Properties of Solids and Liquids Formula Sheet — JEE Main Physics

Every key Properties of Solids and Liquids formula, definition and theorem for JEE Main Physics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 22 sub-topics

Elastic behaviour
Stress-strain relationship
Hooke's law
Young's modulus
Bulk modulus
Shear modulus of rigidity
Poisson's ratio
Elastic energy
Pressure due to a fluid column
Pascal's law and its applications
Effect of gravity on fluid pressure
Viscosity
Stokes' law
Terminal velocity
Streamline and turbulent flow
Critical velocity
Bernoulli's theorem and its applications
Surface energy and surface tension
Angle of contact
Excess of pressure across a curved surface
Application of surface tension ideas to drops, bubbles
Capillary rise

Elasticity — Stress, Strain & Moduli

Stress & strain

stress = \frac{F}{A}, strain = \frac{change in dimension}{original dimension}

F =

deforming force,

A =

area it acts over; stress: N

m^{- 2}

=Pa (scalar); strain: dimensionless

Comparative: types of stress & strain

Type	Stress	Strain	Restored by
Longitudinal	$F / A$ (normal)	$Δ L / L$	Young's $Y$
Volumetric	$Δ P$ (normal, all sides)	$Δ V / V$	Bulk $B$
Shear	$F_{∥} / A$ (tangential)	$ϕ = Δ x / L$	Rigidity $η$

Three strains

▸Longitudinal $\frac{Δ L}{L}$ (normal stress, solids).
▸Volumetric $\frac{Δ V}{V}$ (normal stress, all states).
▸Shear $ϕ = \frac{Δ x}{L} = tan θ$ (tangential stress, solids only — fluids have $η_{s h e a r} = 0$ ).

Hooke's law

stress = E \times strain

valid only within the proportional limit;

E =

modulus of elasticity (slope of the linear region)

Stress-strain curve with proportional limit P, elastic limit E, yield Y, ultimate U and fracture F points — Beyond E: permanent (plastic) set; breaking stress at fracture.

🚫 Examiner Trap · Stress, strain & the curve

(1) Proportional limit

\neq =

elastic limit: Hooke holds only up to P, but elastic recovery lasts up to E. (2) Stress always uses the ORIGINAL area A (engineering stress). (3) A larger Y means LESS strain for a given stress (stiffer), not more. (4) Breaking stress is a material property; breaking FORCE scales with A, so a thicker wire holds more load at the same breaking stress.

Elastic moduli

Three moduli + compressibility

Young's

Y

(longitudinal)

Y = \frac{F / A}{Δ L / L} = \frac{F L}{A Δ L}

Bulk

B

(volumetric)

B = - \frac{Δ P}{Δ V / V}

Rigidity

η

(shear)

η = \frac{F _{∥} / A}{ϕ}

Compressibility

C = \frac{1}{B}

Comparative: the three moduli

Modulus	Measures resistance to	States
$Y$	length change	solids
$B$	volume change	solids, liquids, gases
$η$	shape (shear)	solids only

Poisson's ratio & relations — not in JEE Main 2026 syllabus

σ = - \frac{Δ r / r}{Δ L / L}; Y = 3 B (1 - 2 σ) = 2 η (1 + σ)

σ =

Poisson's ratio (lateral strain

\div

longitudinal strain);

σ \approx 0.1

–

0.3

for most materials, theoretical range

- 1 < σ < 0.5

; volume unchanged on stretching

\Rightarrow σ = 0.5

, and in general

\frac{Δ V}{V} = (1 - 2 σ) \frac{Δ L}{L}

Elastic potential energy — not in JEE Main 2026 syllabus

U = \frac{1}{2} F Δ L = \frac{1}{2} \times stress \times strain \times V

energy density

u = \frac{U}{V} = \frac{1}{2} stress \times strain = \frac{1}{2} Y (strain)^{2}

;

V =

volume

⚡ Shortcut · Wire as a spring

A stretched wire behaves like a spring of constant

k = \frac{Y A}{L}

, so

F = k Δ L

and

U = \frac{1}{2} k (Δ L)^{2}

. Springs in this form combine in series/parallel like ordinary springs.

🎯 Exam · Elongation under its own weight

A uniform rod/rope of length L, mass M, density

ρ

hung from one end: tension grows from

0

at the free end to Mg at the top, so

Δ L = \frac{M gL}{2 A Y} = \frac{ρ g L ^{2}}{2 Y}

— HALF the stretch a load Mg at the end would give. With an extra load m at the end:

Δ L = \frac{( m + M /2 ) gL}{A Y}

Fluid Statics & Buoyancy

Density & pressure

ρ = \frac{M}{V}, P = \frac{F}{A}

ρ =

density,

M =

mass,

V =

volume; relative density

= ρ / ρ_{w a t er}

; pressure is a scalar

Hydrostatic pressure

P = P_{0} + ρ g h

P_{0} =

atmospheric (surface) pressure,

h =

depth below surface; gauge pressure

P_{g} = P - P_{0} = ρ g h

; same at all points at one depth in one connected fluid

Pressure units

▸ $1 atm = 1.013 \times 1 0^{5} Pa = 76 cm Hg = 760 torr$ .
▸ $1 bar = 1 0^{5}$ Pa; $1 torr = 1 mm Hg$ .
▸Atmosphere falls $\approx$ exponentially with height; drops $\times \frac{1}{10}$ per $\sim 20$ km.

🎯 Exam · Pascal's law — hydraulic lift

Pressure applied to a confined fluid is transmitted undiminished. Force multiplier:

F = \frac{A}{a} f

(

a, A =

small/large piston areas,

f =

applied force).

🚫 Examiner Trap · Pressure & Pascal

(1) Hydrostatic

P = ρ g h

depends on DEPTH only, not on the shape or amount of liquid (hydrostatic paradox). (2) Use total

P = P_{0} + ρ g h

for force on a wall/base; use gauge

ρ g h

for differences. (3) The hydraulic lift multiplies FORCE, not energy or work — the small piston moves farther (

f d_{1} = F d_{2}

Accelerating fluid

tan θ = \frac{a}{g}

free surface tilts at angle

θ

to the horizontal for horizontal acceleration a; vertical acceleration a (up) gives

P = P_{0} + ρ (g + a) h

, and in free fall

ρ g h \to 0

Rotating liquid

y = \frac{ω ^{2} x ^{2}}{2 g}; P_{2} - P_{1} = \frac{1}{2} ρ ω^{2} (r_{2}^{2} - r_{1}^{2})

vessel spun at

ω

about its vertical axis: the free surface is a paraboloid,

y =

height above the lowest (axis) point at distance x; rim-to-centre height

= ω^{2} R^{2} /2 g

. Horizontal tube rotating about one end: pressure rises outward (

r =

distance from the axis)

🎯 Exam · U-tube / manometer

Pressures are equal at the same level in ONE continuous liquid. Two immiscible liquids in a U-tube, measured from their interface:

ρ_{1} h_{1} = ρ_{2} h_{2}

. A manometer reads a gauge pressure

Δ P = ρ g Δ h

(

Δ h =

level difference).

Buoyancy — Archimedes

Upthrust

F_{B} = V_{s u b} ρ_{ℓ} g, W_{a pp} = W - F_{B}

V_{s u b} =

submerged volume,

ρ_{ℓ} =

fluid density; upthrust = weight of fluid displaced, acts at the centre of buoyancy

Block floating partly submerged with weight Mg down and buoyant force F_B up balancing — Float: $M g = F_{B}$ , so $\frac{V _{s u b}}{V} = \frac{ρ _{b o d y}}{ρ _{ℓ}}$ .

Comparative: floatation by density

Condition	Behaviour	Fraction submerged
$ρ_{b o d y} < ρ_{ℓ}$	floats partly	$ρ_{b o d y} / ρ_{ℓ}$
$ρ_{b o d y} = ρ_{ℓ}$	floats fully immersed	$1$ (neutral)
$ρ_{b o d y} > ρ_{ℓ}$	sinks (rests on base)	$1$ (fully)

🚫 Examiner Trap · Buoyancy & floatation

(1)

F_{B}

uses the FLUID density

ρ_{ℓ}

and the SUBMERGED volume, never the body's density/volume. (2) A floating body displaces its own WEIGHT of fluid; a fully sunk body displaces its own VOLUME. (3) Ice melting in a glass of water leaves the level UNCHANGED (it displaced its meltwater's weight). (4) In a lift accelerating up or down, BOTH Mg and

F_{B}

scale with the effective g, so the fraction submerged of a floating body does NOT change; the apparent weight of a sunk body does.

Fluid Dynamics & Viscosity

Equation of continuity

A v = constant (A_{1} v_{1} = A_{2} v_{2})

A =

cross-section area,

v =

flow speed; mass conservation for an incompressible fluid (volume flow rate

A v

constant)

Bernoulli's principle

P + \frac{1}{2} ρ v^{2} + ρ g h = constant

P =

static pressure,

\frac{1}{2} ρ v^{2} =

dynamic pressure,

ρ g h =

potential head; energy/volume for ideal (non-viscous, incompressible, streamline) flow

Venturi pipe wide-to-narrow with manometer columns showing higher pressure where the pipe is wide and lower where narrow — At a constriction: faster flow, lower pressure.

Velocity of efflux (Torricelli)

v = 2 g h, x = 2 h (H - h)

h =

depth of hole below surface,

H =

total liquid height,

x =

horizontal range on the floor; range max when

h = H /2

🚫 Examiner Trap · Continuity & Bernoulli

(1) Narrower pipe

\Rightarrow

FASTER flow (

v \propto 1/ A

)

\Rightarrow

LOWER pressure — students invert this. (2) Bernoulli needs a STREAMLINE and ideal flow; it fails for viscous/turbulent flow. (3) Torricelli's

v = 2 g h

uses depth of the HOLE below the free surface, not the hole's height above ground. (4) 'Static pressure' is the lower term when speed is high, not the total.

Bernoulli applications

▸Venturimeter: $v_{1} = A_{2} \frac{2Δ P}{ρ ( A _{1}^{2} - A _{2}^{2} )}$ , flow rate $Q = A_{1} v_{1}$ ; $Δ P = ρ_{m} g Δ h$ from its manometer.
▸Wind over a roof: $Δ P = \frac{1}{2} ρ_{ai r} v^{2}$ pushes the roof UP, lift force $= \frac{1}{2} ρ v^{2} A$ ; aerofoil lift works the same way.
▸Atomiser/sprayer, Magnus effect (spinning ball curves), two boats or papers pulled together by fast flow between them.

Viscosity

Newton's viscous drag

F = - η A \frac{d v}{d x}

η =

coefficient of viscosity,

d v / d x =

velocity gradient;

η

in Pa

\cdot

s (SI) or poise (

1 p o i se = 0.1 P a \cdot

Poiseuille / Stokes

Q = \frac{π P r ^{4}}{8 η l}; F = 6 π ηr v

Q =

volume flow rate (

\propto r^{4}

!),

P =

pressure difference,

l =

tube length; Stokes drag on a sphere of radius r at speed v

Sphere falling in viscous liquid: weight down balanced by buoyancy and viscous drag up at terminal velocity — At terminal speed: $m g = F_{B} + 6 π ηr v_{T}$ .

Terminal velocity & Reynolds no.

v_{T} = \frac{2 r ^{2} ( ρ - σ ) g}{9 η}; R_{e} = \frac{ρ v D}{η}

ρ =

sphere density,

σ =

fluid density,

D =

pipe diameter;

R_{e} < 2000

laminar,

> 3000

turbulent (NCERT quotes

< 1000

streamline,

> 2000

turbulent)

Critical velocity

v_{c} = \frac{R _{e} η}{ρ D}

v_{c} =

speed above which streamline flow turns turbulent; higher for a more viscous liquid and a narrower pipe (

R_{e} =

critical Reynolds number,

ρ =

liquid density)

Comparative: streamline vs turbulent

Feature	Streamline (laminar)	Turbulent
Reynolds $R_{e}$	$< 2000$	$> 3000$
Paths	smooth, non-crossing	chaotic, mixing
Speed	below critical $v_{c}$	above critical $v_{c}$
Energy loss	low	high

🚫 Examiner Trap · Viscosity & terminal velocity

(1)

η

of a LIQUID falls as T rises;

η

of a GAS rises with T — opposite trends. (2)

v_{T} \propto r^{2}

, so a bigger drop falls faster. (3) Use

(ρ - σ)

v_{T}

: subtract fluid density for buoyancy; if

ρ < σ

the body rises (e.g. bubble). (4) Poiseuille

Q \propto r^{4}

— halving the radius cuts flow to

1/16

Surface Tension & Capillarity

Surface tension

S

: Force per unit length in the surface, equal to the surface energy per unit area:

S = \frac{F}{L} = \frac{energy}{area}

. For a film with two surfaces,

S = \frac{F}{2 ℓ}

. Units: N

m^{- 1}

= J

m^{- 2}

Key facts

▸S decreases with temperature ( $S = 0$ at the critical temperature); work to create new area $W = S Δ A$ .
▸Free drops & bubbles are spherical (least area for a given volume).
▸Detergents/soaps lower S and the contact angle $\Rightarrow$ better wetting; impurities can raise or lower S.

Angle of contact

cos θ = \frac{S _{s a} - S _{s l}}{S _{l a}}

S_{s a}, S_{s l}, S_{l a} =

solid-air, solid-liquid, liquid-air tensions; acute (water-glass, concave, wets); obtuse (mercury-glass, convex, non-wetting)

Excess pressure inside a liquid drop is 2S/r and inside a soap bubble (two films) is 4S/r — Excess pressure across a curved surface.

🚫 Examiner Trap · Surface tension & energy

(1) A SOAP bubble has TWO surfaces

\Rightarrow

excess pressure

4 S / r

and work

S \cdot 2Δ A

; a liquid drop or an air bubble in liquid has ONE

\Rightarrow 2 S / r

. (2) When n small drops merge into one big drop, energy is RELEASED (area falls); splitting one drop ABSORBS energy. (3) S is force per unit length of the contact LINE, independent of how much surface area exists.

Excess pressure

Δ P_{d ro p} = \frac{2 S}{r}, Δ P_{b u bb l e} = \frac{4 S}{r}, Δ P_{ai r b u bb l e} = \frac{2 S}{r}

r =

radius; soap bubble has two surfaces, hence

4 S / r

; smaller radius

\Rightarrow

larger excess pressure

Comparative: curved-surface excess pressure

Geometry	Surfaces	Excess $Δ P$
Liquid drop	1	$2 S / r$
Air bubble in liquid	1	$2 S / r$
Soap bubble (in air)	2	$4 S / r$

Capillary rise

Jurin's law

h = \frac{2 S cos θ}{ρ g r}

r =

tube radius,

θ =

contact angle; rises (

θ

acute, concave); falls (

θ

obtuse, convex, e.g. mercury);

h \propto 1/ r

Liquid rising in a narrow capillary tube to height h with a concave meniscus and contact angle theta — Narrower tube $\Rightarrow$ greater rise ( $h \propto 1/ r$ ).

⚡ Shortcut · Short-tube rule

If the tube is shorter than the calculated h, the liquid does NOT overflow — it rises to the top and the MENISCUS radius

R = \frac{r}{cos θ}

grows so that

h R = \frac{2 S}{ρ g}

still holds (

h^{'} R^{'} = h R

). In a freely falling lift (

g_{e ff} = 0

) the liquid climbs to the top of any tube length.

Drops merging / splitting

R = n^{1/3} r; Δ E = 4 π S (n r^{2} - R^{2}) = 4 π R^{2} S (n^{1/3} - 1)

n drops of radius r form one drop of radius R (volume conserved);

Δ E =

surface energy released (absorbed when one drop splits into n). All of it into heat:

Δ T = \frac{3 S}{ρ c} (\frac{1}{r} - \frac{1}{R})

; into kinetic energy:

v = \frac{6 S}{ρ} (\frac{1}{r} - \frac{1}{R})

🎯 Exam · Film between two plates

A liquid film of thickness d and area A that wets two plates (

θ = 0

) has a curved edge of radius

d /2

, so the pressure inside is lower by

\frac{2 S}{d}

and the force needed to pull the plates apart is

F = \frac{2 A S}{d}

. For a drop of volume V squeezed to area A:

d = V / A

, so

F = \frac{2 A ^{2} S}{V}

★ Remember · Why it rises

Pressure just below a concave meniscus is

less

than atmospheric by

\frac{2 S}{r _{m e n}}

(

r_{m e n} = r / cos θ

) — this deficit drives the capillary rise until

ρ g h

balances it.

Thermal Expansion & Thermal Stress

Temperature scales

\frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273.15}{100}

general rule:

\frac{X - LFP}{UFP - LFP}

is the same on every scale, where LFP, UFP

=

that scale's readings for melting ice and boiling water (works for a faulty or arbitrary scale too); a CHANGE

Δ C = Δ K = \frac{5}{9} Δ F

; F vs C is a straight line of slope

9/5

;

- 4 0^{\circ}

= - 4 0^{\circ}

Linear / areal / volume expansion

Δ L = Lα Δ T, Δ A = A β Δ T, Δ V = Vγ Δ T

α, β, γ =

linear, areal, volume expansion coefficients (per

^{\circ}

C); for an isotropic solid

β = 2 α

γ = 3 α

Comparative: the three expansion coefficients

Coeff.	Governs	Isotropic ratio
$α$	length $Δ L / L$	$α$
$β$	area $Δ A / A$	$2 α$
$γ$	volume $Δ V / V$	$3 α$

Density change

ρ = \frac{ρ _{0}}{1 + γ Δ T} \approx ρ_{0} (1 - γ Δ T)

ρ_{0} =

density before heating; mass is fixed, volume rises, so density decreases as temperature rises

★ Remember · Anomalous expansion

Between

0^{\circ}

C and

4^{\circ}

C water

contracts

on heating (

γ < 0

); densest at

4^{\circ}

C — why lakes freeze top-down and aquatic life survives.

Liquid: apparent vs real expansion

γ_{re a l} = γ_{a pp} + γ_{v esse l}

γ_{v esse l} = 3 α_{v esse l}

; the vessel also expands, so observed (apparent) expansion is smaller than the real expansion

🎯 Exam · Thermal stress

A rod rigidly clamped at both ends and heated develops a compressive stress

Y α Δ T

(force

Y A α Δ T

) — independent of length.

Y =

Young's modulus.

🚫 Examiner Trap · Thermal expansion

(1) A HOLE/cavity expands like the surrounding solid, NOT inward — it gets bigger on heating. (2) Use

γ = 3 α

only for ISOTROPIC solids. (3) For a liquid you always measure APPARENT expansion; add

γ_{v esse l}

to get the real value. (4) Thermal stress depends on

Y, α, Δ T

only — NOT on cross-section or length.

Applications

▸Bimetallic strip bends on heating (different $α$ ) — used in thermostats.
▸Gaps in rails and bridges; compensated pendulums built from two metals.
▸Pendulum clock: $\frac{Δ T _{p}}{T _{p}} = \frac{1}{2} α Δ θ$ , so a clock warmer than its calibration runs SLOW, losing $\frac{1}{2} α Δ θ \times 86400$ s per day.
▸Metal scale: true length $=$ reading $\times (1 + α_{sc a l e} Δ θ)$ . A scale that is warmer than its calibration temperature has expanded, so it reads SHORT.

Calorimetry, Latent Heat & Change of State

Heat & specific heat

Q = m c Δ T; c = \frac{Q}{m Δ T}, C = \frac{Q}{n Δ T}

m =

mass,

n =

moles,

c =

specific heat (J k

g^{- 1} K^{- 1}

C =

molar heat capacity;

1 cal = 4.18

Water equivalent

W = \frac{m c}{c _{w a t er}}

W =

mass of water needing the same heat for the same temperature rise; unit: g (or kg). Numerically

W = m c

when c is in

cal g^{- 1}^{\circ} C^{- 1}

Latent heat

Q = m L

L =

specific latent heat (J k

g^{- 1}

); water:

L_{f u s i o n} = 80 cal/g = 336 kJ/kg

L_{v a p} = 540 cal/g \approx 2.26 MJ/kg

;

c_{w a t er} = 1 cal g^{- 1}^{\circ} C^{- 1} = 4200 J kg^{- 1} K^{- 1}

c_{i ce} \approx c_{s t e am} \approx 0.5 cal g^{- 1}^{\circ} C^{- 1}

; temperature stays CONSTANT during a change of state

🎯 Exam · Principle of calorimetry

Heat lost

=

heat gained (no loss to surroundings). Final mixture temperature lies between the two:

T_{L} \leq T \leq T_{H}

Comparative: sensible vs latent heat

	Sensible $Q = m c Δ T$	Latent $Q = m L$
Temperature	changes	constant
What changes	warms/cools	phase changes
On the curve	sloped segment	flat plateau

Heating curve of temperature versus heat: rising segments for solid, liquid, gas separated by flat plateaus at melting and boiling where latent heat is absorbed — Water: plateaus = change of state; slope $= \frac{1}{m c}$ , so liquid water (largest c) rises slowest.

Change of state

▸Melting/boiling absorb heat at constant $T$ (MP, BP); freezing/condensing release it.
▸Sublimation: solid $\to$ vapour directly (e.g. dry ice, iodine).
▸Steeper heating-curve slope $\Rightarrow$ smaller specific heat ( $slope = 1/ m c$ ).

🚫 Examiner Trap · Calorimetry & latent heat

(1) Temperature does NOT change while a substance melts/boils — all heat goes into L. (2) Don't forget the calorimeter/vessel: include its water equivalent in 'heat gained'. (3) If the supplied heat is too small, the mix may END at

0^{\circ}

C with ice + water coexisting — check whether all the ice actually melts. (4) Use consistent units: cal with cal/g, J with J/kg.

💡 Tip · Steam burns

Burns from steam are worse than from boiling water: steam first releases the large

L_{v a p}

(

540

cal/g) on condensing, then cools as water.

Heat Transfer — Conduction, Convection & Radiation

Conduction (steady state)

\frac{Q}{t} = \frac{K A ( T _{1} - T _{2} )}{L}; R_{t h} = \frac{L}{K A}

K =

thermal conductivity,

A =

area,

L =

length,

(T_{1} - T_{2}) =

temperature difference;

R_{t h} =

thermal resistance

Rod conducting heat from a hot reservoir T1 to a cold reservoir T2 over length L — Heat flows down the temperature gradient.

Comparative: the three modes

Mode	Carrier	Needs medium?
Conduction	lattice/electrons	yes (solids)
Convection	bulk fluid motion	yes (fluids)
Radiation	EM waves	no (vacuum OK)

Rod combinations & convection

▸Rods in series: $R_{t h} = R_{1} + R_{2} + \dots$ (same heat current through each).
▸Rods in parallel: $\frac{1}{R _{t h}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots$ (same $Δ T$ across each).
▸Junction of two rods in series (equal A): $T_{J} = \frac{\frac{K _{1}}{L _{1}} T _{1} + \frac{K _{2}}{L _{2}} T _{2}}{\frac{K _{1}}{L _{1}} + \frac{K _{2}}{L _{2}}}$ ; equivalent K in series $= \frac{L _{1} + L _{2}}{L _{1} / K _{1} + L _{2} / K _{2}}$ , in parallel (equal L) $= \frac{K _{1} A _{1} + K _{2} A _{2}}{A _{1} + A _{2}}$ .
▸Convection: bulk transport of heated fluid (natural or forced); needs a medium.

🚫 Examiner Trap · Conduction

(1) The conduction formula is STEADY-STATE only — no temperature still changing. (2) Thermal resistance adds like ELECTRICAL resistance: series adds

R

, parallel adds

1/ R

(don't swap them). (3) Radiation is the ONLY mode that works through a vacuum — that's how the Sun heats Earth.

Radiation

Stefan-Boltzmann

\frac{Q}{A t} = e σ T^{4}; P_{n e t} = e σ A (T^{4} - T_{0}^{4})

e =

emissivity,

T =

body temp (K),

T_{0} =

surroundings;

σ = 5.67 \times 1 0^{- 8}

m^{- 2} K^{- 4}

;

e = 1

for a black body

Wien's displacement law

λ_{m} T = b

λ_{m} =

peak-emission wavelength;

b = 2.9 \times 1 0^{- 3} m \cdot

K; hotter bodies peak at shorter

λ

Kirchhoff & black body

▸Kirchhoff: a good absorber is a good emitter ( $e = a$ at a given T, $λ$ ).
▸Black body: absorbs all incident radiation ( $a = 1$ , $e = 1$ ).

Newton's law of cooling

- \frac{d T}{d t} = k (T - T_{0})

T_{0} =

surrounding temp,

k > 0

const; valid for a small excess temperature; cooling rate

\propto

temperature difference. Solution:

T - T_{0} = (T_{i} - T_{0}) e^{- k t}

, so the EXCESS halves in equal times;

k = \frac{4 e σ A T _{0}^{3}}{m c}

from Stefan's law

🚫 Examiner Trap · Radiation & cooling

(1) Stefan's law uses ABSOLUTE temperature

T^{4}

in KELVIN — never

^{\circ}

C. (2) Net loss uses

(T^{4} - T_{0}^{4})

, not

(T - T_{0})^{4}

. (3) Newton's cooling is the small-

Δ T

approximation of Stefan's law; it needs the EXCESS temperature, and in a T vs t plot the slope (not T) is proportional to the excess.

⚡ Shortcut · Cooling-curve average

For Newton's law over a step, use the AVERAGE excess:

\frac{T _{1} - T _{2}}{t} = k (\frac{T _{1} + T _{2}}{2} - T_{0})

— turns the differential equation into one clean linear equation for exam problems.

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