R = lKA · H = θ₁ − θ₂R

Like Ohm’s law: temperature difference ↔ potential difference, heat current ↔ current.
Unit: K s J⁻¹ (or °C s cal⁻¹); dimension [M⁻¹L⁻²T³θ].
In steady state, θ difference ∝ R: a good conductor (large K) has a small drop across it.

MELTING ICE

13.1.8 Heat conducted into ice

Heat through the rod in time t melts ice at 0 °C:

mL = KA(θ₁ − θ₂)tl

CONVECTION

13.1.9 Heat carried by moving fluid

Hot fluid rises, cool fluid sinks

Natural: density difference under gravity; heating is done from below. Not possible in a freely falling lift or an orbiting satellite.
Forced: fluid is pushed (fan, pump, blood by the heart).
Examples: room heater, sea breeze, trade winds, ventilation.

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Variable state and isothermal surfaces

Before steady state every section is heating up: part of the heat raises its temperature, part is passed on, part is lost from the sides.
Isothermal surface: all points at one temperature. Heat flows at right angles to it: plane surfaces in a rod, cylinders around a hot wire, spheres around a point source.

dQdt = −KAdθdx (any shape, any state)

Searle’s method and Ingen-Hausz experiment

Searle: heat through the rod in time t warms water flowing past the cold end from θ₃ to θ₄.

K = mc(θ₄ − θ₃)lA(θ₁ − θ₂)t

Ingen-Hausz: identical rods coated with wax stand in hot water; wax melts to lengths l₁, l₂, …

K₁ : K₂ : K₃ = l₁² : l₂² : l₃²

Growth of ice on a lake

Heat of freezing escapes up through the ice

Air at −θ °C; water just below the ice is at 0 °C. Thickness y grows by dy in time dt:

dydt = KθρLy

t = ρL2Kθ(y₂² − y₁²)

Time to reach y, 2y, 3y from zero: 1 : 4 : 9. Time for each next equal layer: 1 : 3 : 5.
Ice is a poor conductor, so growth slows down.

Heat flow between shells

Spherical shell, radii r₁ < r₂, and cylindrical shell of length L:

H = 4πKr₁r₂(θ₁ − θ₂)r₂ − r₁ · H = 2πKL(θ₁ − θ₂)ln(r₂ ÷ r₁)

Practise PYQs on this topicHeat Conduction · 55 questions with solutions

CONCEPT

Rods in Series & Parallel

In one line
Treat each rod as a thermal resistance R = l ÷ KA and combine them like resistors.

SERIES

13.2.1 Rods joined end to end

Same heat current through both rods

Same heat current H through every rod; temperature drops add up.

R_s = R₁ + R₂ + … + R_n

l₁ + l₂K_s = l₁K₁ + l₂K₂ (same A)

Equal lengths, two rods:

K_s = 2K₁K₂K₁ + K₂

n rods of equal length: nK_s = 1K₁ + 1K₂ + … + 1K_n.

JUNCTION

13.2.2 Temperature of the interface

Heat current is the same in both parts:

K₁A(θ₁ − θ)l₁ = K₂A(θ − θ₂)l₂

θ = K₁l₂θ₁ + K₂l₁θ₂K₁l₂ + K₂l₁

Equal lengths: θ = K₁θ₁ + K₂θ₂K₁ + K₂. Equal K and l: θ = θ₁ + θ₂2.
With resistances: θ = θ₁ − (θ₁ − θ₂) × R₁ ÷ (R₁ + R₂).
Several rods meeting at a point: heat in = heat out at the junction (like Kirchhoff’s current law).

PARALLEL

13.2.3 Rods side by side

Same temperature difference across each rod

Same temperature difference; heat currents add.

1R_p = 1R₁ + 1R₂ + … · H = H₁ + H₂

K = K₁A₁ + K₂A₂A₁ + A₂

Equal areas: K = K₁ + K₂2; n rods: K = K₁ + … + K_nn.
Temperature gradient is the same in every rod.

TIME

13.2.4 Same heat, series or parallel

For the same heat and the same θ difference, time t ∝ R.
Two identical rods: R_s = 2R, R_p = R2 ⇒ t_s : t_p = 4 : 1.

t_pt_s = R_pR_s · Q_pQ_s = R_sR_p (same t)

METHOD

13.2.5 Solving a network of rods

Step 1: write R = l ÷ KA for each rod.
Step 2: combine series and parallel parts.
Step 3: H = Δθ ÷ R_total, then find each drop as H × R.

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Ice melted by a conducting rod

A rod carries heat from a source into ice at 0 °C; all of it melts ice:

dmdt = KA(θ − 0)lL

Rods first in series and then in parallel (identical rods): R_p = R_s4, so the same heat flows 4 times faster in parallel.

Practise PYQs on this topicHeat Conduction · 55 questions with solutions

CONCEPT

Thermal Radiation, Emissivity & Black Body

In one line
Every body above 0 K emits heat as electromagnetic waves; a good absorber is also a good emitter.

THERMAL RADIATION

13.3.1 Heat without a medium

Electromagnetic waves, mostly infra-red: λ from about 7.8 × 10⁻⁷ m to 4 × 10⁻⁴ m.
Travel at 3 × 10⁸ m s⁻¹, even through vacuum; follow reflection, refraction, interference, polarisation.
Intensity ∝ 1d² from a point source; they exert radiation pressure.
Felt as heat only when absorbed.

a + r + t = 1

13.3.2 Radiation falling on a body

Incident Q is reflected, absorbed and transmitted

Q = Q_a + Q_r + Q_t

a = Q_aQ, r = Q_rQ, t = Q_tQ · a + r + t = 1

a, r, t are pure ratios: no unit, no dimension.
Opaque body: t = 0 ⇒ a + r = 1, so a good reflector is a bad absorber.
Perfect reflector: r = 1. Perfect transmitter: t = 1.
Perfectly black body: a = 1, r = t = 0.

EMISSIVE POWER

13.3.3 How much a surface radiates

Emissive power e: energy radiated per unit time per unit area (all wavelengths). Unit W m⁻²; dimension [MT⁻³].
Spectral emissive power e_λ: the same per unit wavelength range around λ.

e = ∫e_λdλ

EMISSIVITY

13.3.4 Compared with a black body

ε = eE (E: black body at the same T)

Perfect black body ε = 1; highly polished body ε → 0; practical bodies 0 < ε < 1.
For the same surface, emissivity = absorptivity (ε = a).
Black surfaces absorb and emit most, white or shiny ones least: black cools fastest.

BLACK BODY

13.3.5 Perfect absorber and emitter

Absorbs all radiation falling on it, of every wavelength: a = 1.
When hot it emits all wavelengths: it is the best emitter.
Lamp black and platinum black come close (they absorb about 96%).

Ferry’s black body: the hole acts as the black body

Ferry’s black body: double-walled sphere, blackened inside, with a small hole. Radiation entering the hole is reflected again and again until absorbed.
The sun (about 6000 K) behaves nearly as a black body.

PREVOST

13.3.6 Exchange of heat

Every body above 0 K keeps emitting and absorbing radiation all the time.
Emits more than it absorbs ⇒ cools. Absorbs more ⇒ warms. Equal ⇒ thermal equilibrium.

KIRCHHOFF’S LAW

13.3.7 Good absorber, good emitter

e₁a₁ = e₂a₂ = … = E (black body)

At a given temperature, e ÷ a is the same for all surfaces, so e = aE.
A body that absorbs a wavelength strongly also emits that wavelength strongly when hot.

COLOUR OF HOT BODIES

13.3.8 Hotter means bluer

As T rises the peak shifts to shorter λ: dull red (≈ 525 °C) → cherry red → orange → yellow → white (≈ 1600 °C).
A blue flame is hotter than a yellow one; a blue star is hotter than a red star.

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Kirchhoff’s law in daily life

Desert sand absorbs well by day and emits well by night: hot days, cold nights.
Sodium vapour absorbs the same D₁, D₂ lines it emits when hot.
Green glass absorbs red strongly, so when heated it glows red.
A black spot on a polished ball glows brightly when the ball is heated in the dark.

e_λa_λ = (E_λ)_black

Fraunhofer lines and diathermanous media

The photosphere emits all wavelengths; cooler gases of the chromosphere absorb some: dark lines in the sun’s spectrum.
They appear bright during a total solar eclipse (the gases then emit the same lines).
Diathermanous medium: lets heat radiation pass without warming (dry air, rock salt). Athermanous: absorbs it (wood, glass, moist air).
Thermal radiation is studied with a rock-salt prism; glass absorbs infra-red.

Practise PYQs on this topicThermal Radiation: Black Body, Stefan & Wien · 110 questions with solutions

CONCEPT

Stefan–Boltzmann Law & Rate of Cooling

In one line
A black body radiates energy per unit area per second in proportion to the fourth power of its absolute temperature.

STEFAN’S LAW

13.4.1 E ∝ T⁴

E = σT⁴ (black body)

T must be in kelvin: 827 °C ⇒ 1100 K.
σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; dimension [MT⁻³θ⁻⁴].

e = εσT⁴ (ordinary body)

POWER

13.4.2 Energy radiated by a whole body

P = Qt = εσAT⁴ · Q = εσAT⁴t

Sphere: A = 4πr² ⇒ P ∝ r²T⁴. Wire (filament): A = πdL.
Same power from two black spheres: r₁²T₁⁴ = r₂²T₂⁴.
Ratio questions: P₂ ÷ P₁ = (A₂ ÷ A₁)(T₂ ÷ T₁)⁴.

Change	Power becomes
T doubled	16 times
T × 1.5 (50% rise)	≈ 5 times (≈ 400% rise)
r doubled	4 times
r and T both doubled	64 times
16 times the power	T doubled

Small rise: % change in E = 4 × % change in T (only when the change is small).

NET LOSS

13.4.3 Body in surroundings at T₀

The body also absorbs radiation from the surroundings (Prevost).

e = εσ(T⁴ − T₀⁴) · H = εσA(T⁴ − T₀⁴)

RATE OF COOLING

13.4.4 Fall of temperature per second

Heat lost at rate H = mc × (rate of fall of temperature):

dTdt = εσA(T⁴ − T₀⁴)mc = εσA(T⁴ − T₀⁴)ρVc

Same body, two temperatures: rate ∝ T⁴ − T₀⁴. Work in units of 100 K to keep numbers small.
Same material, same T: rate of cooling ∝ AV; for spheres ∝ 1r.
Same mass and material: thin plate cools fastest, then cube, sphere slowest.
Equal volume: rate of loss of heat ∝ area, A_sphere < A_cube.

FACTORS

13.4.5 What makes a body cool faster

Higher emissivity, larger area, higher own temperature: faster.
Larger mass, larger specific heat, hotter surroundings: slower.

SUN AND EARTH

13.4.6 Solar constant

Sun’s power spreads over a sphere of radius r

Power of the sun P = 4πR²σT⁴ spreads over a sphere of radius r.

S = σR²T⁴r²

S ≈ 1.4 kW m⁻² (2 cal cm⁻² min⁻¹).
Energy received on earth ∝ R²T⁴ of the sun: T doubled ⇒ 16 times.

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Rate of cooling of different bodies

Same	Rate of loss of heat	Rate of cooling
spheres of one material	∝ r²	∝ 1r
T, T₀, mass	∝ A	∝ AV
T, T₀, m, A	same	∝ 1c

For equal mass and volume: plate has the largest area (cools fastest), sphere the least.

Temperature of the sun

T = (rR)^1/2(Sσ)^1/4

With r = 1.5 × 10⁸ km, R = 7 × 10⁵ km and S = 1.4 kW m⁻²: T ≈ 5800 K.
Light from the sun takes about 8 min 20 s to reach the earth.

Practise PYQs on this topicThermal Radiation: Black Body, Stefan & Wien · 110 questions with solutions

CONCEPT

Wien’s Displacement Law & Energy Spectrum

In one line
As a black body gets hotter, its peak wavelength λ_m moves to shorter wavelengths: λ_mT stays constant.

SPECTRUM

13.5.1 Energy distribution of a black body

E_λ against λ for T₁ < T₂ < T₃

Energy is not shared equally among wavelengths.
Intensity rises with λ, peaks at λ_m, then falls: least at very short and very long λ.
A black body emits all wavelengths.
At every wavelength, intensity rises with temperature: curves never cross.

area under the curve = ∫E_λdλ = E = σT⁴

Area under the curve ∝ T⁴ (Stefan’s law).

WIEN’S LAW

13.5.2 Peak shifts with temperature

λ_mT = b · λ_m ∝ 1T

b = 2.898 × 10⁻³ m K (Wien’s constant).
λ₁λ₂ = T₂T₁.
The sun: λ_m ≈ 4800 Å ⇒ T ≈ 6000 K. Used to find temperatures of stars.
Peak frequency ν_m ∝ T: ν_m–T graph is a straight line through the origin.

COLOUR

13.5.3 Hot bodies and stars

Blue star: small λ_m, hottest. Red star: large λ_m, coolest.
Order of temperature: blue > white > yellow > red.

WAVELENGTH DIFFERENCE

13.5.4 Two bodies, peaks Δλ apart

If T_A = nT_B, then λ_A = λ_Bn.

λ_B − λ_Bn = Δλ ⇒ λ_B = nΔλn − 1

STEFAN + WIEN

13.5.5 Power from the peak wavelength

T ∝ 1λ_m and E ∝ T⁴, so:

E ∝ 1λ_m⁴ · P ∝ Aλ_m⁴ ∝ r²λ_m⁴

P₂P₁ = (λ₁λ₂)⁴ (same body)

λ_m becomes λ2 ⇒ T doubles ⇒ power 16 times.
λ_m becomes 2λ3 ⇒ power (32)⁴ = 8116 times.
Intensity from λ_m: first T = b ÷ λ_m, then E = σT⁴.

METHOD

13.5.6 Ratio questions

Step 1: change °C to K.
Step 2: Wien gives the temperature ratio from λ_m.
Step 3: Stefan gives the power ratio: (area ratio) × (T ratio)⁴.

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Planck’s law

Walls of the cavity act as oscillators that emit energy in packets hν (photons).

E_λdλ = 8πhcλ⁵ × 1e^{hc ÷ λkT} − 1 dλ

Short λ (λ ≪ hc ÷ kT): reduces to Wien’s distribution law.
Long λ: reduces to the Rayleigh–Jeans law; the earth’s infra-red spectrum fits Planck’s law at all λ.
Energy at the peak: E_max ∝ T⁵.

Green-house effect

Glass lets the short waves from the sun in but stops the long infra-red waves given out by plants and soil, so heat is trapped.

Practise PYQs on this topicThermal Radiation: Black Body, Stefan & Wien · 110 questions with solutions

CONCEPT

Newton’s Law of Cooling

In one line
For a small temperature difference, the rate of cooling is proportional to the excess temperature θ − θ₀.

THE LAW

13.6.1 From Stefan’s law

If T − T₀ = ΔT is small, T⁴ − T₀⁴ ≈ 4T₀³ΔT.

dθdt = −k(θ − θ₀) · k = 4εσAT₀³mc

θ₀: temperature of the surroundings. k depends on the area, nature of the surface, mass and specific heat.
Rate of loss of heat: H = mcdθdt ∝ (θ − θ₀).
Excess doubled ⇒ rate doubled. At θ = θ₀ the rate is zero: a body never cools below its surroundings.

RATIO OF RATES

13.6.2 Same body, two temperatures

R₁R₂ = θ₁ − θ₀θ₂ − θ₀

Two rates and two temperatures given ⇒ solve for θ₀.

AVERAGE FORM

13.6.3 The exam formula

Body cools from θ₁ to θ₂ in time t: use the average temperature.

θ₁ − θ₂t = k[θ₁ + θ₂2 − θ₀]

Step 1: find k from the first cooling.
Step 2: use the same k for the second cooling.
Next equal time: unknown final θ appears on both sides; solve the linear equation.

Equal falls of temperature take longer and longer: t₁ < t₂ < t₃.
Hot water loses heat faster than warm water; milk added to hot tea reduces the rate of cooling.

EXACT FORM

13.6.4 Exponential cooling

θ − θ₀ = (θ_i − θ₀)e^−kt

log_e(θ − θ₀) = −kt + C

In equal time intervals the excess falls by the same factor: θ₂ − θ₀θ₁ − θ₀ = θ₃ − θ₀θ₂ − θ₀.

GRAPHS

13.6.5 Cooling curves

(a) θ falls exponentially to θ₀ · (b) straight line with negative slope · (c) straight line through the origin

Rate of cooling against θ (not θ − θ₀): straight line cutting the rate axis at −kθ₀.

COMPARING LIQUIDS

13.6.6 Same container, same start

Same surface, same temperatures ⇒ same rate of loss of heat.

(m₁c₁ + W)Δθt₁ = (m₂c₂ + W)Δθt₂

W = water equivalent of the calorimeter. Larger heat capacity ⇒ cools more slowly.

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Specific heat of a liquid by cooling

Equal volumes of water and liquid in identical calorimeters cool through the same range, θ₁ to θ₂, in times t₁ and t₂.

m_wc_w + Wt₁ = m_lc_l + Wt₂

Same mass of two liquids from the same start: t ∝ c, the liquid with the larger specific heat cools more slowly.

When the law fails

Valid only for a small excess temperature (about 30–40 °C).
For a large difference use Stefan’s law: rate ∝ T⁴ − T₀⁴.
A body can never cool below its surroundings by radiation.

Practise PYQs on this topicNewton's Law of Cooling · 20 questions with solutions

EXAM-DAY CARD

Just see this before the exam

H = KAΔθl

conduction

R = lKA

series, parallel

a + r + t = 1

radiation

E = σT⁴

Stefan

λ_mT = b

Wien

dθdt ∝ (θ − θ₀)

Newton

FORMULA SHEET

Whole chapter on one page

Conduction

H = KA(θ₁ − θ₂)l

gradient = θ₁ − θ₂l

R = lKA, H = ΔθR

mL = KAΔθtl

Combination

series: R_s = R₁ + R₂

K_s = 2K₁K₂K₁ + K₂ (equal l)

parallel: K = K₁A₁ + K₂A₂A₁ + A₂

θ = K₁l₂θ₁ + K₂l₁θ₂K₁l₂ + K₂l₁

Radiation

a + r + t = 1

ε = eE, black body: a = ε = 1

ea = E (Kirchhoff)

opaque: t = 0

Stefan

P = εσAT⁴

P ∝ r²T⁴

net: εσA(T⁴ − T₀⁴)

S = σR²T⁴r²

Wien

λ_mT = b = 2.9 × 10⁻³ m K

P ∝ r²λ_m⁴

area under E_λ–λ = σT⁴

blue star hotter than red

Newton

dθdt = −k(θ − θ₀)

θ₁ − θ₂t = k[θ₁ + θ₂2 − θ₀]

θ − θ₀ = (θ_i − θ₀)e^−kt

k = 4εσAT₀³mc

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Using °C in E = σT⁴

Series rods: K_s = K₁ + K₂

Net loss ∝ (T − T₀)⁴

50% rise in T ⇒ 200% more power

CORRECT ✓

Always change to kelvin first

R_s = R₁ + R₂, K_s = 2K₁K₂K₁ + K₂

Net loss ∝ T⁴ − T₀⁴

(1.5)⁴ ≈ 5 ⇒ ≈ 400% more

Heat Transfer: Conduction, Radiation & Cooling

Conduction, Thermal Conductivity & Thermal Resistance

13.1.1 Three ways heat travels

13.1.2 How solids pass on heat

13.1.3 When temperatures stop changing

13.1.4 Temperature gradient

13.1.5 Heat through a rod

13.1.6 Thermal conductivity

13.1.7 Heat flow like electric current

13.1.8 Heat conducted into ice

13.1.9 Heat carried by moving fluid

Corner

Rods in Series & Parallel

13.2.1 Rods joined end to end

13.2.2 Temperature of the interface

13.2.3 Rods side by side

13.2.4 Same heat, series or parallel

13.2.5 Solving a network of rods

Corner

Thermal Radiation, Emissivity & Black Body

13.3.1 Heat without a medium

13.3.2 Radiation falling on a body

13.3.3 How much a surface radiates

13.3.4 Compared with a black body

13.3.5 Perfect absorber and emitter

13.3.6 Exchange of heat

13.3.7 Good absorber, good emitter

13.3.8 Hotter means bluer

Corner

Stefan–Boltzmann Law & Rate of Cooling

13.4.1 E ∝ T4

13.4.2 Energy radiated by a whole body

13.4.3 Body in surroundings at T0

13.4.4 Fall of temperature per second

13.4.5 What makes a body cool faster

13.4.6 Solar constant

Corner

Wien’s Displacement Law & Energy Spectrum

13.5.1 Energy distribution of a black body

13.5.2 Peak shifts with temperature

13.5.3 Hot bodies and stars

13.5.4 Two bodies, peaks Δλ apart

13.5.5 Power from the peak wavelength

13.5.6 Ratio questions

Corner

Newton’s Law of Cooling

13.6.1 From Stefan’s law

13.6.2 Same body, two temperatures

13.6.3 The exam formula

13.6.4 Exponential cooling

13.6.5 Cooling curves

13.6.6 Same container, same start

Corner

Just see this before the exam

Whole chapter on one page

Avoid these = easy marks

13.4.1 E ∝ T⁴

13.4.3 Body in surroundings at T₀