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PhysicsCh 10 Thermal Properties of Matter

Thermal Properties of Matter: Class 11 Physics · Chapter 10

Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 120 min
  • 1 simulation
  • 29 practice questions
  • 35 MCQs

What you will learn in Thermal Properties of Matter

  • Distinguish temperature from heat and convert temperatures between the Celsius, Fahrenheit and Kelvin scales
  • Explain how a constant-volume gas thermometer leads to absolute zero and the Kelvin scale, using PV = μRT
  • Calculate linear, area and volume expansion and thermal stress, and explain the anomalous expansion of water
  • Use specific heat capacity, molar specific heat and latent heat to solve calorimetry problems, including changes of state
  • Read heating curves and phase diagrams, and explain how pressure shifts melting and boiling points
  • Calculate steady heat flow by conduction through single and joined rods, and explain convection and radiation in everyday settings
  • Apply Wien's displacement law and the Stefan–Boltzmann law to hot bodies and stars
  • State Newton's law of cooling, derive its exponential form and use it to estimate cooling times

Original companion text aligned to NCERT Physics Class 11, Chapter 10, pp. 202–225. Not affiliated with or endorsed by NCERT.

10.1 Introduction

A mechanic fits a steel ring gear onto a flywheel by warming the ring until it slides over, then letting it cool into a tight grip. On Chennai's Marina beach the afternoon breeze blows in from the sea, but late at night it drifts out. A pressure cooker matters far more in Leh than in Kochi, and a puff of steam can hurt more than a splash of boiling water.

Each puzzle depends on how matter responds when energy flows in or out because of a temperature difference. This chapter shows how temperature is measured, how matter expands, how much energy warming and changes of state take, how heat travels, and how fast a cup of chai cools.

10.2 Temperature and heat

Dip a finger into bath water and you know at once whether it is hot or cold, but you could not tell 38 °C from 41 °C, and you could hardly test molten iron or liquid nitrogen with your hand. Our sense of hot and cold is also easily fooled, as the activity below shows. Physics replaces it with temperature (ताप): a number, read from an instrument on an agreed scale, that says how hot a body is.

Put a pan of just-boiled milk and a frozen gel pack from a lunch bag on the same kitchen shelf on a May afternoon. The milk cools, the pack thaws and warms, and after an hour or two both are at room temperature (Fig. 10.1). Energy has flowed out of the milk into the room, and out of the room into the pack. Energy that passes between a system and its surroundings because their temperatures differ is called heat (ऊष्मा). Once the temperatures match, the net flow stops, and we say the bodies are in thermal equilibrium.

2/3Heat leaves the milk and enters the cold pack.

Fig. 10.1Heat always flows from the hotter body to the colder one, and stops when their temperatures match.

A pan of milk at 95 degrees Celsius and a frozen gel pack at minus 5 degrees sit on a shelf in a room at 32 degrees. Step 1 shows the starting temperatures. Step 2 adds arrows: heat flows out of the milk into the room and from the room into the cold pack. Step 3 shows both at 32 degrees, with no arrows, in thermal equilibrium.

Heat is energy in transit, so its SI unit is the joule (J). A body does not contain heat, any more than a bank account contains deposits: it contains internal energy, and heat is one way of changing it. Energy that crosses a boundary for any other reason, such as the work done by a stirrer churning water, is not heat (Chapter 11 develops this). The SI unit of temperature is the kelvin (K); in daily life we also use the degree Celsius (°C). The rest of the chapter follows what heat does to matter: it warms it, swells it, and melts or boils it.

Quick check
1 of 3
A small flask and a large bucket both hold water at 60 °C. Which statement is correct?
Show answer

Answer (C) The bucket has more internal energy, but the temperatures are equal

Temperature does not depend on the amount of water, but internal energy does. "Heat held" is not a meaningful quantity, and both containers are hotter than the room, so both lose heat to it.

10.3 Measurement of temperature

A digital fever thermometer reads the electrical resistance of a tiny bead that changes with temperature; a laboratory thermometer reads the length of a thread of coloured alcohol; an infrared 'gun' reads the radiation a surface gives off. Each relies on a thermometric property, a measurable quantity that changes steadily and reproducibly with temperature. For a liquid-in-glass thermometer the liquid, usually mercury or alcohol, must expand almost linearly, so that equal steps along the thread mean equal steps of temperature.

To turn a thread length into a number you need a scale, and a scale needs two fixed points: physical events that happen at one fixed temperature every time and are easy to reproduce. The traditional pair is the ice point, where pure ice and water coexist, and the steam point, where pure water boils, both at standard atmospheric pressure. The Celsius scale labels them 0 °C and 100 °C and divides the gap into 100 degrees. The Fahrenheit scale labels them 32 °F and 212 °F, so the same gap holds 180 degrees (Fig. 10.2).

2/4Water boils at 100 °C or 212 °F: 100 Celsius degrees span 180 Fahrenheit degrees.

Fig. 10.2One temperature, three numbers. A Celsius degree is the same size as a kelvin and 1.8 times a Fahrenheit degree.NCERT Fig. 10.4 · p. 204

Three parallel rulers for the Celsius, Fahrenheit and Kelvin scales, aligned so equal temperatures sit one above another, with a break before the ice point. Step 1: ice point, 0 °C, 32 °F, 273.15 K. Step 2: steam point, 100 °C, 212 °F, 373.15 K, with brackets showing 100 and 180 degrees between the two points. Step 3: body temperature, 37 °C, 98.6 °F, 310.15 K. Step 4: absolute zero, −273.15 °C, −459.67 °F, 0 K.

Both scales are linear, so a temperature lies the same fraction of the way from the ice point to the steam point on each:

Celsius–Fahrenheit conversion
tF−32180=tC100
(10.1)

Rearranged, tF=95tC+32. A graph of tF against tC is a straight line with slope 9/5 that cuts the tF axis at 32. Remember that changes convert with the slope alone: a rise of 10 Celsius degrees is a rise of 18 Fahrenheit degrees.

Example

Reading a fever

A digital thermometer shows 102.2 °F. (a) What is this in degrees Celsius? (b) At what temperature do a Celsius and a Fahrenheit thermometer show the same number?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
The temperature of a cup of tea falls by 20 Celsius degrees. By how many Fahrenheit degrees does it fall?
Show answer

Answer (C) 36

Differences convert with the slope only: 20×9/5=36. Adding 32 (to get 68) is the mistake of converting a temperature instead of a change.

10.4 Ideal-gas equation and absolute temperature

Calibrate a mercury thermometer and an alcohol thermometer at 0 °C and 100 °C, then dip both into water that the mercury one calls 50 °C. The alcohol one reads slightly differently, because the two liquids do not expand in exactly proportional ways, and neither has a better claim to be right. Gases settle the question. At low density, hydrogen, helium and nitrogen all respond to temperature in the same way, so thermometers built on them agree with one another, and a gas becomes the natural standard.

For a fixed amount of gas, three quantities describe its state: pressure P, volume V and temperature. In 1662 Robert Boyle reported that at a steady temperature the product PV stays constant: Boyle's law. In the 1780s Jacques Charles found that at steady pressure the volume grows in step with temperature, so that V/T is constant, provided T is counted from the right zero (we find it below): Charles' law. Together they make PV/T constant for a given sample. Doubling the amount of gas doubles PV/T, so it is proportional to the number of moles μ:

Ideal-gas equation
PV=μRT
(10.2)
where
μ
the number of moles of gas
R
the universal gas constant, 8.31 J mol−1 K−1
T
the absolute temperature, in kelvin

Real gases follow Eq. (10.2) closely when their density is low; a gas that would follow it exactly is called an ideal gas. Hold the volume fixed and the equation gives P∝T: the pressure of the trapped gas becomes the thermometric property. This is the principle of the constant-volume gas thermometer.

Plot pressure against Celsius temperature for such a thermometer and you get a straight line (Fig. 10.3). A different gas or a different amount gives a different slope. Real gases liquefy long before their pressure falls anywhere near zero, but if you extend each line downwards they all reach zero pressure at the same temperature, −273.15 °C. An ideal gas would exert no pressure there, and no lower temperature makes sense. This limit is absolute zero.

3/3Every line reaches zero pressure at −273.15 °C: absolute zero.

Fig. 10.3Lines for different low-density gas samples, extended, all reach zero pressure at −273.15 °C.NCERT Fig. 10.3 · p. 204

Pressure against Celsius temperature. Step 1: measured points for one gas sample between 0 and 100 °C lie on a straight line. Step 2: two more samples give straight lines of different slope. Step 3: all three lines, extended as dashed lines, reach zero pressure at the same point, −273.15 °C, marked absolute zero.

Lord Kelvin (William Thomson) proposed a scale whose zero is absolute zero and whose degree is the same size as the Celsius degree. Temperatures on it are written in kelvin, with no degree sign:

Kelvin and Celsius
T=tC+273.15
(10.3)

A temperature difference has the same value in kelvin and in Celsius degrees, so a unit like J kg−1 K−1 works with Celsius changes. But in PV=μRT, and in every law that multiplies or divides by T, you must use kelvin.

Example

Reading a gas thermometer

The bulb of a constant-volume helium thermometer shows a pressure of 50.0 kPa in a triple-point cell of water, at 273.16 K. In the steam above boiling water it shows 68.3 kPa. Find the steam's temperature in kelvin and in degrees Celsius.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
A gas held at constant volume has a pressure of 1.20×105 Pa at 27 °C. At what Celsius temperature will its pressure be 1.60×105 Pa?
°C
Show answer

Answer 127 °C

Use kelvin: T2=300 K×1.60/1.20=400 K, which is 127 ∘C. Scaling 27 °C directly would give a wrong 36 °C.

10.5 Thermal expansion

Electricity cables strung between poles sag visibly more on a May afternoon than on a December morning. Concrete roads are laid as separate slabs with narrow gaps filled with a soft sealant. A thick glass tumbler can crack if boiling water is poured into it straight from the fridge, because its inner surface swells before the outer surface has warmed. Most substances expand when heated and contract when cooled; this change of size with temperature is thermal expansion.

We call the change in a length linear expansion, the change in a surface area expansion and the change in a volume volume expansion (Fig. 10.4). For a rod of length l warmed by a small ΔT, experiments show that the fractional change in length is proportional to ΔT:

Fig. 10.4Linear, area and volume expansion (greatly exaggerated). The dashed outlines show the cold sizes.NCERT Fig. 10.5 · p. 205

Three panels with the growth exaggerated: a rod longer than its dashed cold outline by delta l, a square tile grown in both directions by delta A, and a cube grown in all three directions by delta V.

Linear expansion
Δll=αl ΔT
(10.4)

The constant αl, the coefficient of linear expansion or linear expansivity, depends on the material and has the unit K−1. Table 10.1 lists typical values. Copper expands about five times as much as Pyrex glass for the same warming, which is why Pyrex dishes survive going from a fridge into an oven. Invar, an iron–nickel alloy, hardly changes at all, so it is used in measuring tapes and precision instruments.

Table 10.1Typical coefficients of linear expansion near room temperature
Materialαl (10−5 K−1)
Aluminium2.3
Brass1.9
Copper1.7
Steel1.2
Concrete1.2
Window glass0.9
Pyrex glass0.33
Invar0.12

For volume we define the coefficient of volume expansion in the same way:

Volume expansion
ΔVV=αV ΔT
(10.5)

Like αl, αV is a property of the material, but it is not strictly constant. For a solid such as copper it is tiny at very low temperatures, grows as the solid warms and levels off only at high temperatures. Over everyday ranges we treat both coefficients as constants. Table 10.2 shows that liquids expand far more than solids. Ethanol expands about six times as much as mercury, which is why an alcohol thermometer's thread moves so far for each degree.

Table 10.2Typical coefficients of volume expansion near 20 °C
SubstanceαV (K−1)
Steel3.6×10−5
Window glass2.7×10−5
Mercury1.8×10−4
Water2.1×10−4
Glycerine5.0×10−4
Petrol9.5×10−4
Ethanol1.1×10−3
Air (constant pressure)3.4×10−3

Water: the odd one out

Water is the famous exception. Its density peaks near 4 °C (more precisely 3.98 °C, at 999.97 kg m⁻³), so between 0 °C and 4 °C water shrinks as it warms and swells as it cools. Freezing adds a further jump of about 9 % in volume, which is why ice floats and why a full glass bottle left in a freezer cracks. Both quirks come from hydrogen bonds, which lock the molecules of ice into an open, roomy framework; fragments of that framework already appear in water just above freezing.

Picture a lake in Ladakh as winter sets in (Fig. 10.5). Density decides which water lies on top. While the surface is warmer than 4 °C, chilling makes it heavier, so it sinks and the lake turns over until all of it is near 4 °C. Water chilled further is now lighter than the 4 °C water beneath, so the coldest layer floats, reaches 0 °C and freezes into a lid. Ice conducts heat poorly, so the lid slows further cooling, and the bed of the lake stays near 4 °C all winter, with its fish alive. A liquid that grew steadily denser on cooling would freeze solid from the bed upwards.

4/4Ice forms on top and insulates the 4 °C water below.

Fig. 10.5Water is densest near 4 °C, so a lake freezes from the top and its bottom water stays liquid.NCERT Fig. 10.7 · p. 206

A lake in cross-section, with a small density curve peaking at 4 °C. Step 1: surface water at 10 °C cools and sinks while warmer water rises. Step 2: the whole lake is at 4 °C. Step 3: a colder layer below 4 °C stays on top. Step 4: ice covers the lake, the water just under it is at 0 °C and the bottom water, with a fish, is at 4 °C.

Gases

Gases expand much more than solids or liquids, and their expansivity depends strongly on temperature. For an ideal gas it follows directly from the gas equation.

DerivationVolume expansivity of an ideal gas

Keep the pressure fixed and let the temperature change by ΔT. The gas equation PV=μRT gives P ΔV=μR ΔT. Divide this by PV=μRT:

ΔVV=ΔTT

Comparing with Eq. (10.5):

Ideal gas at constant pressure
αV=1T
(10.6)

At 0 °C (273 K) this gives 3.7×10−3 K−1, and at room temperature (about 300 K) 3.3×10−3 K−1. That is roughly fifteen times the value for water or mercury and about a hundred times that for steel. Unlike a liquid's, a gas's expansivity clearly falls as it gets warmer.

Linking volume and linear expansion

DerivationWhy αV=3αl

Take a cube of side l made of an isotropic solid, one that expands equally in every direction. Warming it by ΔT lengthens each edge by Δl=αll ΔT, so

ΔV=(l+Δl)3−l3≃3l2 Δl
(10.7)

The full expansion is 3l2Δl+3l(Δl)2+(Δl)3. Even for a large warming Δl/l is only about 10−3, so the last two terms are at least a thousand times smaller than the first, and we drop them. Dividing by V=l3:

ΔVV=3Δll=3αl ΔT
(10.8)
αV=3αl
(10.9)

The same reasoning for a thin sheet gives an area expansivity of 2αl, as Example 10.1 shows.

Example 10.1NCERT Example 10.1 · p. 207

Area expansivity of a steel tile

A thin rectangular steel tile has sides a and b (Fig. 10.6). Show that its coefficient of area expansion, (ΔA/A)/ΔT, is 2αl.

Fig. 10.6The extra area is two thin strips plus a tiny corner square.NCERT Fig. 10.8 · p. 207

A rectangle with sides a and b is drawn with its expansion exaggerated. A thin strip of width delta b runs along the top, of area a times delta b; a thin strip of width delta a runs down the right side, of area b times delta a; and a tiny square of area delta a times delta b sits in the top right corner.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.2NCERT Example 10.2 · p. 207

Shrink-fitting a ring gear

A steel ring gear has an inner diameter of 299.70 mm at 30 ∘C. It must slide over a flywheel rim of diameter 300.00 mm. To what temperature must the ring be heated? Take αl=1.2×10−5 K−1 for steel.

Try it yourself first. Then reveal the solution one step at a time.

Thermal stress

Now stop the expansion. Fix a steel rod between two immovable walls and warm it (Fig. 10.7). The walls must push it back by exactly the length it would have gained, so the rod carries a compressive strain Δl/l=αl ΔT. By Hooke's law (Chapter 8) the thermal stress in the rod is

Thermal stress
FA=Yαl ΔT
where
Y
the Young's modulus of the material

The stress depends only on the material and the temperature change, not on the rod's length or thickness. The force, F=AYαl ΔT, grows with the cross-section and can be enormous. That is why railway lines were once laid with gaps between rails, why modern welded rails are laid at a carefully chosen temperature and anchored to heavy concrete sleepers, and why bridges rest on rollers and expansion joints.

3/3The walls push it back: compressive stress Yαl ΔT.

Fig. 10.7A clamped rod cannot grow, so the walls squeeze it: thermal stress Yαl ΔT.

A steel rod fits exactly between two rigid walls. Step 1: cool, no force. Step 2: heated, a dashed ghost shows it would grow by delta l past the right wall. Step 3: the walls stop the growth and push inwards on both ends with forces F, giving compressive stress Y alpha delta T.

Example

Stress in a bridge girder

A steel girder with a cross-section of 120 cm2 is fixed rigidly at both ends on a winter morning. By the afternoon it is 25 K warmer. Find the thermal stress and the force on each support. Take Y=2.0×1011 Pa and αl=1.2×10−5 K−1.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
A steel plate has a circular hole in it. When the plate is heated, the diameter of the hole
Show answer

Answer (B) increases, as if the hole were filled with steel

Thermal expansion scales every length in the plate by the same factor, including the rim of the hole. The steel does not "flow inward".

10.6 Specific heat capacity

Why does an empty steel tawa get too hot to touch within a minute on the gas, while a pan of water on the next burner takes several minutes just to get warm? The heat needed to warm something depends on more than the flame. A heater that supplies energy at a steady rate lets you compare amounts of heat simply by timing them.

Fig. 10.8Same heater, different samples: more mass or a larger specific heat means a slower rise.

Temperature rise is plotted against heating time for one steady heater. Three straight lines start at the origin. The steepest is 200 g of coconut oil, rising about twice as fast as 200 g of water. The middle line is 200 g of water. The shallowest is 400 g of water, rising half as fast as 200 g of water.

Three things matter, then: how much material there is (m), how far it is warmed (ΔT) and what it is made of. The heat capacity S of a body is the heat it needs per kelvin of warming:

Heat capacity
S=ΔQΔT
(10.10)

Heat capacity belongs to an object: a full bucket has more than a cup. To get a property of the material, divide by the mass. The specific heat capacity s is the heat needed per kilogram per kelvin:

Specific heat capacity
s=Sm=1mΔQΔT
(10.11)

Equivalently ΔQ=ms ΔT, which holds while the substance stays in one state. The SI unit of s is J kg−1 K−1. It depends on the substance and, slightly, on its temperature.

Definition
Specific heat capacity

The heat needed to raise the temperature of unit mass of a substance by one kelvin, without a change of state; s=ΔQ/(m ΔT).

ExampleWater: about 4186 J kg−1 K−1.

When the amount of substance is counted in moles μ, the heat needed per mole per kelvin is the molar specific heat capacity C, in J mol−1 K−1:

Molar specific heat capacity
C=1μΔQΔT
(10.12)

For a gas the answer depends on how you heat it. Warmed in a rigid container, a gas uses all the heat to raise its internal energy; this gives CV, the molar specific heat at constant volume. Warmed at constant pressure, it also expands and pushes back its surroundings, so it needs more heat for each kelvin; this gives CP. In Table 10.5, CP−CV comes out close to 8.3 J mol−1 K−1, the gas constant R, for every gas. Chapter 11 explains why.

Table 10.4Specific heat capacities near room temperature and 1 atm
Substances (J kg−1 K−1)
Water4186
Ethanol2440
Ice (near 0 °C)2100
Aluminium900
Glass840
Dry sand830
Iron450
Copper385
Mercury140
Lead128
Table 10.5Molar specific heats of some gases near room temperature (J mol⁻¹ K⁻¹)
GasCPCVCP−CV
Helium20.812.58.3
Argon20.812.58.3
Nitrogen29.120.88.3
Oxygen29.421.18.3
Carbon dioxide37.128.88.3

Compare a day in coastal Mumbai with one in Jaisalmer, in the Thar desert. In Mumbai the night is barely cooler than the afternoon; in Jaisalmer the temperature often swings by more than 15 °C. The difference is the sea. Water's specific heat is about five times that of sand or rock, so the Arabian Sea soaks up the day's sunshine with only a small rise in temperature and gives the heat back slowly through the night, while dry sand heats and cools fast. The same property makes water the coolant in car engines and the filling of a rubber hot-water bottle.

Example

Heating bath water

A 2.0 kW electric geyser heats 15 L of water from 18 ∘C to 45 ∘C. Ignoring heat losses, how long does it take?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
Equal masses of copper and aluminium receive the same amount of heat. Which warms more?
Show answer

Answer (D) Copper, because its specific heat is smaller

ΔT=ΔQ/(ms). Copper (385 J kg−1 K−1) has a smaller s than aluminium (900), so it warms more than twice as much. Equal masses do not mean equal warming.

10.7 Calorimetry

Wrap a system so well that no heat can get in or out, and we call it isolated. Energy cannot leave it, so when heat flows between its parts, every joule lost by the hot parts turns up in the cold ones. Calorimetry rests on this bookkeeping, also called the method of mixtures:

Principle of calorimetry
heat lost by hot parts=heat gained by cold parts

Calorimetry means the measurement of heat, and a calorimeter is the vessel used for it (Fig. 10.9). Each part has a job:

  • A thin cup of copper or aluminium. It takes little heat itself and quickly matches the temperature of its contents.
  • A stirrer of the same metal, so that the water, the cup and any sample reach one common temperature.
  • An outer jacket, with glass wool or foam packed into the gap, to slow heat leaking in from or out to the room.
  • A lid with a hole for the thermometer.

The cup warms along with its contents, so it must be counted. A cup of mass mc and specific heat sc absorbs as much heat per kelvin as a mass mcsc/sw of water would; this mass is the cup's water equivalent. In practice some heat still leaks to the room, so careful experiments are quick and start a little below room temperature.

2/3Heat flows from the block into the water and the cup.

Fig. 10.9Heat lost by the hot block equals heat gained by the water and the cup, if none escapes.

A metal cup of water with a stirrer and thermometer inside an insulated jacket. Step 1: a metal block at 98 °C hangs above water at 24 °C. Step 2: the block is in the water and arrows show heat flowing from it into the water and cup. Step 3: everything reads a steady 28 °C.

Example 10.3NCERT Example 10.3 · p. 209

Finding the specific heat of a metal

A 0.120 kg metal cylinder is heated in a water bath to 98.0 ∘C and quickly lowered into an aluminium calorimeter of mass 0.080 kg holding 0.200 kg of water at 24.0 ∘C. The final steady temperature is 28.0 ∘C. Find the specific heat of the metal. Take sw=4186 and sAl=900 J kg−1 K−1.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
In an insulated flask of negligible heat capacity, 200 g of water at 80 °C is mixed with 300 g of water at 20 °C. What is the final temperature, in °C?
°C
Show answer

Answer 44 °C

Heat lost = heat gained: 200(80−T)=300(T−20), so 500T=22 000 and T=44 ∘C. It is nearer 20 °C because there is more cold water.

10.8 Change of state

Ice cubes in a glass of nimbu-paani turn to water; water in a kettle turns to steam; on a winter morning, the water vapour in your breath turns back into tiny droplets. Each is a change of state, a transition between solid, liquid and gas, and each happens when the substance takes heat from, or gives heat to, its surroundings. Changes of state share a feature that you can check for yourself.

Names first: solid to liquid is melting (or fusion), and liquid to solid is freezing. Your graph shows what is special about them. The flat step means that for as long as both ice and water are present, they sit together in thermal equilibrium at one fixed temperature, the melting point. Each pure substance has its own. Because it shifts a little with pressure, data tables quote the normal melting point, measured at 1 atm=1.013×105 Pa.

Two states in equilibrium share one temperature and one pressure; what differs between them is the density, that is, the volume per mole. Ice floating in water at 0 °C is the everyday example: same temperature, same pressure, but the ice is about 9 % less dense.

SimulationIce to steam, particle by particle

Water molecules in a closed container, drawn as particles, with a thermometer and a temperature slider that heats or cools the sample. Ice particles vibrate in a lattice, liquid particles slide past one another and steam particles fly freely. While the sample melts or boils, the thermometer holds at 0 °C or 100 °C and a label reports the percentage that has changed state.

Fig. 10.10Move the temperature slider to heat or cool the sample, and watch the thermometer pause at each change of state.
Try this
  • Start with ice at −20 °C and set the slider to 30 °C. Where does the thermometer pause, and what does the label report while it waits?
  • Now set 120 °C. Compare the pause at 100 °C with the pause at 0 °C. Which change of state needs more heat?
  • Cool from 120 °C back to 50 °C. What happens at 100 °C on the way down?

Pressure changes melting points, and for ice the effect runs the 'wrong' way. Ice is less dense than water, so squeezing it favours the liquid and lowers the melting point, by about 0.007 °C for each extra atmosphere. A neat experiment shows the result.

2/3Ice melts under the wire and refreezes above it.

Fig. 10.11Pressure melts the ice under the wire; the water refreezes above it.NCERT Fig. 10.10 · p. 211

A block of ice rests on two supports with a wire over it, weighted at both ends. Step 1: the wire sits on top. Step 2: a close-up circle shows high pressure under the wire, a thin film of meltwater flowing round it, and refreezing above it. Step 3: the wire has dropped out below the block, which is whole, with a faint line where the wire passed.

Ice skating is often explained by the same effect, with the blade gliding on water melted by pressure. A skater's weight on a narrow blade does lower the melting point, but only slightly, while skating works well even at −20 °C. Measurements show that most of the slippery layer comes from frictional heating and from a thin film at the surface of ice that is liquid-like even below 0 °C.

Carry on heating and a second flat step appears near 100 °C. Now liquid is turning into vapour, which is vaporisation; in boiling, bubbles of vapour form inside the liquid itself. Liquid and vapour coexist in equilibrium at a fixed temperature, the boiling point, until the liquid is used up.

Ask anyone who has cooked rajma in Leh. About 3500 m up, the air pressure is roughly two-thirds of its sea-level value and water boils near 88 °C, so dal simmers for ages. A pressure cooker does the opposite: its weighted valve holds the pressure inside at about twice atmospheric, water boils near 120 °C, and food cooks in a fraction of the time. The boiling point at 1 atm is the normal boiling point.

Some solids skip the liquid stage altogether. Camphor (कपूर) left in an open puja box slowly disappears, and solid carbon dioxide, dry ice, turns straight into gas at −78.5 °C; the white fog around it is water vapour from the air condensing in the cold. A direct change from solid to vapour is sublimation; iodine and naphthalene balls also sublime.

3/4Lower pressure, lower boiling point: Leh 88 °C, a pressure cooker 120 °C.

Fig. 10.12A phase diagram for water (not to scale). Each curve is a set of conditions where two states coexist.NCERT Box: Triple point · p. 211

Schematic phase diagram of water, pressure against temperature. Step 1: fusion, vaporisation and sublimation curves separate the solid, liquid and vapour regions; the vaporisation curve ends at the critical point. Step 2: the triple point where they meet, 273.16 K and 611.66 Pa. Step 3: points on the vaporisation curve at 0.65 atm and 88 °C, 1 atm and 100 °C, and 2 atm and 120 °C. Step 4: an upward arrow by the left-leaning fusion curve: more pressure, lower melting point.

Quick check
1 of 3
Water boils at about 88 °C in Leh mainly because
Show answer

Answer (C) the air pressure there is lower

The boiling point is set by the pressure on the liquid. Lower pressure at altitude means water boils at a lower temperature; cold air only makes it take longer to reach that point.

10.8.1 Latent heat

Look again at the flat steps of Fig. 10.13: the heater is still pouring energy in, yet the thermometer does not move. That energy goes into rearranging the molecules, loosening the solid's framework or pulling the liquid's molecules apart, and it is called latent heat, from the Latin for hidden. The heat needed for a change of state is proportional to the mass that changes:

Latent heat
Q=mL
(10.13)
where
L
the latent heat of the substance for that change, in J kg−1
Definition
Latent heat

The heat absorbed or released per unit mass when a substance changes state at constant temperature and pressure; L=Q/m.

ExampleMelting 1 kg of ice at 0 °C takes 3.34×105 J.

The latent heat L depends on pressure and is quoted at 1 atm. For solid to liquid it is the latent heat of fusion Lf; for liquid to vapour, the latent heat of vaporisation Lv. The same amount comes back out when the change reverses: freezing water releases Lf per kilogram and condensing steam releases Lv. Table 10.7 gives values for a few substances.

5/5Above 100 °C the steam warms; its slope is steep again.

Fig. 10.13Heating curve of water (not to scale). Flat parts are changes of state; slopes are 1/(ms).NCERT Fig. 10.12 · p. 213

Temperature against heat supplied to water, not to scale. Step 1: a steep line as ice warms from −20 °C to 0 °C. Step 2: a flat step at 0 °C while it melts. Step 3: a less steep line as water warms to 100 °C. Step 4: a long flat step at 100 °C while it boils. Step 5: a steep line as steam warms above 100 °C.

On a graph of temperature against heat supplied, the flat steps are the changes of state and the sloping parts have slope ΔT/ΔQ=1/(ms). The ice and steam lines are steeper than the water line because ice and steam have about half the specific heat of liquid water.

Table 10.7Melting and boiling points and latent heats at 1 atm (latent heats in 10⁵ J/kg)
SubstanceMelting point (°C)LfBoiling point (°C)Lv
Water03.3410022.6
Ethanol−1141.1788.5
Mercury−390.113572.9
Nitrogen−2100.26−1962.0
Lead3270.2317498.6
Copper10852.1256247

For water, Lf=3.34×105 J/kg and Lv=2.26×106 J/kg. Turning a kilogram of boiling water into steam takes almost seven times as much heat as melting a kilogram of ice, and more than five times as much as warming that water from 0 °C to 100 °C. Turn that around: a kilogram of steam condensing on skin dumps 2.26×106 J there before it has even begun to cool, which water already at 100 °C cannot do. A puff of steam from an idli steamer or a pressure-cooker whistle therefore burns worse than a splash of boiling water.

Example 10.4NCERT Example 10.4 · p. 213

Measuring the latent heat of fusion

In an insulated flask of negligible heat capacity, 80 g of ice at 0 ∘C is added to 400 g of water at 40.0 ∘C. When all the ice has melted, the mixture is at 20.0 ∘C. Find the latent heat of fusion of ice. Take sw=4186 J kg−1 K−1.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.5NCERT Example 10.5 · p. 214

From freezer ice to steam

How much heat turns 2.0 kg of ice at −10 ∘C into steam at 100 ∘C at 1 atm? Use sice=2100, sw=4186 J kg−1 K−1, Lf=3.34×105 and Lv=2.26×106 J/kg.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
How much heat, in kilojoules, is released when 50 g of steam at 100 °C condenses to water at 100 °C? (Lv=2.26×106 J/kg)
kJ
Show answer

Answer 113 kJ

Q=mLv=0.050×2.26×106=1.13×105 J=113 kJ.

10.9 Heat transfer

Heat is energy on the move from hot to cold, between two bodies or between two parts of one body. It travels in three distinct ways, and a pan of milk on a gas stove shows all of them at once (Fig. 10.14): conduction through the steel ladle resting in it, convection in the milk as it circulates, and radiation that you feel on your hand held beside the flame.

3/3The flame warms your hand directly, with no medium needed.

Fig. 10.14Three ways heat travels: conduction, convection and radiation.NCERT Fig. 10.13 · p. 214

A pan of milk on a gas flame with a steel ladle in it and a hand to one side. Step 1: an arrow along the ladle shows conduction. Step 2: loops in the milk show hot milk rising in the middle and cooler milk sinking at the sides: convection. Step 3: wavy arrows from the flame reach the hand directly: radiation.

10.9.1 Conduction

Leave a steel ladle in a pot of simmering dal and soon its handle is too hot to hold, although nothing has flowed up it. Energy has been handed along the steel from particle to particle: fast-jiggling particles at the hot end bump their slower neighbours, and in metals free electrons carry most of the energy, which is why metals are such good conductors. This is conduction, the transfer of heat through a material from hotter to cooler regions with no bulk motion of the material. Liquids conduct less well than solids, and gases very poorly.

To put numbers on conduction, use a standard arrangement (Fig. 10.15): a bar with a uniform cross-section A and a length L, its sides lagged so that heat can leave only through the ends, which are pressed against two large reservoirs at TC and TD (TC>TD). Leave it long enough and the bar settles into a steady state in which nothing changes with time: the temperature falls evenly from one end to the other, and heat leaves the cold end exactly as fast as it enters the hot end. Experiments on such bars give a simple rule. The rate of flow, the heat current H, doubles if you double the cross-section or the temperature difference, and halves if you double the length:

Heat conduction
H=KA TC−TDL
(10.14)
where
K
the thermal conductivity of the material
H
the heat current, in watts

3/3The same heat current H crosses every section of the bar.

Fig. 10.15In the steady state the temperature falls linearly along the bar, and the same heat current crosses every section.NCERT Fig. 10.14 · p. 215

A lagged bar joins a hot reservoir at T C to a cold one at T D, above a graph of temperature against position. Step 1: soon after the start only the hot end is warm; the curve drops steeply then flattens. Step 2: in the steady state the graph is a straight line from T C to T D. Step 3: arrows show the same heat current H along the bar.

The constant K is the thermal conductivity (ऊष्मा चालकता) of the material. Its SI unit is J s−1 m−1 K−1, that is, W m−1 K−1. It varies a little with temperature, but over ordinary ranges we treat it as constant. Table 10.8 shows its enormous range: copper conducts heat about 15 000 times better than still air.

Table 10.8Thermal conductivities near room temperature
MaterialK (W m−1 K−1)
Silver429
Copper401
Aluminium237
Brass109
Steel50
Lead35
Ice2.2
Glass1.0
Concrete0.8
Water0.60
Wood0.12
Thermocole0.035
Air (still)0.026

Designers use both ends of the table. A steel tumbler of hot chai is uncomfortable to hold while a clay kulhad of the same chai is fine, because the kulhad's wall conducts heat to your fingers far more slowly. Ice-cream sellers pack their stock in thermocol boxes, and a razai keeps you warm, mainly because both trap air, one of the poorest conductors there is. At the other extreme, a laptop moves heat from its processor through copper heat pipes to aluminium fins, so that the chip never overheats.

DerivationRods in series: thermal resistance

Rewrite Eq. (10.14) as TC−TD=H L/(KA). The quantity R=L/(KA) is the thermal resistance of the bar: the temperature difference plays the part of a voltage and H the part of a current. Join two bars end to end and lag them. In the steady state the same H must cross both, or the junction would warm up or cool down, so the two temperature drops simply add:

Thermal resistances in series
T1−T2=H(R1+R2),R=LKA

Resistances in series add, just as you will find for electric circuits in Class 12. For bars side by side between the same two reservoirs, the heat currents add instead.

Example 10.6NCERT Example 10.6 · p. 215

Where is the junction hottest?

A copper rod 20.0 cm long and a steel rod 10.0 cm long, both of cross-section 4.0 cm2, are joined end to end and lagged. The free end of the copper is held in steam at 100 ∘C and the free end of the steel in melting ice at 0 ∘C. Find the steady temperature of the junction and the heat current. Take KCu=400 and Ksteel=50 W m−1 K−1.

Try it yourself first. Then reveal the solution one step at a time.

Example 10.7NCERT Example 10.7 · p. 216

Two rods of equal length

Two rods of equal length L and cross-section A, with conductivities K1 and K2, are joined end to end and lagged. The free end of rod 1 is at T1 and that of rod 2 at T2, with T1>T2. (i) Find the junction temperature T0. (ii) Find the conductivity K′ of a single rod of length 2L that would carry the same heat current. (iii) Evaluate both, and the heat current, for an aluminium-alloy rod (K1=200 W m−1 K−1) and a steel rod (K2=50 W m−1 K−1), each 0.25 m long with A=4.0 cm2, T1=400 K and T2=300 K.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveCompound rod lab

Two lagged rods joined end to end between a hot and a cold reservoir. Menus choose each material (copper, aluminium, brass, steel or glass); sliders set the lengths and end temperatures. A graph shows temperature along the rods, straight within each rod with a kink at the junction, and readouts give the junction temperature, both thermal resistances, the equivalent conductivity and the heat current.

Fig. 10.16Pick two materials, set their lengths and the end temperatures, and read the junction temperature and the heat current.
Quick check
1 of 3
A window pane of area 1.0 m2 and thickness 4.0 mm has its faces at temperatures 5.0 K apart. With K=1.0 W m−1 K−1, what is the heat current, in watts?
W
Show answer

Answer 1250 W

H=KA ΔT/L=1.0×1.0×5.0/0.0040=1250 W. Thin glass lets heat through fast, which is why double glazing traps a layer of air.

10.9.2 Convection

Sprinkle a pinch of haldi into a pan of water on the stove and watch the specks: they rise up the middle and sink down the sides, round and round. The water itself is moving and carrying heat with it. This is convection (संवहन), and it can happen only in liquids and gases. Water at the bottom is warmed, expands and becomes less dense, so buoyancy pushes it up; denser, cooler water flows down to replace it and is warmed in turn. Such a convection current spreads heat through the pan far faster than conduction alone. Because it relies on buoyancy, this natural convection needs gravity: on the International Space Station a candle burns as a dim, round glow instead of a tall flame.

In forced convection a fan or pump makes the fluid move. A ceiling fan, a convection oven and the fan inside a laptop all push air past hot surfaces to carry heat away faster. Your body runs a forced-convection system too: the heart pumps blood, which collects heat from busy muscles and organs and carries it to the skin, keeping the whole body at nearly the same temperature.

Stand on Marina beach at four in the afternoon and the wind blows steadily in off the Bay of Bengal; stand there after midnight and a gentler breeze drifts out to sea (Fig. 10.17). The cause is a convection loop driven by uneven heating. Under the afternoon sun the sand gets much hotter than the sea surface, because the sea's large specific heat and its constant stirring keep its temperature nearly steady. The hot sand warms the air above it; that air turns buoyant and climbs, spreads seawards high up, sinks over the cooler water and returns along the surface as the sea breeze. After dark the sand loses its heat quickly while the sea stays warm, so the loop runs backwards and a land breeze blows off the shore.

Fig. 10.17By day the land is warmer and the breeze blows in from the sea; at night the loop reverses.NCERT Fig. 10.17 · p. 217

Two coastal panels. Day: the sun shines, warm air rises over the land, flows out to sea high up, sinks over the cooler sea and blows inland at the surface as a sea breeze. Night: the moon is out, air rises over the warmer sea and the surface wind blows from land to sea as a land breeze.

Scale the loop up to a whole hemisphere and you get the trade winds, the steady north-easterlies that carried sailing ships across the tropics. Sunlight is strongest at the equator, so air there rises and drifts towards the poles at high altitude; without rotation it would sink near the poles and flow straight back. But the Earth spins. A point on the equator moves east at about 1670 km/h and a point near a pole hardly moves at all, and air travelling across this spinning surface is turned aside. The high-level flow therefore sinks at around 30° latitude, and the return flow near the ground is bent into a wind from the north-east.

10.9.3 Radiation

Step out of the shade into sunshine and your skin warms at once. Nothing material links you to the Sun across 150 million kilometres of nearly empty space, so neither conduction nor convection can be responsible. The energy arrives as radiation: electromagnetic waves, oscillating electric and magnetic fields that travel even through vacuum at the speed of light, 3×108 m/s. Sunlight takes about 8.3 minutes to reach us. The energy carried this way is called radiant energy; you will study electromagnetic waves in Class 12.

You radiate too. Every object, solid, liquid or gas, emits electromagnetic waves just because it is warm, and this thermal radiation grows rapidly with temperature: your body gives off invisible infrared, a toaster's elements glow orange, and the Sun shines white. Radiation that lands on a surface is partly bounced back and partly soaked up. Matt black surfaces soak up nearly all of it and, at the same temperature, also send out the most; polished, pale surfaces reflect most of it and emit little.

That is why village houses in Rajasthan are often lime-washed white and why white 'cool-roof' paint keeps top-floor rooms cooler, while the absorber plates of rooftop solar water heaters and the pots in a box-type solar cooker are painted matt black.

A vacuum flask, or thermos, keeps tea hot or ice cold for hours by defeating each mode in turn (Fig. 10.18):

  • Conduction and convection: the bottle is double-walled glass with a vacuum between the walls, so neither can carry heat across the gap.
  • Radiation: the glass surfaces facing the gap are silvered. Radiation from the hot inner wall is reflected back inwards, and radiation arriving from outside is reflected away.
  • The leaks that remain: heat can still conduct through the stopper and the neck, so the stopper is cork or plastic and the bottle rests on insulating pads.

3/3Silvered walls bounce radiation back towards the tea.

Fig. 10.18A vacuum flask stops conduction and convection with a vacuum, and radiation with silvered walls.

Cross-section of a vacuum flask: double glass walls with a vacuum between them, silvered faces, a stopper and base pads, holding tea. Step 1: an arrow trying to cross the vacuum is crossed out, so no conduction. Step 2: an upward arrow is stopped by the stopper and the empty gap has no air to circulate, so no convection. Step 3: wavy radiation arrows bounce off the silvered walls back into the tea.

Quick check
1 of 4
Why does a land breeze blow from the land towards the sea at night?
Show answer

Answer (B) The land cools faster than the sea, so air rises over the warmer sea

With its smaller specific heat, land loses heat faster after sunset. Air rises over the now-warmer sea and surface air flows out from the land to replace it. The first option has the temperatures the wrong way round.

10.9.4 Blackbody radiation

Send the glow of a heater coil through a prism and you get not one colour but a smear: thermal radiation is spread over a continuous range of wavelengths, most of them invisible. How the energy is shared out depends on the temperature. Fig. 10.19 plots the energy that a perfect emitter radiates per unit area per unit wavelength at three temperatures. Each curve climbs to a peak and then tails off; as the temperature rises, the whole curve rises and its peak slides to shorter wavelengths.

3/3At 6000 K it peaks near 485 nm; the peaks march to shorter wavelengths.

Fig. 10.19Hotter bodies radiate more at every wavelength, and their peak moves towards shorter wavelengths.NCERT Fig. 10.18 · p. 218

Energy per unit wavelength against wavelength from 0 to 3000 nm, with the visible band shaded and a legend of line styles. Step 1: a low dotted 4000 K curve peaking near 725 nm. Step 2: a dashed 5000 K curve, about three times taller, peaking near 580 nm. Step 3: a much taller solid 6000 K curve peaking near 485 nm, with a dotted line joining the three peaks as they move to shorter wavelengths.

The wavelength at the peak, λm, is inversely proportional to the absolute temperature. This is Wien's displacement law:

Wien's displacement law
λmT=b
(10.15)
where
b
Wien's constant, 2.9×10−3 m K

Wien's law explains the colour sequence of anything heated strongly, such as an electric stove coil that glows dull red, then orange, and a filament that glows yellow-white: as the temperature rises, the peak moves out of the infrared towards the visible. It also lets astronomers read a star's surface temperature from its colour. Sirius, whose spectrum peaks near 290 nm in the ultraviolet, has a surface at about 10 000 K; red Betelgeuse, peaking near 830 nm, is at roughly 3500 K; the Sun, peaking near 500 nm, is at about 5800 K. Even you radiate: skin at 34 °C peaks near 9.4 μm in the infrared, which is what a thermal camera sees.

The shape of these curves is universal: it is fixed by the temperature alone, and how big the perfect emitter is, what shape it has and what it is made of make no difference. Such a perfect emitter is called a blackbody. A small hole in the wall of a hot furnace is a very good one. Explaining the curves defeated nineteenth-century physics. Max Planck succeeded in 1900 only by assuming that energy is emitted in small packets, or quanta, and so began quantum physics, which you will meet in Class 12.

InteractiveBlackbody spectrum lab

A graph of blackbody radiation against wavelength with the visible band shaded. A temperature slider (1000 to 12 000 K) and buttons for a heater coil, a lamp filament, Betelgeuse, the Sun and Sirius redraw the curve beside a dashed Sun curve. Readouts give the Wien peak wavelength, the power per square metre, that power compared with the Sun's, and the approximate glow colour.

Fig. 10.20Drag the temperature, or pick a star or a lamp, and watch the peak move and the total power soar.

How much power does a hot body radiate in all? For a perfect radiator of surface area A at absolute temperature T, the energy emitted per second is

Stefan–Boltzmann law
H=AσT4
(10.16)
where
σ
the Stefan–Boltzmann constant, 5.67×10−8 W m−2 K−4

Josef Stefan found this law from experimental data in 1879, and Ludwig Boltzmann derived it from theory in 1884. Doubling the absolute temperature multiplies the radiated power by 16. Real surfaces radiate less than a blackbody at the same temperature, so we write

H=AeσT4
(10.17)

where the emissivity e, between 0 and 1, depends on the surface: above 0.95 for lamp black and human skin, below 0.1 for polished silver. Because a good absorber is a good emitter, the same e also gives the fraction of incoming radiation absorbed. An orange-hot heater rod with a surface area of 120 cm2, at 1100 K and with e≈0.9, radiates about 0.9×0.012×5.67×10−8×11004≈900 W, most of a 1 kW heater's output.

A body also absorbs radiation from its surroundings. If it is at T and its surroundings are at Ts, the net rate at which it loses energy by radiation is

Net radiation loss
H=eσA (T4−Ts4)
(10.18)
Example

How much power reaches the Earth?

Treat the Sun as a blackbody of radius 6.96×108 m with a surface temperature of 5770 K. (a) Find the total power it radiates. (b) Find the power per square metre arriving at the Earth, 1.50×1011 m away, on a surface facing the Sun.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
A star's spectrum peaks at 400 nm. Estimate its surface temperature in kelvin. (b=2.9×10−3 m K)
K
Show answer

Answer 7250 K

T=b/λm=2.9×10−3/4.00×10−7≈7250 K, hotter than the Sun, whose peak is near 500 nm.

10.10 Newton's law of cooling

A steel tumbler of filter coffee left on the counter loses its heat quickly at first, then more and more slowly, until it is at room temperature. How does the rate of cooling depend on how hot it is? You can find out with a thermometer and a watch.

Isaac Newton stated the rule in 1701, in an anonymous paper on measuring high temperatures. According to Newton's law of cooling, a body loses heat at a rate proportional to the difference between its temperature T2 and the temperature T1 of its surroundings:

Newton's law of cooling
−dQdt=k(T2−T1)
(10.19)

The constant k>0 is set by how much surface is exposed and what it is like: shiny or dull, in still air or a breeze. The law is an approximation that works when the body is not much hotter than its surroundings, a few tens of kelvin at most, as for a cup of tea on a table or a room losing heat through its walls. Then every route by which heat escapes grows roughly in step with the temperature difference. Radiation shows why: T24−T14=(T2−T1)(T2+T1)(T22+T12)≈4T13(T2−T1) when T2 is close to T1, so even Eq. (10.18) becomes linear.

DerivationFrom the law to an exponential

Let the body have mass m and specific heat s. When its temperature changes by dT2 in a time dt, its heat content changes at the rate

dQdt=ms dT2dt
(10.20)

Put this into Eq. (10.19) and separate the variables:

dT2T2−T1=−kms dt=−K dt
(10.21)
where
K
the cooling constant k/ms

Integrate, keeping the surroundings at a fixed T1:

ln⁡(T2−T1)=−Kt+c
(10.22)

or, taking exponentials,

T2=T1+C′e−Kt
(10.23)
where
C′
ec, the excess temperature at t=0

The excess temperature decays exponentially: it falls by the same fraction in every equal interval of time. It halves every (ln⁡2)/K, whatever its starting value.

To test the law, the surroundings must stay at a known, steady temperature. A practical arrangement stands a copper calorimeter of hot water inside a double-walled vessel with water between its walls. One thermometer reads the calorimeter water, T2, and another reads the water in the walls, T1. A graph of ln⁡(T2−T1) against time is a straight line of slope −K (Fig. 10.21), as Eq. (10.22) predicts.

3/3The log of the excess against time is a straight line of slope −K.

Fig. 10.21The excess temperature decays exponentially, so its logarithm falls in a straight line.NCERT Fig. 10.20 · p. 220

Step 1: temperature against time for cooling water, a curve that falls steeply at first and levels off towards room temperature, shown dashed. Step 2: a steep tangent early and a gentle tangent later show that the cooling rate falls with the excess temperature. Step 3: the graph of the natural log of the excess temperature against time, a straight line sloping down with slope minus K.

InteractiveCooling lab

A temperature–time graph and a graph of the log of the excess temperature against time. Sliders set the starting temperature, the room temperature and the cooling constant K; Play runs the clock. The curve bends towards room temperature while the log graph stays straight, and a readout gives the halving time.

Fig. 10.22Set the starting temperature, the room temperature and the cooling constant, press Play, and compare the two graphs.
Example 10.8NCERT Example 10.8 · p. 221

How long to cool a cup of chai?

Chai cools from 80 ∘C to 70 ∘C in 4.0 minutes in a room at 30 ∘C. Find the time it needs to drop from 60 ∘C to 50 ∘C. Solve (a) using average temperatures, as is usual for small intervals, and (b) with the exponential law.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
For a body obeying Newton's law of cooling, a graph of ln⁡(T2−T1) against time is
Show answer

Answer (C) a straight line with a negative slope

Eq. 10.22 is ln⁡(T2−T1)=−Kt+c: a straight line of slope −K. The levelling curve is the graph of T2 itself.
Recap

Thermal Properties of Matter: key points and summary

  1. Temperature measures how hot a body is; heat is energy transferred because of a temperature difference, and the flow stops at thermal equilibrium (Section 10.2).
  2. A temperature scale needs a thermometric property and fixed points; tF=95tC+32 (Section 10.3).
  3. Low-density gases obey PV=μRT. Constant-volume gas thermometers extrapolate to absolute zero, −273.15 °C, the zero of the Kelvin scale: T=tC+273.15 (Section 10.4).
  4. Δl/l=αl ΔT and ΔV/V=αV ΔT; for an isotropic solid αV=3αl and the area expansivity is 2αl; for an ideal gas at constant pressure αV=1/T (Section 10.5).
  5. Water is densest near 4 °C, so lakes freeze from the top down (Water: the odd one out).
  6. A rod prevented from expanding carries a thermal stress Yαl ΔT (Thermal stress).
  7. ΔQ=ms ΔT; molar specific heat C=ΔQ/(μ ΔT), and for gases CP−CV≈R (Section 10.6).
  8. In an isolated system, heat lost by the hot parts equals heat gained by the cold parts (Section 10.7).
  9. During a change of state the temperature stays constant and Q=mL; pressure shifts melting and boiling points, and all three states coexist at the triple point (Section 10.8, Section 10.8.1).
  10. Conduction: H=KA(TC−TD)/L in the steady state; thermal resistances L/KA add in series (Section 10.9.1).
  11. Convection moves heat by the bulk flow of a fluid, natural or forced; radiation travels as electromagnetic waves and needs no medium (Section 10.9.2, Section 10.9.3).
  12. Wien's law λmT=b; Stefan–Boltzmann law H=AeσT4, with net loss eσA(T4−Ts4) (Section 10.9.4).
  13. Newton's law of cooling, −dQ/dt=k(T2−T1), gives T2=T1+C′e−Kt for small temperature differences (Section 10.10).
Practice · 29 questions

Practice questions with answers (NCERT “Exercises”)

  1. NumericalEasyNCERT Exercises 10.1

    Two fixed points of the International Temperature Scale are the triple point of hydrogen, 13.8033 K, and the triple point of mercury, 234.3156 K. Express each in degrees Celsius and in degrees Fahrenheit.

    Show solution

    Hydrogen: tC=13.8033−273.15=−259.35 ∘C and tF=95(−259.35)+32=−434.82 ∘F. Mercury: tC=234.3156−273.15=−38.83 ∘C and tF=95(−38.83)+32=−37.90 ∘F.

  2. NumericalMediumNCERT Exercises 10.2

    Two absolute temperature scales, P and Q, both have their zero at absolute zero. Scale P gives the triple point of water the value 400, and scale Q gives it the value 150.

    1. Find the relation between TP and TQ.

      Show solution

      Both are proportional to the Kelvin temperature, so TP/TQ=400/150, that is, TP=83TQ.

    2. What does scale Q read when scale P reads 600?

      Show solution

      TQ=38×600=225.

  3. NumericalMediumNCERT Exercises 10.3

    A resistance thermometer obeys R=R0[1+α(T−T0)]. It reads 80.0 Ω in a triple-point cell (273.16 K) and 129.0 Ω at the freezing point of zinc (692.68 K). What temperature, in kelvin, corresponds to a reading of 100.0 Ω?

    K
    Show solution

    Linear interpolation: T=273.16+100.0−80.0129.0−80.0(692.68−273.16). T=273.16+0.408×419.52≈444.4 K.

  4. Short answerMediumNCERT Exercises 10.4

    Answer these questions about fixed points and the Kelvin scale.

    1. Why is the triple point of water a better fixed point than the ice point or the steam point?

      Show solution

      The triple point occurs at one unique temperature and pressure. The ice and steam points shift with the pressure and with dissolved air or impurities, so they are harder to reproduce exactly.

    2. The Kelvin scale needs only one experimental fixed point. What serves as the other?

      Show solution

      Absolute zero, 0 K, is the natural second fixed point; the triple point is given the value 273.16 K.

    3. A student writes T=tC+273.16 "because the triple point is 273.16 K". Explain the mistake.

      Show solution

      0 °C is the melting point of ice at 1 atm, not the triple point. The ice point lies 0.01 K below the triple point, so with the triple point at 273.16 K the ice point, 0 °C, is 273.15 K.

    4. Suppose an absolute scale uses degrees as large as Fahrenheit degrees. What number does it give the triple point of water?

      Show solution

      Each kelvin is 1.8 such degrees, so the triple point is 273.16×1.8=491.69 on this scale (the Rankine scale).

  5. NumericalHardNCERT Exercises 10.5

    Two constant-volume gas thermometers, A (nitrogen) and B (helium), are read at the triple point of water and at the melting point of tin. A reads 0.800×105 Pa and 1.473×105 Pa; B reads 0.300×105 Pa and 0.5548×105 Pa.

    1. What melting point of tin does thermometer A give, in kelvin?

      K
      Show solution

      TA=273.16×1.473/0.800≈503.0 K.

    2. What does thermometer B give, in kelvin?

      K
      Show solution

      TB=273.16×0.5548/0.300≈505.2 K.

    3. Neither thermometer is faulty. Why do they disagree, and how could the disagreement be removed?

      Show solution

      Real gases are not ideal at these densities, nitrogen less so than helium. Repeat with less and less gas in each bulb and extrapolate the readings to zero pressure; both converge on about 505.1 K.

  6. NumericalMediumNCERT Exercises 10.6

    A steel measuring tape is accurate at 20.0 ∘C. On a 40.0 ∘C afternoon it shows a copper pipe to be 85.0 cm long. Take αsteel=1.20×10−5 K−1 and αCu=1.70×10−5 K−1.

    1. What is the actual length of the pipe that afternoon, in cm?

      cm
      Show solution

      The tape has expanded, so each marked centimetre is 1+1.20×10−5×20=1.00024 cm. Actual length =85.0×1.00024=85.020 cm.

    2. What is the length of the pipe at 20.0 ∘C, in cm?

      cm
      Show solution

      The copper itself contracts on cooling: 85.020/(1+1.70×10−5×20)≈84.991 cm.

  7. NumericalMediumNCERT Exercises 10.7

    A brass bush (a short tube) has an outer diameter of 6.012 cm at 30 ∘C. It must slip into a hole of diameter 6.000 cm in a steel housing, which stays at 30 ∘C. Take αbrass=1.9×10−5 K−1, constant over the range.

    1. To what temperature, in °C, must the bush be cooled?

      °C
      Show solution

      ΔT=−0.012/(6.012×1.9×10−5)≈−105 K, so the bush must reach about 30−105=−75 ∘C.

    2. Is dry ice, at −78.5 ∘C, cold enough?

      Show solution

      Just: it gives about 3 K to spare, a very tight fit. Workshops often use liquid nitrogen (−196 °C) for a comfortable clearance.

  8. NumericalEasyNCERT Exercises 10.8

    A hole of diameter 6.00 cm is drilled in an aluminium plate at 25 ∘C. The plate is then heated to 175 ∘C. By how much does the diameter of the hole change, in cm? (αAl=2.3×10−5 K−1)

    cm
    Show solution

    Δd=αd ΔT=2.3×10−5×6.00×150≈0.021 cm. The hole grows, by about 0.2 mm.

  9. NumericalMediumNCERT Exercises 10.9

    A steel wire 2.5 m long and 1.5 mm in diameter is stretched between two rigid posts at 32 ∘C with almost no tension. On a winter night it cools to 2 ∘C. Find the tension that develops, in newtons. (Y=2.0×1011 Pa, α=1.2×10−5 K−1)

    N
    Show solution

    Stress =Yα ΔT=2.0×1011×1.2×10−5×30=7.2×107 Pa. Area =π(0.75×10−3)2=1.77×10−6 m2, so the tension is 7.2×107×1.77×10−6≈127 N. The length of the wire does not enter.

  10. NumericalMediumNCERT Exercises 10.10

    An aluminium rod 40.0 cm long and a copper rod 60.0 cm long, of the same diameter, are joined end to end at 25 ∘C and heated to 225 ∘C with both ends free. (αAl=2.3×10−5, αCu=1.7×10−5 K−1)

    1. By how many centimetres does the joined rod lengthen?

      cm
      Show solution

      Δl=40.0(2.3×10−5)(200)+60.0(1.7×10−5)(200)=0.184+0.204=0.388 cm.

    2. Is a thermal stress set up along the rod?

      Show solution

      No: with both ends free each rod expands without restraint, so no stress develops along the length. (At the joint itself a small stress can arise because the two rods widen by slightly different amounts.)

  11. NumericalMediumNCERT Exercises 10.11

    Petrol has a volume expansivity of about 9.5×10−4 K−1. A tanker is filled at 20 ∘C and its load is sold at 35 ∘C.

    1. By what percentage does the density of the petrol change?

      %
      Show solution

      Δρ/ρ≈−αV ΔT=−9.5×10−4×15≈−0.014, a fall of about 1.4 %.

    2. Petrol is sold by volume. Who gains from selling it at 35 °C: the seller or the buyer?

      Show solution

      The seller: the same mass of petrol now fills about 1.4 % more litres, so each litre sold contains less fuel.

  12. NumericalMediumNCERT Exercises 10.12

    A 4.0 kW lathe machines a 6.0 kg copper block for 3.0 minutes. Assume 40 % of the power ends up heating the block. By how much does its temperature rise, in kelvin? (sCu=0.39 J g−1 K−1)

    K
    Show solution

    Heat into the block: 0.40×4000×180=2.88×105 J. Heat capacity: 6000 g×0.39=2340 J/K. Rise: 2.88×105/2340≈123 K, which is why machinists pour coolant on the work.

  13. NumericalMediumNCERT Exercises 10.13

    A 1.2 kg iron block at 400 ∘C is placed on a large slab of ice at 0 ∘C. What is the greatest mass of ice it can melt, in grams? (siron=0.45 J g−1 K−1, Lf=334 J/g)

    g
    Show solution

    Heat given out cooling to 0 °C: 1200×0.45×400=2.16×105 J. Mass melted: 2.16×105/334≈647 g, about 0.65 kg.

  14. NumericalMediumNCERT Exercises 10.14

    A 0.25 kg metal block at 120 ∘C is dropped into a calorimeter of water equivalent 0.020 kg that holds 200 cm3 of water at 25 ∘C. The final temperature is 35 ∘C.

    1. Find the specific heat of the metal, in J kg−1 K−1.

      J/(kg K)
      Show solution

      Heat gained: (0.200+0.020)×4186×10=9209 J. Heat lost: 0.25×s×85. So s=9209/21.25≈433 J kg−1 K−1.

    2. If some heat escapes to the surroundings, how does the true specific heat compare with your answer?

      Show answer

      Answer (A) It is larger

      Escaping heat makes the measured temperature rise of the water too small, so the heat gained, and hence s, is underestimated. The true value is larger.
  15. Long answerHardNCERT Exercises 10.15

    Measured molar specific heats at constant volume near room temperature, in J mol−1 K−1, are: helium 12.5, argon 12.5, nitrogen 20.8, oxygen 21.1, carbon monoxide 20.8, chlorine 25.6.

    1. Why are the values for the diatomic gases larger than those for the monatomic gases by about 8.3, which is R?

      Show solution

      A monatomic gas stores energy only as translational kinetic energy, giving 32R=12.5. A diatomic molecule can also rotate about two axes, which adds about R, giving 52R≈20.8. Chapter 12 explains this with the equipartition of energy.

    2. What does chlorine's larger value suggest?

      Show solution

      Chlorine molecules are heavy and loosely bonded, so their vibration has a low frequency and is already partly excited at room temperature. Vibration provides extra ways to store energy, raising CV above 52R.

  16. NumericalMediumNCERT Exercises 10.16

    A 55 kg cyclist finishes a hill climb with a core temperature of 102.6 ∘F. Resting in the shade, she cools to 99.0 ∘F in 25 minutes. Assume all the heat leaves by evaporation of sweat, the body's specific heat equals water's (4186 J kg−1 K−1), and the latent heat of evaporation of sweat is 2.42×106 J/kg. Find the average rate of evaporation, in grams per minute.

    g/min
    Show solution

    Fall: 3.6×5/9=2.0 K. Heat lost: 55×4186×2.0=4.60×105 J. Sweat evaporated: 4.60×105/2.42×106=0.190 kg in 25 min, about 7.6 g/min.

  17. NumericalHardNCERT Exercises 10.17

    A vaccine carrier is a cubical box of side 40 cm with polyurethane-foam walls 4.0 cm thick. It holds 5.0 kg of ice at 0 ∘C on a day when it is 40 ∘C outside. Take K=0.025 W m−1 K−1 for the foam and Lf=3.34×105 J/kg, and treat the heat-flow area as six faces of 40 cm×40 cm.

    1. Find the heat current into the box, in watts.

      W
      Show solution

      A=6×0.402=0.96 m2, so H=0.025×0.96×40/0.040=24 W.

    2. How much ice is left after 6.0 hours, in kg?

      kg
      Show solution

      Heat in 6.0 h: 24×21 600=5.18×105 J, which melts 5.18×105/3.34×105=1.55 kg. About 3.45 kg remains, enough to keep the vaccines cold for the rest of the day.

  18. NumericalHardNCERT Exercises 10.18

    A large steel cauldron in a langar kitchen has a flat base of area 0.20 m2 and thickness 1.2 cm. On a large burner it boils away water at 4.0 kg per minute. Estimate the temperature of the underside of the base, in contact with the flame, in °C. (Ksteel=50 W m−1 K−1, Lv=2.26×106 J/kg; the water side is at 100 ∘C.)

    °C
    Show solution

    Heat current: H=(4.0/60)×2.26×106≈1.51×105 W. Temperature drop across the base: ΔT=HL/(KA)=1.51×105×0.012/(50×0.20)≈181 K, so the underside is at about 281 ∘C.

  19. Short answerMediumNCERT Exercises 10.19 (a)–(c)

    Explain each of the following.

    1. A highly polished, shiny surface is a poor emitter of thermal radiation.

      Show solution

      It reflects most of the radiation falling on it, so it absorbs little; a poor absorber is a poor emitter (small e), so it radiates only a fraction of what a blackbody at the same temperature would.

    2. On a cold morning a steel tumbler feels much colder than a wooden chakla (rolling board), although both are at room temperature.

      Show solution

      Both are colder than your skin, but steel conducts heat away from your fingers far faster than wood, so the skin cools quickly. Touch senses the rate of heat loss, not temperature.

    3. An optical pyrometer calibrated for blackbody radiation reads too low for a red-hot iron bar lying in the open, but reads correctly when the same bar is inside the furnace.

      Show solution

      In the open the bar has e<1, so it radiates less than a blackbody at its temperature and the pyrometer reads low. Inside the furnace the bar sits in a cavity at the same temperature, and its own emission plus the wall radiation it reflects adds up to blackbody radiation, so the reading is correct.

  20. Short answerMediumNCERT Exercises 10.19 (d)–(e)

    Explain each of the following.

    1. The Moon is about as far from the Sun as the Earth, yet its nights fall to about −170 °C. What does the Earth's atmosphere do that the Moon lacks?

      Show solution

      Both radiate infrared into space. The Earth's water vapour and carbon dioxide absorb much of it and send part back down (the greenhouse effect), and winds carry heat to the night side. Without them the average surface temperature would be about −18 °C instead of about 15 °C.

    2. A radiator fed with steam at 100 °C delivers far more heat per kilogram of fluid than one fed with water at 100 °C.

      Show solution

      Each kilogram of steam that condenses in a radiator gives out about 2.26×106 J of latent heat at 100 °C. Hot water that cools by 20 K gives out only 4186×20≈8.4×104 J per kilogram, about 27 times less.

  21. NumericalMediumNCERT Exercises 10.20

    A body cools from 90 ∘C to 60 ∘C in 6.0 minutes in surroundings at 25 ∘C. Find the time, in minutes, that it needs to go from 50 ∘C down to 35 ∘C. Use Newton's law with average temperatures, then check with the exponential law.

    min
    Show solution

    Averages: first interval, rate 30/6.0=5.0 K/min at an average excess of 50 K, so K=0.10 min−1. Second interval, average excess 17.5 K, rate 1.75 K/min, time 15/1.75≈8.6 min. Exponential: K=ln⁡(65/35)/6.0=0.103 min−1, so t=ln⁡(25/10)/0.103≈8.9 min. The averaging method is slightly low because these intervals are not small.

  22. Quick answerMedium

    At what temperature does a Fahrenheit thermometer read exactly twice the number shown by a Celsius thermometer?

    Show answer

    Answer (B) 160 ∘C

    2x=95x+32 gives x/5=32, so x=160 ∘C, which is 320 ∘F. −40 is where the readings are equal.
  23. Quick answerMedium

    Assertion (A): The volume expansivity of an ideal gas at constant pressure is the same at every temperature. Reason (R): For an ideal gas at constant pressure, αV=1/T.

    Show answer

    Answer (D) A is false but R is true

    R is correct (Eq. 10.6), and it shows that αV falls as T rises, so A is false.
  24. Quick answerMedium

    Assertion (A): A burn from steam at 100 °C is usually worse than one from water at 100 °C. Reason (R): Steam has a larger specific heat capacity than liquid water.

    Show answer

    Answer (C) A is true but R is false

    A is true, but R is false: steam's specific heat (about 2.0 kJ kg−1 K−1) is about half that of water. The real reason is the latent heat, 2.26×106 J/kg, released when steam condenses on the skin.
  25. Quick answerMedium

    Two blackbody spheres of radii 1 cm and 2 cm are at 1000 K and 500 K respectively. What is the ratio of the power radiated by the small sphere to that radiated by the large sphere?

    Show answer

    Answer (C) 4 : 1

    P1P2=(12)2(1000500)4=164=4.
  26. Quick answerHard

    The peak of a blackbody's spectrum shifts from 600 nm to 400 nm. By about what factor does its total radiated power increase?

    Show answer

    Answer (D) 5.1

    Wien: T2/T1=600/400=1.5. Stefan–Boltzmann: 1.54≈5.1.
  27. Quick answerHard

    Three rods of equal length and cross-section, made of copper (K=400), aluminium (K=200) and steel (K=50 W m−1 K−1), are joined in series. What is the equivalent conductivity of the combined rod, in W m−1 K−1?

    W/(m K)
    Show answer

    Answer 109 W/(m K)

    3K′=1400+1200+150=0.0275, so K′≈109 W m−1 K−1, far below the average of the three: the steel dominates.
  28. Quick answerMedium

    Equal masses of ice at 0 °C and water at 50 °C are mixed in an insulated flask. What is the final state? (sw=4186 J kg−1 K−1, Lf=3.34×105 J/kg)

    Show answer

    Answer (A) A mixture of ice and water at 0 °C

    Cooling to 0 °C, each kilogram of water gives out 4186×50=2.09×105 J, less than the 3.34×105 J needed to melt a kilogram of ice. Only about 63 % of the ice melts, and the mixture stays at 0 °C.
  29. Quick answerEasy

    Assertion (A): Lakes in very cold places freeze from the top downwards. Reason (R): Water is densest at about 4 °C.

    Show answer

    Answer (A) Both A and R are true, and R explains A

    Both are true and R explains A: water colder than 4 °C is less dense, so it stays at the surface and freezes there, while 4 °C water stays at the bottom.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Exercises 10.1Practice Q1
NCERT Exercises 10.2Practice Q2
NCERT Exercises 10.3Practice Q3
NCERT Exercises 10.4Practice Q4
NCERT Exercises 10.5Practice Q5
NCERT Exercises 10.6Practice Q6
NCERT Exercises 10.7Practice Q7
NCERT Exercises 10.8Practice Q8
NCERT Exercises 10.9Practice Q9
NCERT Exercises 10.10Practice Q10
NCERT Exercises 10.11Practice Q11
NCERT Exercises 10.12Practice Q12
NCERT Exercises 10.13Practice Q13
NCERT Exercises 10.14Practice Q14
NCERT Exercises 10.15Practice Q15
NCERT Exercises 10.16Practice Q16
NCERT Exercises 10.17Practice Q17
NCERT Exercises 10.18Practice Q18
NCERT Exercises 10.19 (a)–(c)Practice Q19
NCERT Exercises 10.19 (d)–(e)Practice Q20
NCERT Exercises 10.20Practice Q21

Quick check: MCQs on Thermal Properties of Matter

  1. 10.2 Temperature and heat3 questions
  2. 10.3 Measurement of temperature2 questions
  3. 10.4 Ideal-gas equation and absolute temperature2 questions
  4. Thermal stress3 questions
  5. 10.6 Specific heat capacity3 questions
  6. 10.7 Calorimetry2 questions
  7. 10.8 Change of state3 questions
  8. 10.8.1 Latent heat3 questions
  9. 10.9.1 Conduction3 questions
  10. 10.9.3 Radiation4 questions
  11. 10.9.4 Blackbody radiation4 questions
  12. 10.10 Newton's law of cooling3 questions
Glossary · 15

Key terms (English–Hindi)

absolute zeroपरम शून्यparam shoonya
The lowest possible temperature, 0 K or −273.15 °C, where an ideal gas would exert no pressure.First seen in the chapter →
coefficient of linear expansionरैखिक प्रसार गुणांकrekhik prasaar gunaank
Fractional increase in length per kelvin, αl=Δl/(l ΔT).First seen in the chapter →
coefficient of volume expansionआयतन प्रसार गुणांकaayatan prasaar gunaank
Fractional increase in volume per kelvin, αV=ΔV/(V ΔT); for an isotropic solid αV=3αl.First seen in the chapter →
convectionसंवहनsanvahan
Heat transfer by the bulk flow of a fluid, driven by buoyancy (natural) or by a pump or fan (forced).First seen in the chapter →
emissivityउत्सर्जकताutsarjakta
The ratio of the radiation a surface emits to that of a blackbody at the same temperature, between 0 and 1.First seen in the chapter →
heatऊष्माushma
Energy transferred because of a temperature difference, measured in joules.First seen in the chapter →
latent heatगुप्त ऊष्माgupt ushma
Heat per unit mass taken in or given out in a change of state at constant temperature, L=Q/m.First seen in the chapter →
molar specific heat capacityमोलर विशिष्ट ऊष्मा धारिताmolar vishisht ushma dhaarita
Heat needed per mole per kelvin; for a gas CP exceeds CV.First seen in the chapter →
regelationपुनर्हिमायनpunarhimaayan
Melting of ice under pressure and refreezing when the pressure is removed.First seen in the chapter →
specific heat capacityविशिष्ट ऊष्मा धारिताvishisht ushma dhaarita
Heat needed to warm 1 kg of a substance by 1 K, in J kg⁻¹ K⁻¹.First seen in the chapter →
sublimationऊर्ध्वपातनoordhvapaatan
A direct change from solid to vapour, as in dry ice or camphor.First seen in the chapter →
temperatureतापtaap
A number on an agreed scale that says how hot a body is; heat flows from higher to lower temperature.First seen in the chapter →
thermal conductivityऊष्मा चालकताushma chaalakta
The constant K in H=KA ΔT/L; how well a material conducts heat, in W m⁻¹ K⁻¹.First seen in the chapter →
thermal stressतापीय प्रतिबलtaapiya pratibal
Stress in a body that is heated or cooled but not allowed to change size; Yαl ΔT for a clamped rod.First seen in the chapter →
triple pointत्रिक बिंदुtrik bindu
The single temperature and pressure at which solid, liquid and vapour coexist; for water 273.16 K and 611.66 Pa.First seen in the chapter →