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Class XI ⚛️ Physics Ch 10 of 14

Thermal Properties of Matter

CUET unit: Thermal Properties of Matter

📌 Snapshot

  • Heat, temperature, expansion, calorimetry and heat transfer explained with solved numericals and traps.
  • The difference between heat and temperature, why railway tracks have gaps, and why a lake freezes from the top down.
  • Specific heat, latent heat and heating curves, then conduction, convection, radiation and Newton's law of cooling.
  • Six worked examples with the common wrong step shown, a formula sheet and thirteen NTA style MCQs.

📖 Detailed Notes

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 10.

Why does steam burn worse than boiling water?

Water at 100 °C and steam at 100 °C have the same temperature, yet steam leaves a far worse burn. A steel railway track is laid with small gaps, a sealed bottle of water cracks in a freezer, and a thermos keeps tea hot for hours. All of this comes from three questions: how much heat does a body hold, how does it expand, and how does heat travel? Temperature tells you how hot something is. It does not tell you how much heat it can deliver, and that gap is what this chapter closes.

Heat, temperature and the thermometer

Temperature is a measure of how hot or cold a body is. Heat is the energy that flows between two bodies because their temperatures differ. A body does not "contain" heat. It contains internal energy, and heat is that energy in transit. The SI unit of heat is the joule, and of temperature the kelvin.

A thermometer uses a property that changes steadily with temperature, such as the length of a mercury thread, and is marked at two fixed points: the ice point and the steam point of water.

Scale Ice point Steam point Number of divisions
Celsius (°C) 0 100 100
Fahrenheit (°F) 32 212 180
Kelvin (K) 273.15 373.15 100

The scales are linked by

$\dfrac{t_{F} - 32}{180} = \dfrac{t_{C}}{100}$ and T = $t_{C}$ + 273.15

A normal body temperature of 98.6 °F converts to $(98.6 - 32) \times 100/180 = 37$ °C. The two scales give the same number at −40, because putting $t_{F} = t_{C} = x$ in the first relation gives $100x - 3200 = 180x$.

Absolute zero and the ideal gas equation

Fill a rigid container with a low-density gas and measure its pressure as you change the temperature. The pressure falls linearly as the gas cools. Extend the straight line backwards and it reaches zero pressure at −273.15 °C. The remarkable thing is that every low-density gas gives the same line endpoint, whichever gas you use. This point is taken as absolute zero, and the Kelvin scale starts there. A Kelvin degree is the same size as a Celsius degree, so the two scales differ only in their zero.

The gas laws combine into the ideal gas equation

$$PV = \mu RT$$

where $\mu$ is the number of moles and R = 8.31 J mol⁻¹ K⁻¹ is the universal gas constant. Temperature here must always be in kelvin. The triple point of water, 273.16 K at a pressure of 4.58 mm of Hg, is the unique state in which ice, water and vapour coexist in equilibrium.

Beyond the textbook. For the exam: absolute zero is −273.15 °C, found by extrapolating the pressure–temperature lines of gases. In reality: real gases liquefy before they get that cold, so the line is never actually followed to zero, and absolute zero can be approached but never reached. Since 2019 the kelvin is defined by fixing the value of the Boltzmann constant at $1.380649 \times 10^{-23}$ J K⁻¹, so the triple point is now a measured value, 273.16 K, rather than a definition. Why it matters: the number 273.16 is still the right one for the triple point, but the Celsius offset is 273.15.

Thermal expansion

Heat a solid and its atoms vibrate with larger amplitude, pushing each other slightly further apart. The solid gets bigger. For a rod of length l,

$$\dfrac{\Delta l}{l} = \alpha_{l}\,\Delta T$$

where $\alpha_{l}$ is the coefficient of linear expansion, a property of the material. Heat a flat sheet and its area grows as well, and a cube grows in volume:

Expansion Relation Coefficient
Linear $\Delta l/l = \alpha_{l}\Delta T$ $\alpha_{l}$ (steel about $1.2 \times 10^{-5}$ K⁻¹)
Area $\Delta A/A = 2\alpha_{l}\Delta T$ 2$\alpha_{l}$
Volume (solid) $\Delta V/V = \alpha_{V}\Delta T$ $\alpha_{V} = 3\alpha_{l}$
Volume (ideal gas at constant P) $\Delta V/V = \alpha_{V}\Delta T$ $\alpha_{V} = 1/T$, about $3.7 \times 10^{-3}$ K⁻¹ at 0 °C

The factor 2 for area and 3 for volume come from adding the fractional change in each dimension: the tiny $(\alpha_{l}\Delta T)^{2}$ terms are ignored. Gases expand far more than solids, since their molecules are free, and liquids lie in between. A hole in a metal plate expands like the metal around it, as though it were filled with the same metal.

Water behaves oddly. Between 0 °C and 4 °C water contracts on heating, so its volume is smallest, and its density highest, at 4 °C. Above 4 °C it expands normally. This is why a lake cools from the top, with the surface water sinking as it reaches 4 °C. Once the whole lake is at 4 °C, further cooling makes the surface water lighter, so it stays on top and freezes there. The ice then acts as a lid, and fish live in the roughly 4 °C water below.

Pause and predict: a metal ring with a small hole is heated. Does the hole get larger or smaller?

Check your answer: larger. Every length in the plate expands, including the diameter of the hole, exactly as if the hole were a disc of the same metal.

Thermal stress. If a rod is clamped rigidly so it cannot expand, it is squeezed instead. The strain it avoids is $\alpha_{l}\Delta T$, so the stress is

$$\sigma = Y\alpha_{l}\Delta T$$

Worked example: a rail with no gaps. A steel rail (Y = $2 \times 10^{11}$ N m⁻², $\alpha_{l}$ = $1.2 \times 10^{-5}$ K⁻¹) is rigidly fixed at both ends. The temperature rises by 10 °C. Find the thermal stress.

Step What we do Result
1 The rail cannot lengthen, so the "prevented" strain is $\alpha_{l}\Delta T$ $1.2 \times 10^{-5} \times 10 = 1.2 \times 10^{-4}$
2 Stress = Young's modulus × strain $2 \times 10^{11} \times 1.2 \times 10^{-4}$
3 Simplify $2.4 \times 10^{7}$ N m⁻²

The common wrong step is to include the rail's length in the formula. The strain is a ratio, so length cancels, and a 1 m rail and a 1 km rail suffer the same stress. This is why long tracks are laid with gaps, or welded under controlled conditions.

Specific heat and calorimetry

Heat a kilogram of water and a kilogram of iron by the same amount and they warm up differently. The heat capacity of a body is the heat per unit temperature rise, $S = \Delta Q/\Delta T$, in J K⁻¹. The specific heat capacity is the heat per unit mass per unit temperature rise:

$s = (1/m)(\Delta Q/\Delta T)$, so $\Delta Q = m s \Delta T$

Its SI unit is J kg⁻¹ K⁻¹. The molar specific heat $C = (1/\mu)(\Delta Q/\Delta T)$ is the heat per mole, and for gases it has two values, $C_{p}$ at constant pressure and $C_{V}$ at constant volume.

Substance Specific heat (J kg⁻¹ K⁻¹)
Water 4186
Ice about 2100
Aluminium 900
Copper 386
Lead 128

Water has an unusually large specific heat. A lot of heat is needed to warm it, and a lot is released when it cools. That is why water works as a coolant in car radiators and as a hot-water bottle, and why places near the sea have milder weather than inland ones.

Calorimetry is measuring heat by mixing. A calorimeter is an insulated metal vessel. In an isolated system the principle is simple: heat lost by the hot body = heat gained by the cold body.

Worked example: ice in warm water. 50 g of ice at 0 °C is dropped into 200 g of water at 40 °C in an insulated vessel. Find the final temperature. Take $s_{\text{water}} = 4186$ J kg⁻¹ K⁻¹ and $L_{f} = 3.33 \times 10^{5}$ J kg⁻¹.

Step What we do Result
1 Is there enough heat to melt all the ice? Heat the water gives up in cooling to 0 °C $0.200 \times 4186 \times 40 = 33\,488$ J
2 Heat needed to melt all the ice $0.050 \times 3.33 \times 10^{5} = 16\,650$ J
3 The water has more, so all the ice melts. Heat left over $33\,488 - 16\,650 = 16\,838$ J
4 This warms all 0.250 kg of water from 0 °C $\Delta T = 16\,838 / (0.250 \times 4186)$
5 Final temperature about 16 °C

The wrong step is to skip the latent heat and take a weighted mean, $(50 \times 0 + 200 \times 40)/250 = 32$ °C. That treats the ice as if it warmed up directly, without first absorbing the heat of melting.

Change of state and latent heat

When ice melts or water boils, heat goes in and the temperature does not change. The heat is spent on loosening the bonds between molecules, not on speeding them up. This is the latent heat: the heat per unit mass for a change of state at constant temperature,

$$Q = mL$$

with unit J kg⁻¹. For water, the latent heat of fusion is $L_{f} = 3.33 \times 10^{5}$ J kg⁻¹ and the latent heat of vaporisation is $L_{v} = 22.6 \times 10^{5}$ J kg⁻¹. The temperature at which solid and liquid coexist is the melting point, and the temperature at which liquid and vapour coexist is the boiling point. Some substances, such as dry ice and iodine, change directly from solid to vapour. This is sublimation.

The heating curve of ice shows this. Sloping parts are the heating of one phase, and the flat parts at 0 °C and 100 °C are the changes of state. A steeper slope means a smaller specific heat.

Worked example: from ice to steam. How much heat turns 100 g of ice at −10 °C into steam at 100 °C? Take $s_{\text{ice}} = 2100$, $s_{\text{water}} = 4186$ J kg⁻¹ K⁻¹, $L_{f}$ = $3.33 \times 10^{5}$ and $L_{v} = 22.6 \times 10^{5}$ J kg⁻¹.

Stage Formula Heat (J)
Ice from −10 °C to 0 °C $m s \Delta T = 0.1 \times 2100 \times 10$ 2100
Melt the ice $m L_{f} = 0.1 \times 3.33 \times 10^{5}$ 33 300
Water from 0 °C to 100 °C $0.1 \times 4186 \times 100$ 41 860
Boil the water $m L_{v} = 0.1 \times 22.6 \times 10^{5}$ 226 000
Total about $3.03 \times 10^{5}$ J

Notice that boiling alone takes about three quarters of the total. Skipping a stage, or using $m s \Delta T$ across a flat part of the curve, is the usual error. This also explains the opening puzzle: 1 kg of steam at 100 °C releases $22.6 \times 10^{5}$ J more heat on condensing than 1 kg of water at 100 °C, so steam burns are far more severe.

Boiling point depends on pressure. A liquid boils when its vapour pressure equals the pressure above it. In a pressure cooker the pressure is raised, so water boils at about 120 °C and food cooks faster. On a mountain the pressure is lower, so water boils below 100 °C and cooking takes longer. Evaporation differs from boiling: it happens only at the surface and at any temperature, and it cools the liquid left behind, which is why sweating cools you.

Beyond the textbook. For the exam: ice melts under the pressure of a skate blade and refreezes behind it, a process called regelation, which makes skating possible. In reality: the pressure of a skater lowers the melting point of ice by only about a degree at most, which is not enough on its own on very cold ice. Friction heating and a thin, slippery layer of water-like molecules on the ice surface also play a large part. Why it matters: regelation is the answer to write in the exam, but it is only one contributor to the slipperiness.

How heat travels: conduction, convection, radiation

Conduction. Heat passes along a body from the hotter parts to the colder parts without any movement of matter. Fix the two ends of a rod at different temperatures and wait. In the steady state the heat current is

$$H = \dfrac{KA(T_{C} - T_{D})}{L}$$

where K is the thermal conductivity in W m⁻¹ K⁻¹. The current grows with the area A and the temperature difference, and falls with the length L. Metals conduct well (copper 385, aluminium 205, steel about 50 W m⁻¹ K⁻¹), while water (0.6) and air (0.024) are poor conductors. Trapped air in wool, feathers and double glazing keeps heat in for this reason.

Worked example: two rods in series. Two rods of the same length and cross-section are joined end to end. Rod 1 has conductivity 2K and rod 2 has conductivity K. The free end of rod 1 is held at 100 °C and the free end of rod 2 at 0 °C. Find the temperature of the joint in the steady state.

Step What we do Result
1 In the steady state, the same heat current flows through both rods $H_{1} = H_{2}$
2 Write each with the joint at temperature T $2KA(100 - T)/L = KA(T - 0)/L$
3 Cancel $KA/L$ $2(100 - T) = T$
4 Solve $T = 200/3 \approx 66.7$ °C

The wrong step is to say the joint sits at the average, 50 °C. The better conductor needs a smaller temperature drop to carry the same heat, so the joint stays close to its end temperature.

Convection is heat transfer by the actual movement of the fluid, so it happens only in liquids and gases. In natural convection gravity does the driving: warm fluid is lighter and rises, cool fluid sinks. By day, land warms faster than the sea, the air above it rises, and cooler sea air flows in as the sea breeze. At night it reverses. In forced convection a pump or fan moves the fluid, as in a car radiator.

Radiation carries heat as electromagnetic waves, so it needs no medium and travels at $3 \times 10^{8}$ m s⁻¹. That is how the Sun warms the Earth through empty space. A perfect radiator, called a black body, absorbs all the radiation falling on it. The power radiated is given by the Stefan–Boltzmann law:

$$H = Ae\sigma T^{4}$$

with $\sigma = 5.67 \times 10^{-8}$ W m⁻² K⁻⁴ and e the emissivity (e = 1 for a black body, less for real surfaces). The temperature must be in kelvin. A body at T in surroundings at $T_{s}$ loses net power $e\sigma A(T^{4} - T_{s}^{4})$.

A hot body does not radiate a single wavelength. The spectrum is continuous, and the wavelength of the peak moves to shorter values as the body gets hotter. This is Wien's displacement law:

$$\lambda_{m}T = 2.9 \times 10^{-3}\ \text{m K}$$

That is why iron glows dull red, then yellow, then white as it heats. A star whose peak is at 500 nm has a surface temperature of $2.9 \times 10^{-3} / 5 \times 10^{-7} = 5800$ K.

Worked example: doubling the temperature. A black body at 727 °C is heated to 1727 °C. By what factor does the radiated power change, and what happens to $\lambda_{m}$?

Step What we do Result
1 Convert to kelvin $T_{1} = 1000$ K, $T_{2} = 2000$ K
2 $H \propto T^{4}$, so the ratio is $(T_{2}/T_{1})^{4}$ $2^{4} = 16$
3 $\lambda_{m} \propto 1/T$, so $\lambda_{m}$ halves 2.9 µm → 1.45 µm

The wrong step is to put the Celsius values in: $(1727/727)^{4}$ is about 32, double the right answer.

A Dewar flask (thermos) blocks all three paths. The double walls are silvered to reflect radiation and the gap between them is evacuated to stop conduction and convection.

Beyond the textbook. For the exam: convection happens when a fluid is heated from below, as warm fluid rises. In reality: natural convection needs gravity. On the International Space Station a candle flame is nearly spherical and weak, because the hot gases do not rise and fresh oxygen reaches the flame only by slow diffusion. Why it matters: it shows that "hot air rises" is really a statement about weight.

Newton's law of cooling

A cup of tea cools quickly at first and then more and more slowly. For a small temperature difference, the rate of cooling is proportional to the excess temperature over the surroundings:

$$-\dfrac{dT}{dt} = k(T - T_{s})$$

Integrating gives $T - T_{s} = (T_{0} - T_{s}) e^{-kt}$, so a graph of $\ln(T - T_{s})$ against t is a straight line with a negative slope. The law is accurate only for small differences, and it also fits forced convection well. The constant k is not the thermal conductivity K.

Worked example: tea on a table. A body cools from 80 °C to 64 °C in 5 minutes in surroundings at 20 °C. How long does it then take to cool from 64 °C to 52 °C?

Step What we do Result
1 Excess temperatures: start 60, after 5 min 44 $\ln(44/60) = -0.310$
2 Find the constant $k = 0.310 / 5 = 0.062$ min⁻¹
3 Excess at 52 °C is 32, from 44 $\ln(32/44) = -0.318$
4 Time $t = 0.318 / 0.062 \approx 5.1$ min

The time is a little over 5 minutes because cooling has slowed: the excess temperature is now smaller, so even a smaller drop of 12 °C takes slightly longer. The wrong step is to assume equal temperature drops take equal times.

Exam traps

The trap Why it's tempting The truth
"Steam and water at 100 °C give equal burns." Same temperature Steam carries an extra $22.6 \times 10^{5}$ J kg⁻¹ of latent heat
"Celsius to kelvin: add 273.16." 273.16 K is the triple point Add 273.15
"Stress in a clamped rod depends on its length." Longer rods expand more $\sigma = Y\alpha_{l}\Delta T$, independent of length
"Water is densest at 0 °C." Ice forms at 0 °C Maximum density is at 4 °C
"Temperature rises during melting." Heat is still being supplied Temperature stays fixed while the state changes
"Use Celsius in the Stefan law." Temperatures are usually given in Celsius $H \propto T^{4}$ needs kelvin, and net loss is $T^{4} - T_{s}^{4}$, not $(T - T_{s})^{4}$
"Convection can happen in a solid." Heat flows through solids too Convection needs a fluid; in solids heat moves by conduction
"Newton's law holds for any temperature difference." It looks like a general law It is valid only for small differences

Formula sheet

Quantity Formula SI unit When to use
Scale conversion $(t_{F} - 32)/180 = t_{C}/100$, $T = t_{C} + 273.15$ °F, °C, K Changing scales
Ideal gas $PV = \mu RT$ J Gas state problems
Linear expansion $\Delta l = l\alpha_{l}\Delta T$ m Rods, wires
Area, volume expansion $\Delta A = 2A\alpha_{l}\Delta T$, $\Delta V = 3V\alpha_{l}\Delta T$ m², m³ Sheets, solids
Gas volume coefficient $\alpha_{V} = 1/T$ K⁻¹ Ideal gas at constant pressure
Thermal stress $\sigma = Y\alpha_{l}\Delta T$ N m⁻² Clamped rod
Heat and specific heat $Q = ms\Delta T$ J Warming without change of state
Latent heat $Q = mL$ J Melting, boiling
Calorimetry Heat lost = heat gained J Mixing
Conduction $H = KA(T_{C} - T_{D})/L$ W Steady flow through a bar
Stefan–Boltzmann $H = Ae\sigma T^{4}$ W Radiation from a surface
Wien $\lambda_{m}$T = $2.9 \times 10^{-3}$ m K Peak wavelength
Newton's cooling $\ln(T - T_{s}) = -kt + c$ K Cooling under small difference

60-second revision

Topic Lock this in
Heat vs temperature Heat is energy in transit, temperature measures hotness; SI units J and K
Scales $(t_{F} - 32)/180 = t_{C}/100$, $T = t_{C} + 273.15$, −40 is common
Expansion $\alpha_{V} = 3\alpha_{l}$, area $2\alpha_{l}$, gas $\alpha_{V} = 1/T$
Water Densest at 4 °C, lakes freeze from the top
Thermal stress $Y\alpha_{l}\Delta T$, no length in it
Specific heat $Q = ms\Delta T$, water 4186 J kg⁻¹ K⁻¹
Latent heat $Q = mL$, temperature constant, $L_{f}$ = $3.33 \times 10^{5}$ and $L_{v} = 22.6 \times 10^{5}$ J kg⁻¹
Boiling point Rises with pressure (cooker), falls on hills
Conduction $H = KA\,\Delta T/L$
Radiation $H = Ae\sigma T^{4}$ in kelvin, $\lambda_{m}T = 2.9 \times 10^{-3}$ m K
Newton Small differences only, ln graph is a straight line

Where this chapter connects

🎯 Practice MCQs

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Q1. The temperature at which a Fahrenheit thermometer and a Celsius thermometer show the same reading is

▸ Show answer & explanation

Answer: B

Putting $t_{F} = t_{C} = x$ in $(t_{F} - 32)/180 = t_{C}/100$ gives $100x - 3200 = 180x$, so $x = -40$. Option C has the wrong sign, and 32 is simply the Fahrenheit ice point.

Q2. Constant volume gas thermometers filled with different low-density gases all give the same value of absolute zero because

▸ Show answer & explanation

Answer: B

For a fixed volume, the pressure of a low-density gas falls linearly with temperature, and every such line extrapolates to zero pressure at −273.15 °C. Option A is false because Boyle's law holds only at constant temperature, and C is false because gases liquefy at different temperatures.

Q3. A brass rod 2.0 m long at 20 °C is heated to 120 °C. Take $\alpha$ for brass as $1.9 \times 10^{-5}$ K⁻¹. The increase in length is

▸ Show answer & explanation

Answer: C

$\Delta l = l\alpha\Delta T = 2.0 \times 1.9 \times 10^{-5} \times 100 = 3.8 \times 10^{-3}$ m = 3.8 mm. Option B forgets the length of 2.0 m, and D doubles it.

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