Week of 3–8 Oct 2026 · Module Test 7 on Sat 10 Oct

Heat, temperature and expansion

This week's physics: Thermal Properties of Matter 10.1–10.3 and 10.5, the laws of thermodynamics, and Experiment 6, measuring surface tension by capillary rise and seeing what detergent does to it. Each topic has a short explanation, a model you can play with, the formula, the trap and tap-to-check questions. A 5-minute recap is at the end.

9 interactive models 28 solved problems Experiment 6 virtual lab 5-minute recap ↓ Full chapter: Thermal Properties → Full chapter: Thermodynamics →

★ Start here

This week's plan

The coaching schedule, day by day, with a link to the part of this page that covers it. Tick a day when you have finished it; your ticks are saved in this browser.

Sat 3 Oct Done

Experiment 6: surface tension by capillary rise

Measure h and r, find T = rhρg/2. Then add detergent and watch the rise drop.

Go to the lab →

Mon 5 Oct Done

10.1–10.3 Temperature, heat and scales

Heat versus temperature, thermometers, fixed points, and converting between °C, °F and K.

Temperature and heat → Scales →

Tue 6 Oct Done

Problems on 10.1–10.3

Scale conversions, faulty thermometers and mixed concept questions.

Problem set A →

Wed 7 Oct Done

10.5 Thermal expansion and its applications

Linear, area and volume expansion, thermal stress, bimetallic strips and water's odd behaviour.

Expansion → Applications →

Thu 8 Oct Done

Problems on thermal expansion

Gaps in rails, holes in plates, clamped rods, clocks and density changes.

Problem set B → Mixed set C →

Extra · Ch 11 Done

Laws of thermodynamics

Zeroth, first and second laws: Q = ΔU + W, the four processes, heat engines and the Carnot limit.

The three laws → Problem set D →

Sat 10 Oct Done

Module Test 7

All questions come from the new NCERT textbook, on this week's syllabus. Read the recap the night before.

5-minute recap →

10.1–10.2 Mon 5 Oct

Temperature and heat are not the same thing

We use "hot" and "heat" loosely in daily life, but physics separates them. Temperature tells you how hot a body is. Heat is energy that moves from one body to another because their temperatures are different.

Temperature

Temperature decides which way energy will flow: always from the hotter body to the colder one.

It measures how fast the molecules jiggle on average, not how many molecules there are.

Heat

Heat is energy in transit. A body does not "contain heat"; it contains internal energy.

SI unit: joule (J). Older unit: calorie, 1 cal = 4.186 J.

Thermal equilibrium

When two bodies reach the same temperature, the net heat flow stops.

If A and B are each in equilibrium with a thermometer C, then A and B are at the same temperature. This is why thermometers work.

	Temperature	Heat
What it is	Degree of hotness; linked to the average kinetic energy of molecules	Energy transferred because of a temperature difference
SI unit	kelvin (K)	joule (J)
Depends on the amount of matter?	No. A cup and a bucket of 50 °C water are both at 50 °C	Yes. The bucket can give out much more heat as it cools
Measured with	Thermometer	Calorimeter (10.7, coming later)

Interactive

Heat flows from hot to cold, not from "more energy" to "less energy"

Temperature of A

Mass of A

Temperature of B

Mass of B

A now–

B now–

Common final T–

Heat moved–

Both blocks are water. Predict the final temperature first, then put them in contact.

Heat lost by the hot body = heat gained by the cold body

\[ Q=mc\,\Delta T,\qquad m_Ac\,(T_A-T_f)=m_Bc\,(T_f-T_B)\ \Rightarrow\ T_f=\frac{m_AT_A+m_BT_B}{m_A+m_B} \]

Heat stops flowing when both reach the same final temperature T_f. For the same material, T_f is a mass-weighted average, so it lies closer to the heavier body's temperature. For different materials use \(T_f=\dfrac{m_Ac_AT_A+m_Bc_BT_B}{m_Ac_A+m_Bc_B}\). (Q = mcΔT is studied fully in 10.6, specific heat.)
Symbols and assumptions: Q = heat (J); m = mass (kg); c = specific heat capacity (water: 4186 J kg⁻¹ K⁻¹); T_A, T_B = starting temperatures; T_f = common final temperature; no heat lost to the surroundings and no change of state.

Trap"A body at a higher temperature always contains more heat" is false. A small red-hot nail is hotter than a bucket of warm water, so heat flows from the nail into the bucket, yet the bucket stores far more internal energy.

How it is asked

T / FTwo bodies with equal amounts of internal energy are put in contact. Heat cannot flow between them. True or false?

False. Heat flow depends only on the temperature difference. If their temperatures differ, heat flows even if their energies are equal.

Concept1 kg of water at 50 °C is mixed with 2 kg of water at 50 °C. What is the final temperature?

50 °C. They are already in thermal equilibrium, so no heat flows. Temperature does not add up with mass.

ConceptWhich law lets a thermometer compare the temperatures of two bodies that never touch each other?

The zeroth law of thermodynamics: two bodies each in thermal equilibrium with a third (the thermometer) are in equilibrium with each other.

10.3 Mon 5 Oct

Measuring temperature: thermometers and scales

A thermometer uses some property that changes steadily with temperature: the length of a mercury or alcohol column, the resistance of a platinum wire, or the pressure of a gas kept at constant volume. To turn that property into numbers, we mark two fixed points that are easy to reproduce:

Ice point: pure ice melting at 1 atm.
Steam point: pure water boiling at 1 atm.

The gap between the two points is divided into equal parts. Every scale differs only in the numbers it writes at the two points.

Scale	Ice point	Steam point	Divisions between them
Celsius (°C)	0	100	100
Fahrenheit (°F)	32	212	180
Kelvin (K)	273.15	373.15	100
Any scale X	L (lower fixed point)	U (upper fixed point)	U − L

One rule converts every scale

\[ \frac{C-0}{100}=\frac{F-32}{180}=\frac{K-273.15}{100}=\frac{X-L}{U-L} \]

Each fraction says how far you are from the ice point towards the steam point, so it is the same on every scale. Short form: \(\dfrac{C}{5}=\dfrac{F-32}{9}=\dfrac{K-273}{5}\).
Symbols and assumptions: C, F, K, X = readings of the same temperature on each scale; L, U = readings of scale X at the ice and steam points; both fixed points at 1 atm.

Temperature changes (differences)

\[ \Delta C=\Delta K,\qquad \Delta F=\tfrac95\,\Delta C \]

A rise of 1 °C is the same as a rise of 1 K and a rise of 1.8 °F. The "−32" and "−273" drop out of differences.
Symbols and assumptions: ΔC, ΔK, ΔF = the same temperature change measured on each scale.

Interactive

One thermometer, four scales

Temperature

Scale X: ice point L

Scale X: steam point U

Celsius–

Fahrenheit–

Kelvin–

Your scale X–

Drag the mercury, or use the slider.

Special point

At −40 the Celsius and Fahrenheit readings are equal.

\(\tfrac{x}{5}=\tfrac{x-32}{9}\Rightarrow x=-40\)

Special point

At 574.25 the Fahrenheit and Kelvin readings are equal (taking K = C + 273).

\(\tfrac{x-273}{5}=\tfrac{x-32}{9}\Rightarrow x=574.25\)

Graph

A graph of F against C is a straight line with slope 9/5 and intercept 32.

\(F=\tfrac95C+32\)

Where does the Kelvin scale start?

In a constant-volume gas thermometer, the gas pressure falls in a straight line as the gas is cooled. Real gases turn liquid before the pressure reaches zero, but if you extend every line, they all meet the temperature axis at the same point: −273.15 °C. This is absolute zero, the zero of the Kelvin scale, so \(T(\mathrm K)=t(^\circ\mathrm C)+273.15\). (This idea belongs to 10.4, but you need it to understand the Kelvin scale.)

Interactive

Three gas samples, one absolute zero

Gas temperature

T in kelvin–

P (middle gas)–

P ÷ T(K)–

Change the temperature: P ÷ T stays fixed only when T is in kelvin.

TrapConvert temperatures with the full rule, but convert temperature differences without the 32 or 273. A rise of 45 °F is a rise of 25 °C and 25 K, not (45 − 32) × 5/9.

How it is asked

NumNormal body temperature is 98.6 °F. What is it in °C and in K?

C = (98.6 − 32) × 5/9 = 37 °C; K = 37 + 273 = 310 K.

ScaleA scale X reads 20 °X at the ice point and 220 °X at the steam point. What does it read at 50 °C?

(X − 20)/200 = 50/100, so X = 120 °X.

ConceptWhy is mercury a good thermometer liquid?

It expands uniformly, is opaque and shiny (easy to read), does not wet glass, and stays liquid from −39 °C to 357 °C. Alcohol is used for very low temperatures because it freezes only at about −114 °C.

NEET-style MCQs: temperature and scales

10.5 Wed 7 Oct

Thermal expansion: things grow when heated

Why it happens: heating makes atoms vibrate with bigger swings. The force between atoms is not symmetric (it is easier to pull atoms apart than to squash them together), so bigger swings push the average spacing outwards. Every length of the body, its area and its volume all grow.

Linear · α

A rod grows in proportion to its length and to the temperature rise.

\(\Delta L=\alpha L\,\Delta T\)

Area · β

A sheet grows in two directions, so its coefficient is twice as big.

\(\Delta A=\beta A\,\Delta T,\ \ \beta=2\alpha\)

Volume · γ

A block grows in three directions, so its coefficient is three times as big.

\(\Delta V=\gamma V\,\Delta T,\ \ \gamma=3\alpha\)

Coefficients of thermal expansion

\[ \alpha=\frac{\Delta L}{L\,\Delta T},\qquad \beta=\frac{\Delta A}{A\,\Delta T},\qquad \gamma=\frac{\Delta V}{V\,\Delta T},\qquad \alpha:\beta:\gamma=1:2:3 \]

All three have the unit K⁻¹ (same number per °C). The new length is \(L'=L(1+\alpha\Delta T)\). Percentage shortcut: if the length grows by x %, the area grows by 2x % and the volume by 3x % (for small changes).
Symbols and assumptions: L, A, V = original length, area, volume; ΔT = temperature rise; isotropic solid (expands equally in every direction); αΔT ≪ 1.

Solid	α (×10⁻⁵ K⁻¹)
Aluminium	2.5
Brass	1.8
Copper	1.7
Iron / steel	1.2
Glass (pyrex)	0.32
Invar (alloy)	about 0.1

Liquid or gas	γ (×10⁻⁵ K⁻¹)
Mercury	18.2
Water (20 °C)	20.7
Alcohol (ethyl)	110
Ideal gas at 0 °C	366 (= 1/273 K)

Liquids and gases

Liquids and gases have no fixed shape, so only volume expansion (γ) matters. Liquids expand far more than solids.

Ideal gas at constant pressure: \(\gamma=\dfrac1T\)

Holes grow too

A hole or cavity expands exactly as if it were made of the material around it.

Heating is like a photo enlargement of the whole plate, hole included.

Density falls

Same mass in a bigger volume, so density drops on heating.

\(\rho'=\dfrac{\rho}{1+\gamma\Delta T}\approx\rho(1-\gamma\Delta T)\)

Interactive

Expansion lab: a plate with a hole, and a rod trapped between walls

Temperature change ΔT

––

TrapA heated hole gets bigger, not smaller. Use β = 2α and γ = 3α only for the same material. For liquids in a container, the container expands too, so the overflow depends on (γ_liquid − γ_vessel).

How it is asked

Ratioα of a metal is 2 × 10⁻⁵ K⁻¹. What are β and γ?

β = 4 × 10⁻⁵ K⁻¹ and γ = 6 × 10⁻⁵ K⁻¹.

%Heating increases the edge of a metal cube by 0.1 %. By what percentage do its surface area and volume increase?

Area by 0.2 % and volume by 0.3 % (2× and 3× the linear change).

ConceptA steel ring fits tightly on a steel shaft. Heating which one helps you remove the ring?

Heat the ring. Its hole expands like steel and grows bigger than the shaft.

NEET-style MCQs: expansion

10.5 Wed 7 Oct

Applications: where expansion helps and where it hurts

Railway tracksSmall gaps are left between rails so they can lengthen in summer without buckling.

BridgesOne end rests on rollers or has an expansion joint, so the deck can lengthen freely.

Power linesWires are hung with some slack. They sag more in summer and tighten in winter.

Tight lid or glass stopperPour hot water on the lid or neck; it expands and loosens.

Iron rim on a cart wheelThe rim is heated, slipped on, then cooled. It shrinks and grips the wheel tightly.

Pyrex glasswareVery low α, so it does not crack when heated unevenly.

InvarAlmost no expansion, used for pendulums, measuring tapes and precision instruments.

Bimetallic stripTwo metals bonded together bend when heated. Used in thermostats, electric irons and fire alarms.

Thermal stress: when expansion is blocked

If a rod is clamped between rigid walls, it cannot grow. The walls squeeze it back by exactly the amount it wanted to grow, so the strain is αΔT and the stress follows from Young's modulus.

Thermal stress in a clamped rod

\[ \text{strain}=\alpha\Delta T,\qquad \text{stress}=Y\alpha\Delta T,\qquad F=YA\alpha\Delta T \]

Heating gives compression; cooling gives tension. The stress does not depend on the length or the cross-section of the rod; the force does depend on A.
Symbols and assumptions: Y = Young's modulus (Pa); A = cross-sectional area (m²); α = linear expansion coefficient (K⁻¹); ΔT = temperature change; walls perfectly rigid.

Bimetallic strip

Two strips with different α are riveted together. On heating, the metal with the larger α grows more, so it ends up on the outside (convex side) of the curve. On cooling, the strip bends the other way.

Interactive

A bimetallic strip working as a thermostat

Temperature change ΔT

Δα–

Radius of curve–

Tip moves–

Heater–

Bending of a bimetallic strip

\[ R\approx\frac{d}{(\alpha_1-\alpha_2)\,\Delta T} \]

A bigger difference in α or a bigger ΔT gives a tighter curve (smaller R).
Symbols and assumptions: R = radius of curvature of the strip; d = distance between the middles of the two layers (about the thickness of one layer); α₁ > α₂; layers of similar stiffness.

Two coaching favourites

Pendulum clock in summer

\[ \frac{\Delta T_p}{T_p}=\tfrac12\alpha\,\Delta\theta,\qquad \text{time lost per day}=\tfrac12\alpha\,\Delta\theta\times86400\ \text{s} \]

A hotter pendulum is longer, so it swings more slowly and the clock loses time. A colder one gains time.
Symbols and assumptions: T_p = period (∝ √L); α = coefficient of the pendulum rod; Δθ = rise above the temperature at which the clock is correct.

Metal scale used when hot

\[ \text{true length}=\text{reading}\times(1+\alpha\,\Delta\theta) \]

A hot scale has stretched markings, so it reads less than the true length.
Symbols and assumptions: α = coefficient of the scale; Δθ = temperature above the one at which the scale was calibrated; the object's own expansion is ignored.

TrapThermal stress YαΔT is the same for a short rod and a long rod. On heating, the larger-α metal goes to the convex (outer) side of a bimetallic strip.

How it is asked

NumA steel rod (Y = 2 × 10¹¹ Pa, α = 1.2 × 10⁻⁵ K⁻¹) is clamped between rigid walls and heated by 50 K. Find the thermal stress.

2 × 10¹¹ × 1.2 × 10⁻⁵ × 50 = 1.2 × 10⁸ Pa (compressive).

ConceptA brass–iron strip is cooled. Which way does it bend?

Brass shrinks more, so brass ends up on the inner (concave) side. The strip bends towards the brass side.

ConceptA brass disc fits tightly in a hole in a steel plate. Should you heat or cool the pair to free the disc?

Cool it. Brass has the larger α, so the disc shrinks more than the hole and comes loose.

NEET-style MCQs: stress, strips and clocks

10.5 Wed 7 Oct

Water breaks the rule between 0 °C and 4 °C

The odd part

Warming water from 0 °C to 4 °C makes it shrink. Above 4 °C it expands like other liquids.

Result

Water is densest at 4 °C (about 1000 kg m⁻³). Ice (about 917 kg m⁻³) is lighter, so it floats.

Why it matters

Lakes freeze from the top down. The bottom stays at 4 °C, so fish survive the winter.

Interactive

Density of water near 4 °C, and a lake in winter

Water temperature

Air above the lake

Density–

Volume of 1 kg–

Warm it slightly–

Trap"Water expands on heating" is true only above 4 °C. Water cooled from 10 °C to 0 °C first contracts (down to 4 °C) and then expands.

How it is asked

GraphSketch the volume of 1 kg of water from 0 °C to 10 °C.

A U-shaped curve with its lowest point (minimum volume, maximum density) at 4 °C.

Why?Why do water pipes sometimes burst on very cold nights?

Water expands below 4 °C and expands even more (about 9 %) when it freezes, pushing hard on the pipe walls.

NEET-style MCQs: water

CH 11 Added topic

The laws of thermodynamics

Thermodynamics is the bookkeeping of energy for a system (usually a gas in a cylinder) and its surroundings. Three quantities appear in every question:

Internal energy U: the energy stored inside the gas. For an ideal gas it depends only on temperature, so it is a state function: ΔU depends only on the start and end states.
Heat Q: energy that enters because of a temperature difference.
Work W: energy that leaves when the gas pushes a piston outwards. Q and W are path functions: they depend on how you go from one state to the other.

Zeroth law

If A and B are each in thermal equilibrium with C, then A and B are in equilibrium with each other.

It defines temperature and is the reason a thermometer works. (See temperature and heat.)

First law

Energy is conserved: the heat you give a gas either raises its internal energy or is spent doing work.

\(\Delta Q=\Delta U+\Delta W\)

Second law

No engine turns all its heat into work, and heat never flows from cold to hot by itself.

\(\eta<1,\quad \eta_{\max}=1-\dfrac{T_2}{T_1}\)

The first law and its sign convention

First law of thermodynamics

\[ \Delta Q=\Delta U+\Delta W,\qquad \Delta W=\int P\,dV=\text{area under the }P\text{–}V\text{ graph},\qquad \Delta U=nC_V\,\Delta T \]

Q is + when heat goes into the gas; W is + when the gas does work on the surroundings (expands). ΔU = nC_VΔT holds for an ideal gas in any process, not only at constant volume.
Symbols and assumptions: n = moles; C_V = molar heat capacity at constant volume (3R/2 monatomic, 5R/2 diatomic); C_P = C_V + R; γ = C_P/C_V; quasi-static (slow) processes for the area rule.

Process	What stays fixed	Work W	ΔU	Heat Q
Isochoric	Volume	0	nC_VΔT	Q = ΔU
Isobaric	Pressure	PΔV = nRΔT	nC_VΔT	nC_PΔT
Isothermal	Temperature (PV = const)	nRT ln(V₂/V₁)	0	Q = W
Adiabatic	No heat flow (PV^γ = const)	(P₁V₁ − P₂V₂)/(γ − 1)	−W	0
Cyclic	Returns to start	Area of the loop	0	Q = W
Free expansion	Insulated, into vacuum	0	0	0

Interactive

One mole of gas, four processes: where does the heat go?

Final volume ÷ start volume

Heat in, Q–

Change in U–

Work by gas, W–

End state–

TrapKeep the signs straight: work done on the gas (compression) is negative W. In an adiabatic expansion the gas cools because it pays for its work out of U. In an isothermal process ΔU = 0, but Q is not zero.

The second law: heat engines and refrigerators

The first law says energy is conserved; the second law says which way energy can go. NCERT gives it in two equivalent forms:

Kelvin–Planck: no process can take heat from a reservoir and turn it completely into work. Every engine must throw some heat away to a colder sink.
Clausius: no process can move heat from a colder body to a hotter one without work being done. A refrigerator needs power.

A reversible process can be run backwards, leaving no change anywhere: it must be slow and free of friction and of heat flow across a finite temperature difference. Real processes (free expansion, friction, heat flowing from hot to cold) are irreversible. The Carnot engine, made of two isothermal and two adiabatic steps, is reversible and is the most efficient engine possible between two temperatures.

Heat engine efficiency

\[ \eta=\frac{W}{Q_1}=1-\frac{Q_2}{Q_1},\qquad \eta_{\text{Carnot}}=1-\frac{T_2}{T_1} \]

An engine takes Q₁ from the hot source, does work W and rejects Q₂ to the sink. No engine between T₁ and T₂ can beat Carnot.
Symbols and assumptions: T₁, T₂ = source and sink temperatures in kelvin; W = Q₁ − Q₂ per cycle.

Refrigerator (coaching extra ★)

\[ \beta=\frac{Q_2}{W},\qquad \beta_{\text{Carnot}}=\frac{T_2}{T_1-T_2} \]

A refrigerator uses work W to pull Q₂ out of the cold inside and dumps Q₁ = Q₂ + W into the room. β can be greater than 1.
Symbols and assumptions: β = coefficient of performance; T₂ = inside (cold) temperature, T₁ = room temperature, both in kelvin.

Interactive

Can this engine exist? Test it against the Carnot limit

Hot reservoir T₁

Cold reservoir T₂

Claimed efficiency

Claimed COP β

Carnot limit–

Work W per cycle–

Heat to cold sink Q₂–

Verdict–

TrapAlways put temperatures in kelvin into 1 − T₂/T₁. An engine claiming more than the Carnot efficiency is impossible, and so is any engine with η = 100 %.

How it is asked

ConceptWhich of these are state functions: internal energy, heat, work, temperature?

Internal energy and temperature. Heat and work depend on the path.

Why?Why does a bicycle pump get hot when you pump quickly?

Fast compression is nearly adiabatic (Q ≈ 0). Work is done on the gas (W negative), so ΔU = −W is positive and the temperature rises.

NumFind the maximum efficiency of an engine working between 127 °C and 27 °C.

T₁ = 400 K, T₂ = 300 K, so η = 1 − 300/400 = 25 %.

ConceptAn ideal gas expands freely into a vacuum inside an insulated container. What are Q, W and ΔU?

All zero. Nothing pushes back, so W = 0; the container is insulated, so Q = 0; hence ΔU = 0 and the temperature of an ideal gas is unchanged.

NEET-style MCQs: the first law

NEET-style MCQs: the second law and engines

EXPERIMENT 6 Sat 3 Oct

Surface tension of water by capillary rise, and the effect of detergents

RepeatYou met capillary rise in Fluids II §9.6.5. It is repeated here so this week's syllabus is all on one page. The new part is the effect of detergents, further down.

Aim

To find the surface tension of water by the capillary rise method, and to see how adding a detergent changes it.

Apparatus

Clean glass capillary tubes of different bores, a beaker of water, a clamp stand, a travelling microscope, a pointer (pin) on a stand, and some detergent.

Theory in one line

The surface pulls the water column up along the edge of the tube, and the column's weight pulls it down. At balance:

\[ 2\pi rT\cos\theta=\pi r^{2}h\rho g\ \Rightarrow\ T=\frac{rh\rho g}{2\cos\theta}\approx\frac{rh\rho g}{2} \]

For clean glass and water, θ ≈ 0, so cos θ ≈ 1.

Surface tension from capillary rise

\[ T=\frac{r\,h\,\rho\,g}{2},\qquad h=\frac{2T\cos\theta}{r\rho g}\ \Rightarrow\ h\,r=\text{constant} \]

Thinner tube, higher rise. Lower T (detergent or hot water), lower rise.
Symbols and assumptions: T = surface tension (N m⁻¹); r = inner radius of the tube (m); h = height of the meniscus bottom above the outside water surface (m); ρ = density of water; θ = contact angle (≈ 0 for clean glass); the small volume of water in the curved meniscus is neglected.

Procedure

Clean the capillary tube (with dilute acid, then water) so that grease does not change the contact angle.
Clamp the tube vertically in the beaker. Water rises inside it.
Focus the travelling microscope on the bottom of the concave meniscus inside the tube and note the reading R₁.
Lower the pointer until its tip just touches the water surface outside. Remove the beaker, focus on the pointer tip and note R₂. Then h = R₁ − R₂.
Measure the inner diameter of the tube at the meniscus position with the microscope, in two perpendicular directions, and take the mean radius r.
Repeat with two or three tubes of different bore. Check that h × r comes out nearly the same.
Add a little detergent to the water, stir gently and repeat. Compare the new rise.

Virtual lab

Capillary rise: measure, record, then add detergent

Detergent added

Water temperature

Inner radius r–

Rise h–

h × r–

Surface tension T–

#	Tube	r (mm)	Detergent	Water	h (cm)	T = rhρg/2 (N m⁻¹)
No readings yet. Pick a tube and press "Record this reading".

Result

For clean water at room temperature, T ≈ 0.072 N m⁻¹. Adding detergent lowers T, so the water rises less in the same tube.

Precautions

The tube must be clean, vertical and of uniform bore.
Read the bottom of the meniscus, and measure h from the outside water surface.
Measure the inner radius at the place where the meniscus stood.
Note the water temperature: T falls as temperature rises.

Sources of error

Grease or dust raises the contact angle and lowers the rise.
The bore may not be perfectly circular.
The water in the curved meniscus is ignored. (The correction replaces h by h + r/3.)
Percentage error: \(\dfrac{\Delta T}{T}=\dfrac{\Delta r}{r}+\dfrac{\Delta h}{h}\). Since r is tiny, measuring it is the biggest source of error.

The new part: detergents and surface tension

A detergent (or soap) molecule has two ends: a head that likes water and a long tail that likes oil and grease. Here is what that does:

The molecules crowd at the water surface and at any water–oil boundary, and lower the surface tension there.
Lower surface tension means less energy is needed to make new surface, so grease breaks into tiny drops. The tails stick into the oil, and the heads face the water, so the drops are carried away when you rinse.
Detergent water also wets surfaces better (smaller contact angle), so it spreads into fabric and dirt.
In a capillary, lower T means a smaller rise: h ∝ T.

Detergent molecules surround a grease drop: tails inside the oil, heads out in the water. The drop lifts off and is rinsed away.

Try at home

Sprinkle pepper on water, then touch the middle with a soapy finger. The pepper rushes to the edges.

The soapy patch has lower T, so the cleaner water around it pulls the surface outwards.

Try at home

Float a dry needle on water, then add a drop of detergent nearby. The needle sinks.

Lower T can no longer hold up the needle's weight.

Link to heat

Hot water also has lower surface tension, which is one reason hot soapy water cleans best.

T of water drops from about 0.076 N m⁻¹ at 0 °C to about 0.059 N m⁻¹ at 100 °C.

TrapMeasure h from the outside water level, not from the bottom of the tube. Use the inner radius. Detergent and heating both lower T; dissolved salt slightly raises it.

How it is asked

NumWater rises 3.0 cm in a tube of inner radius 0.5 mm. Find T. (ρ = 1000 kg m⁻³, g = 9.8 m s⁻², θ = 0)

T = rhρg/2 = (0.5 × 10⁻³)(0.03)(1000)(9.8)/2 ≈ 0.0735 N m⁻¹.

DetergentDetergent halves the surface tension of water. What happens to the capillary rise in the same tube?

It halves, because h ∝ T (assuming the contact angle stays near zero).

Why?Why do detergents clean greasy clothes better than plain water?

They lower the surface tension of water and of the water–oil boundary, and their molecules bind oil with their tails and water with their heads. The grease breaks into small drops that the water can carry away.

NEET-style MCQs: the capillary experiment

✎ Tue 6 + Thu 8 Oct + Ch 11

28 practice problems, including mixed concepts

Solve each problem on paper first. Then open the solution and mark yourself honestly. Your marks are saved in this browser, so you can come back to the ones marked "Retry".

Got it: 0 Retry: 0 Not tried: 28

Set A · Temperature, heat and scales (Tue)

A1Convert −40 °F, 212 °F and 0 K to Celsius.

−40 °F = −40 °C (the special equal point). 212 °F = 100 °C. 0 K = −273.15 °C.

A2The temperature of a body rises by 45 °F. What is the rise in °C and in K?

ΔC = (5/9) × 45 = 25 °C, and ΔK = ΔC = 25 K. (Differences ignore the 32.)

A3At what temperature is the Fahrenheit reading twice the Celsius reading?

F = 2C and F = 1.8C + 32, so 2C = 1.8C + 32, giving C = 160 °C and F = 320 °F.

A4A faulty thermometer reads 5° in melting ice and 95° in steam at 1 atm. What is the correct Celsius temperature when it reads 59°?

(59 − 5)/(95 − 5) = C/100, so C = 54/90 × 100 = 60 °C.

A5On a scale X, the ice point is 20 °X and the steam point is 220 °X. What Celsius temperature does 0 °X correspond to?

(0 − 20)/200 = C/100, so C = −10 °C.

A6A constant-volume gas thermometer shows 1.00 atm at 0 °C. What pressure does it show at 100 °C?

P ∝ T in kelvin: P = 1.00 × 373.15/273.15 ≈ 1.37 atm. (Using °C directly would give an impossible answer.)

Set B · Thermal expansion and applications (Thu)

B1Steel rails 10 m long are laid at 20 °C. What gap must be left between rails for summer days of 40 °C? (α = 1.2 × 10⁻⁵ K⁻¹)

ΔL = αLΔT = 1.2 × 10⁻⁵ × 10 × 20 = 2.4 × 10⁻³ m = 2.4 mm.

B2A brass plate has a hole of diameter 10.00 cm at 20 °C. What is the diameter at 120 °C? (α = 1.8 × 10⁻⁵ K⁻¹)

The hole expands like brass: d' = 10.00 × (1 + 1.8 × 10⁻⁵ × 100) = 10.018 cm.

B3A metal sphere's radius increases by 0.2 % when heated. Find the percentage increase in its surface area and in its volume.

Area: 2 × 0.2 = 0.4 %. Volume: 3 × 0.2 = 0.6 %.

B4A solid has γ = 3 × 10⁻⁵ K⁻¹. By what percentage does its density fall when heated by 100 K?

Δρ/ρ ≈ −γΔT = −3 × 10⁻⁵ × 100 = −3 × 10⁻³, a 0.3 % decrease.

B5A steel rod of cross-section 2 cm² is clamped between rigid walls and heated by 50 K. Find the force on the walls. (Y = 2 × 10¹¹ Pa, α = 1.2 × 10⁻⁵ K⁻¹)

Stress = YαΔT = 1.2 × 10⁸ Pa. F = stress × A = 1.2 × 10⁸ × 2 × 10⁻⁴ = 2.4 × 10⁴ N.

B6Rod P is twice as long as rod Q. Both are made of the same material, clamped between rigid walls and heated equally. Compare their thermal stresses.

Equal. Stress = YαΔT does not depend on length.

B7A pendulum clock with a brass rod (α = 2 × 10⁻⁵ K⁻¹) is correct at 20 °C. How much time does it lose per day at 30 °C?

Fractional slow-down = ½αΔθ = ½ × 2 × 10⁻⁵ × 10 = 10⁻⁴. Time lost = 10⁻⁴ × 86 400 ≈ 8.64 s per day.

B8A brass–iron bimetallic strip is heated. Which metal is on the convex side, and what happens when it is cooled below room temperature instead?

Brass (larger α) is on the convex side when heated. On cooling, brass shrinks more, so it goes to the concave side and the strip bends the other way.

Set C · Experiment 6 and mixed concepts

C1Water rises 6.0 cm in a capillary of inner radius 0.24 mm. Find T. (ρ = 1000 kg m⁻³, g = 9.8 m s⁻², θ = 0)

T = rhρg/2 = (0.24 × 10⁻³)(0.060)(1000)(9.8)/2 ≈ 0.071 N m⁻¹.

C2In tube B water rises 3.0 cm. Detergent lowers T from 0.072 to 0.036 N m⁻¹. What is the new rise in the same tube?

h ∝ T, so h = 3.0 × 0.036/0.072 = 1.5 cm.

C3Two tubes of inner radii 0.2 mm and 0.4 mm are dipped in water. The rise in the wider tube is 3.6 cm. What is the rise in the narrower tube?

hr = constant, so h = 3.6 × 0.4/0.2 = 7.2 cm.

C4 · mixedThe water in a capillary experiment is heated from 20 °C to 60 °C. Does the capillary rise increase or decrease? Why?

It decreases. Surface tension falls by about 9 % (from about 0.073 to 0.066 N m⁻¹), while density falls by under 2 %. Since h = 2T/(rρg), the drop in T wins.

C5 · mixedA glass capillary tube is heated, so its bore expands slightly. What happens to the capillary rise of water at the same temperature? (Ignore the change in T.)

It decreases slightly. The bore (a hole) expands like glass, so r increases, and h ∝ 1/r.

C6 · mixedIn the experiment, the percentage errors in r and h are 2 % and 1 %. What is the maximum percentage error in T?

T ∝ rh (ρ and g treated as exact), so the error is 2 % + 1 % = 3 %.

Set D · Laws of thermodynamics (take R = 8.3 J mol⁻¹ K⁻¹)

D1A gas absorbs 500 J of heat and does 200 J of work on its surroundings. Find the change in its internal energy.

ΔU = Q − W = 500 − 200 = 300 J (internal energy increases).

D22 mol of a monatomic ideal gas is heated at constant volume from 300 K to 400 K. Find W, ΔU and Q.

W = 0. ΔU = nC_VΔT = 2 × (3/2)(8.3) × 100 = 2490 J. Q = ΔU = 2490 J.

D3The same gas is heated from 300 K to 400 K at constant pressure instead. Find W, ΔU and Q. What fraction of Q raises U?

W = nRΔT = 2 × 8.3 × 100 = 1660 J. ΔU = 2490 J (same ΔT, same ΔU). Q = 4150 J. Fraction = ΔU/Q = C_V/C_P = 3/5.

D4A gas expands from 2 L to 5 L at a constant pressure of 2 × 10⁵ Pa. How much work does it do?

W = PΔV = 2 × 10⁵ × 3 × 10⁻³ = 600 J.

D5In one cycle a gas absorbs 800 J of heat and rejects 500 J. Find ΔU for the cycle, the net work and the efficiency.

ΔU = 0 for a full cycle. W = 800 − 500 = 300 J. η = 300/800 = 37.5 %.

D6A Carnot engine works between 527 °C and 127 °C and absorbs 1000 J per cycle. Find its efficiency, the work and the heat rejected.

T₁ = 800 K, T₂ = 400 K, so η = 1 − 400/800 = 50 %. W = 500 J and Q₂ = 500 J.

D7 · mixedAn inventor claims an engine working between 400 K and 300 K with 30 % efficiency. Accept or reject?

Reject. The Carnot limit is 1 − 300/400 = 25 %, and no engine can beat it (second law).

D8 · mixedA freezer keeps its inside at 0 °C in a room at 27 °C. What is its maximum coefficient of performance, and what is the least work needed to remove 1000 J from inside?

β = T₂/(T₁ − T₂) = 273/27 ≈ 10.1. W = Q₂/β ≈ 1000/10.1 ≈ 99 J.

This week's plan

Experiment 6: surface tension by capillary rise

10.1–10.3 Temperature, heat and scales

Problems on 10.1–10.3

10.5 Thermal expansion and its applications

Problems on thermal expansion

Laws of thermodynamics

Module Test 7

Temperature and heat are not the same thing

Heat flows from hot to cold, not from "more energy" to "less energy"

Measuring temperature: thermometers and scales

One thermometer, four scales

Where does the Kelvin scale start?

Three gas samples, one absolute zero

NEET-style MCQs: temperature and scales

Thermal expansion: things grow when heated

Expansion lab: a plate with a hole, and a rod trapped between walls

NEET-style MCQs: expansion

Applications: where expansion helps and where it hurts

Thermal stress: when expansion is blocked

Bimetallic strip

A bimetallic strip working as a thermostat

Two coaching favourites

NEET-style MCQs: stress, strips and clocks

Water breaks the rule between 0 °C and 4 °C

Density of water near 4 °C, and a lake in winter

NEET-style MCQs: water

The laws of thermodynamics

The first law and its sign convention

One mole of gas, four processes: where does the heat go?

The second law: heat engines and refrigerators

Can this engine exist? Test it against the Carnot limit

NEET-style MCQs: the first law

NEET-style MCQs: the second law and engines

Surface tension of water by capillary rise, and the effect of detergents

Aim

Apparatus

Theory in one line

Procedure

Capillary rise: measure, record, then add detergent

Result

Precautions

Sources of error

The new part: detergents and surface tension

NEET-style MCQs: the capillary experiment

28 practice problems, including mixed concepts

Set A · Temperature, heat and scales (Tue)

Set B · Thermal expansion and applications (Thu)

Set C · Experiment 6 and mixed concepts

Set D · Laws of thermodynamics (take R = 8.3 J mol⁻¹ K⁻¹)

5-minute recap

Before Module Test 7, I can…