Thermal Properties of Matter ISC Class 11: Notes and Numericals

What Thermal Properties of Matter is about

Weightage: Thermal Properties of Matter belongs to unit 8 of the ISC Class XI Physics syllabus. CISCE weights units 7-9 together (Properties of Bulk Matter; Heat and Thermodynamics; Behaviour of Perfect Gases and Kinetic Theory of Gases) at 20 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Heat, temperature and temperature scales

Heat is energy in transit from a hotter to a colder body (joule). Temperature is the property that decides which way heat flows; two bodies in thermal equilibrium have the same temperature. A temperature scale needs two fixed points: the ice point and the steam point at standard pressure. Celsius: 0 and 100; Fahrenheit: 32 and 212; Kelvin: 273.15 and 373.15. Conversion: $\frac{C}{100}=\frac{F-32}{180}=\frac{K-273.15}{100}$. A temperature difference of 1 °C equals 1 K and 1.8 °F. The Celsius and Fahrenheit scales agree at −40°.

Worked example

Normal body temperature is 98.6 °F. Express it in °C and K. At what temperature do the Celsius and Fahrenheit scales give the same reading?

Try it first, then show the solutionSolution

1. C = (98.6 − 32) × 100/180 = 37 °C; K = 37 + 273.15 = 310.15 K
2. Put C = F = x: x/100 = (x − 32)/180, so 180x = 100x − 3200 and x = −40

Answer: 37 °C, 310.15 K; −40°

Ideal gas equation and absolute temperature

Real gases at low pressure obey Boyle's law (pV constant at fixed T) and Charles's law (V/T constant at fixed p), which combine into the ideal gas equation: pV = nRT, with R = 8.31 J mol⁻¹ K⁻¹. In a constant-volume gas thermometer, pressure varies linearly with temperature; extending the line to zero pressure gives the same temperature, −273.15 °C, for every gas. This is absolute zero, the zero of the absolute (kelvin) scale: T = t + 273.15. Temperatures in gas laws must always be in kelvin.

Worked example

A car tyre has a pressure of 2.0 × 10⁵ Pa at 27 °C. After a long drive its temperature is 47 °C. Find the new pressure, assuming the volume stays constant.

Try it first, then show the solutionSolution

1. $\frac{p_2}{p_1}=\frac{T_2}{T_1}=\frac{320}{300}$
2. p₂ = 2.0 × 10⁵ × 320/300 = 2.13 × 10⁵ Pa

Answer: About $2.13\times 10^5$ Pa (use kelvin, not 47/27)

Thermal expansion of solids

On heating, the atoms vibrate with larger amplitude about positions that drift slightly apart, so solids expand. Linear expansion: $\Delta L=\alpha L\Delta T$, where α is the coefficient of linear expansion (K⁻¹). Area expansion: $\Delta A=\beta A\Delta T$; volume expansion: $\Delta V=\gamma V\Delta T$. For an isotropic solid β = 2α and γ = 3α. Examples: gaps in railway tracks and expansion joints in bridges; the bimetallic strip (two metals of different α riveted together bend on heating, used in thermostats); fitting an iron tyre on a wheel by heating it. If a rod is clamped so that it cannot expand, a thermal stress $Y\alpha \Delta T$ develops.

Worked example

Steel rails 25 m long are laid at 10 °C. What gap must be left between them if the temperature can reach 50 °C? (α = 1.2 × 10⁻⁵ K⁻¹)

Try it first, then show the solutionSolution

1. ΔL = αLΔT = 1.2 × 10⁻⁵ × 25 × 40
2. ΔL = 0.012 m

Answer: At least 12 mm

Expansion of liquids and gases; anomalous expansion of water

A liquid has no shape of its own, so only volume expansion matters. The container also expands, so the apparent expansion seen is less than the real expansion: ${\gamma }_{real}={\gamma }_{apparent}+{\gamma }_{vessel}$. Gases expand far more. For an ideal gas at constant pressure, V ∝ T, so $\gamma =\frac{1}{T}$; at 0 °C, γ = 1/273 K⁻¹. Anomalous expansion of water: from 0 °C to 4 °C water contracts on heating; above 4 °C it expands normally. So water is densest at 4 °C. In winter, as surface water cools to 4 °C it sinks; below 4 °C the colder, lighter water stays on top and freezes; ice, a poor conductor, insulates the water below, which stays at about 4 °C at the bottom, so aquatic life survives.

Worked example

A glass flask of volume 500 cm³ is filled to the brim with mercury at 20 °C. How much mercury overflows when both are heated to 100 °C? (γ of mercury = 1.8 × 10⁻⁴ K⁻¹, γ of glass = 2.7 × 10⁻⁵ K⁻¹)

Try it first, then show the solutionSolution

1. Overflow = V(γ_Hg − γ_glass)ΔT
2. = 500 × (1.8 × 10⁻⁴ − 0.27 × 10⁻⁴) × 80 = 500 × 1.53 × 10⁻⁴ × 80
3. = 6.12 cm³

Answer: About 6.1 cm³ (the apparent expansion)

Specific heat capacity and calorimetry

The heat needed to raise the temperature of mass m by ΔT is Q = mcΔT, where c is the specific heat capacity: the heat needed to raise the temperature of unit mass by 1 K (J kg⁻¹ K⁻¹). Water has an unusually high c, 4200 J kg⁻¹ K⁻¹, so it warms and cools slowly (coastal climates are mild; water is a good coolant). Heat capacity (thermal capacity) of a body = mc (J K⁻¹). Molar specific heat = heat per mole per kelvin. Principle of calorimetry: when hot and cold bodies are mixed in an insulated container, heat lost by the hot bodies = heat gained by the cold ones (conservation of energy). The calorimeter itself (usually copper) also absorbs heat; include its heat capacity unless told to neglect it.

Worked example

200 g of copper at 100 °C is dropped into 300 g of water at 20 °C. Find the final temperature, neglecting the calorimeter. (c of copper 390, of water 4200 J kg⁻¹ K⁻¹)

Try it first, then show the solutionSolution

1. Heat lost by copper = heat gained by water: 0.2 × 390 × (100 − T) = 0.3 × 4200 × (T − 20)
2. 78(100 − T) = 1260(T − 20), so 1338T = 33 000
3. T = 24.7 °C

Answer: About 24.7 °C

Change of state and latent heat

During melting, boiling and their reverses, heat is absorbed or released at constant temperature: it changes the molecular arrangement (potential energy), not the average kinetic energy. Specific latent heat L is the heat needed to change unit mass from one state to another without change of temperature: Q = mL. For water, $L_f$ (fusion) = 3.36 × 10⁵ J kg⁻¹ and $L_v$ (vaporisation) = 2.26 × 10⁶ J kg⁻¹. A heating curve (temperature against heat supplied) has sloping parts (Q = mcΔT) and flat parts (Q = mL). The melting point of ice falls with increased pressure (regelation: a loaded wire passes through a block of ice, which refreezes behind it), and the boiling point rises with pressure (pressure cooker). Steam burns are worse than boiling-water burns because steam gives up its latent heat as it condenses.

Worked example

Find the heat needed to turn 100 g of ice at −10 °C into steam at 100 °C. (c_ice = 2100, c_water = 4200 J kg⁻¹ K⁻¹; L_f = 3.36 × 10⁵, L_v = 2.26 × 10⁶ J kg⁻¹)

Try it first, then show the solutionSolution

1. Ice −10 → 0 °C: 0.1 × 2100 × 10 = 2100 J; melting: 0.1 × 3.36 × 10⁵ = 33 600 J
2. Water 0 → 100 °C: 0.1 × 4200 × 100 = 42 000 J; boiling: 0.1 × 2.26 × 10⁶ = 226 000 J
3. Total = 303 700 J

Answer: About $3.04\times 10^5$ J (three-quarters of it is the latent heat of vaporisation)

Conduction: steady state, temperature gradient and thermal conductivity

In conduction, heat passes through a material from particle to particle without the material moving (free electrons carry most of it in metals). When one end of a rod is heated, its temperatures first rise everywhere (variable state); eventually each point reaches a constant temperature: the steady state, in which no heat is absorbed and the heat entering each cross-section per second equals that leaving. The temperature gradient is the fall of temperature per unit length, $\frac{{\theta }_1-{\theta }_2}{L}$. In the steady state the heat current is $H=\frac{Q}{t}=\frac{KA({\theta }_1-{\theta }_2)}{L}$, where K is the coefficient of thermal conductivity (W m⁻¹ K⁻¹). Metals have high K (copper ≈ 385), insulators low K (air ≈ 0.025). Thermal resistance R = L/(KA); slabs in series add resistances, so for two slabs of equal thickness the junction temperature is $\theta =\frac{K_1{\theta }_1+K_2{\theta }_2}{K_1+K_2}$.

Worked example

(a) A copper rod 0.5 m long with a cross-section of 1 cm² has its ends at 100 °C and 0 °C. Find the heat current. (K = 385 W m⁻¹ K⁻¹) (b) Two slabs of equal thickness and area, with K₁ = 2K₂, are joined face to face; the outer faces are at 100 °C and 0 °C. Find the junction temperature.

Try it first, then show the solutionSolution

1. (a) H = 385 × 10⁻⁴ × 100/0.5 = 7.7 W
2. (b) Equal heat current: K₁(100 − θ) = K₂(θ − 0) → 2(100 − θ) = θ
3. θ = 66.7 °C

Answer: (a) 7.7 W (b) 66.7 °C (nearer the hot face, across the better conductor)

Searle's experiment for the thermal conductivity of a good conductor

A thick, well-lagged metal bar (e.g. copper) is heated at one end by steam; the other end is cooled by water flowing steadily through a copper coil wound round it. Two thermometers in holes a distance d apart along the bar read θ₁ and θ₂; two more read the inflow and outflow water temperatures θ₃ and θ₄. When all four are steady (steady state), the heat conducted through the bar per second equals the heat carried away by the water: $\frac{KA({\theta }_1-{\theta }_2)}{d}=\frac{mc({\theta }_4-{\theta }_3)}{t}$, where m is the mass of water collected in time t. Lagging prevents heat loss from the sides so that the heat flow is parallel to the axis. A thick bar (large A) and a long gap d make the temperature difference large enough to measure. (Lee's disc method for poor conductors is not required.)

Worked example

In Searle's experiment, the bar has a radius of 2 cm, the thermometers 10 cm apart read a difference of 10 K, and water flows at 2.3 g per second, rising 5 K in temperature. Find K. (c of water = 4200 J kg⁻¹ K⁻¹)

Try it first, then show the solutionSolution

1. Heat carried by water per second = 0.0023 × 4200 × 5 = 48.3 W
2. A = π(0.02)² = 1.26 × 10⁻³ m²
3. $K=\frac{48.3\times 0.10}{1.26\times 10^{-3}\times 10}=384W{m}^{-1}{K}^{-1}$

Answer: About 384 W m⁻¹ K⁻¹ (copper)

Convection and radiation: blackbody, Stefan's law and Wien's law

Convection transfers heat by the actual movement of the heated fluid: warm fluid becomes less dense and rises, cooler fluid sinks (natural convection: sea and land breezes, room heaters near the floor); with a fan or pump it is forced convection. Radiation transfers energy as electromagnetic waves, needs no medium and travels at the speed of light (sunlight). A perfect blackbody absorbs all radiation falling on it; a cavity with a small hole is the best approximation, and its cavity radiation depends only on its temperature. Emissivity e (0 to 1) compares a surface with a blackbody; good absorbers are good emitters. Stefan-Boltzmann law: the energy radiated per second per unit area is I = σT⁴ for a blackbody (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴), eσT⁴ for other bodies; net loss to surroundings at T₀ is eσA(T⁴ − T₀⁴). Spectral radiancy R(λ) plotted against λ gives a hump-shaped curve for each temperature; a hotter body gives a higher curve (more total energy, the area) with its peak at a shorter wavelength. Wien's displacement law: ${\lambda }_{max}T=b$, with b ≈ 2.9 × 10⁻³ m K. (Qualitative ideas; deductions from Stefan's law are not required.)

Worked example

The Sun's radiation peaks at about 500 nm. Estimate its surface temperature. At what wavelength does the human body (310 K) radiate most?

Try it first, then show the solutionSolution

1. $T=\frac{b}{{\lambda }_{max}}=\frac{2.9\times 10^{-3}}{500\times 10^{-9}}=5800K$
2. Body: ${\lambda }_{max}=\frac{2.9\times 10^{-3}}{310}=9.4\times 10^{-6}m$

Answer: About 5800 K; about 9.4 μm (infrared, used by thermal cameras)

Newton's law of cooling

Newton's law of cooling: the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the difference is small: $-\frac{dQ}{dt}=k(T-T_0)$, so $-\frac{dT}{dt}=K(T-T_0)$. A cooling curve (temperature against time) falls steeply at first and then more slowly as T approaches T₀. In problems over a range of temperatures, use the average form: $\frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_0\right)$. The law can be seen as a small-difference form of the Stefan-Boltzmann result for radiation (with convection also contributing).

Worked example

A body cools from 80 °C to 60 °C in 10 minutes in a room at 20 °C. How long will it take to cool from 60 °C to 40 °C?

Try it first, then show the solutionSolution

1. First interval: 20/10 = K(70 − 20), so K = 0.04 min⁻¹
2. Second: 20/t = 0.04(50 − 20) = 1.2
3. t = 16.7 min

Answer: About 16.7 minutes (it cools more slowly as it nears room temperature)

Physics in real life: Thermal Properties of Matter

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Calorimetry and Latent Heat
• Conduction and Newton's Law of Cooling

Thermal Properties of Matter formula sheet

Temperature scales

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

Equal at −40 for C and F.

Ideal gas equation

$pV=nRT$

T in kelvin; R = 8.31 J mol⁻¹ K⁻¹.

Thermal expansion

$\Delta L=\alpha L\Delta T;\Delta V=\gamma V\Delta T;\beta =2\alpha ,\gamma =3\alpha$

Ideal gas at constant p: γ = 1/T.

Thermal stress

$\text{stress}=Y\alpha \Delta T$

Rod prevented from expanding.

Heat and specific heat

$Q=mc\Delta T$

Heat capacity = mc.

Latent heat

$Q=mL$

At constant temperature.

Heat conduction

$H=\frac{KA({\theta }_1-{\theta }_2)}{L},R=\frac{L}{KA}$

Steady state; series resistances add.

Searle's method

$\frac{KA({\theta }_1-{\theta }_2)}{d}=\frac{mc({\theta }_4-{\theta }_3)}{t}$

Good conductors.

Stefan-Boltzmann law

$I=e\sigma T^4,P_{net}=e\sigma A(T^4-T_0^4)$

σ = $5.67\times 10^{-8}$ W m⁻² K⁻⁴.

Wien's displacement law

${\lambda }_{max}T=b\approx 2.9\times 10^{-3}mK$

Hotter: peak at shorter λ.

Newton's law of cooling

$-\frac{dT}{dt}=K(T-T_0);\frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_0\right)$

Small temperature differences.

Thermal Properties of Matter: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Thermal Properties of Matter

The question patterns to expect, and the method that works for each.

Tricks to remember Thermal Properties of Matter

Short memory aids. Say each one aloud once, then use it on a question the same day.

• 1, 2, 3 for expansion

  α : β : γ = 1 : 2 : 3 for solids.

  Use it when: Converting between linear, area and volume coefficients.

• Four-degree water

  Water is 'fat' at 4: densest at 4 °C, so the bottom of a frozen lake stays at 4 °C.

  Use it when: Anomalous expansion questions.

• Conduction like current

  H = Δθ/R with R = L/KA, like I = V/R: series resistances add.

  Use it when: Composite slabs.

• Stefan four, Wien one

  Power ∝ T⁴ (Stefan); λ_max ∝ 1/T (Wien).

  Use it when: Radiation numericals.

• Average for cooling

  Newton: drop per minute = K × (average temperature − room temperature).

  Use it when: Cooling-time problems.

Thermal Properties of Matter: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Thermal Properties of Matter

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using °C in gas laws and Stefan's law

  Students substitute 27 and 47 instead of 300 K and 320 K.

  Wrong working

  p₂ = 2.0 × 10⁵ × 47/27.

  Right working

  p₂ = 2.0 × 10⁵ × 320/300 = 2.13 × 10⁵ Pa.

  ✓Do this: Convert to kelvin whenever T appears as a ratio or a power.

  Usually costs: 1-2 marks.

• ✗Mistake: Assuming all the ice melts

  Students write one equation without checking whether there is enough heat to melt the ice.

  Wrong working

  Large ice mass in a little warm water gives a negative final temperature.

  Right working

  Compare the heat available (cooling water to 0 °C) with mL_f first; if it is less, the final temperature is 0 °C with some ice left.

  ✓Do this: Do the 'enough heat?' check before solving.

  Usually costs: 2 marks.

• ✗Mistake: Forgetting latent heat in heating problems

  Students apply mcΔT straight from ice to steam.

  Wrong working

  Heat for ice at −10 °C to steam at 100 °C = m × c × 110.

  Right working

  Add the separate stages: warm ice, melt, warm water, boil.

  ✓Do this: Sketch the heating curve and add each segment.

  Usually costs: 2 marks.

• ✗Mistake: Using the temperature at the start instead of the average in Newton's law of cooling

  The rate changes during the interval, so using the initial excess temperature overestimates it.

  Wrong working

  20/t = K(60 − 20).

  Right working

  20/t = K(50 − 20), using the average 50 °C.

  ✓Do this: Use the mean temperature of each interval.

  Usually costs: 1 mark.

• ✗Mistake: Explaining anomalous expansion vaguely

  Students say 'water freezes at the top because ice is light' without the 4 °C argument.

  Wrong working

  Lakes freeze from the top because cold air is above.

  Right working

  Water cools and sinks until all of it is at 4 °C (densest). Below 4 °C the cooler water is lighter and stays at the top, where it freezes; ice is a poor conductor, so water below stays at about 4 °C.

  ✓Do this: Mention the density maximum at 4 °C and the insulating ice layer.

  Usually costs: 1-2 marks.

Practice questions on Thermal Properties of Matter

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The SI unit of the coefficient of thermal conductivity is:
   (a) W m K⁻¹
   (b) W m⁻¹ K⁻¹
   (c) J m⁻¹ K⁻¹
   (d) W m⁻² K⁻¹

   Hint

   K = HL/(AΔθ).

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   According to the Stefan-Boltzmann law, the energy radiated per second per unit area by a blackbody is proportional to:
   (a) T
   (b) T²
   (c) T³
   (d) T⁴

   Hint

   Fourth power.

   Answer

   (d)

3. Q3 · 1 mark · Very short answerExam level

   Define specific latent heat of fusion.

   Hint

   Solid to liquid.

   Answer

   The heat needed to change unit mass of a solid at its melting point into liquid at the same temperature.

4. Q4 · 1 mark · Very short answerExam level

   Why do lakes freeze from the top downward?

   Hint

   Density at 4 °C.

   Answer

   Water is densest at 4 °C; colder water below 4 °C is lighter and stays at the surface, where it freezes first.

5. Q5 · 2 marks · Short answerExam level

   Convert 50 °C to the Fahrenheit and Kelvin scales.

   Hint

   F = 9C/5 + 32.

   Answer

   122 °F; 323.15 K.

6. Q6 · 2 marks · Short answerHarder exam level

   A metal rod 1 m long expands by 0.6 mm when heated through 50 K. Find its coefficients of linear and volume expansion.

   Hint

   γ = 3α.

   Answer

   α = 0.6 × 10⁻³/(1 × 50) = 1.2 × 10⁻⁵ K⁻¹; γ = 3.6 × 10⁻⁵ K⁻¹.

7. Q7 · 2 marks · Short answerHarder exam level

   A steel rod clamped at both ends is heated by 50 K. Find the thermal stress. (Y = 2 × 10¹¹ Pa, α = 1.2 × 10⁻⁵ K⁻¹)

   Hint

   Stress = YαΔT.

   Answer

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

8. Q8 · 2 marks · Short answerHarder exam level

   Show why a burn from 10 g of steam at 100 °C is far worse than one from 10 g of water at 100 °C, if each cools to skin temperature 40 °C on the skin. (L_v = 2.26 × 10⁶ J kg⁻¹)

   Hint

   Steam first condenses.

   Answer

   Water: 0.01 × 4200 × 60 = 2520 J. Steam: 0.01 × 2.26 × 10⁶ + 2520 = 25 120 J, about ten times more.

9. Q9 · 2 marks · Short answerHarder exam level

   Find the volume of 1 mole of an ideal gas at 0 °C and 1.013 × 10⁵ Pa. (R = 8.31 J mol⁻¹ K⁻¹)

   Hint

   V = nRT/p.

   Answer

   V = 8.31 × 273/1.013 × 10⁵ = 0.0224 m³ (22.4 litres).

10. Q10 · 3 marks · Short answerHarder exam level

    A wall of area 1 m² is made of a 2 cm layer of wood (K = 0.8 W m⁻¹ K⁻¹) and a 3 cm layer of cork (K = 0.2 W m⁻¹ K⁻¹). The outer faces are at 30 °C (wood side) and 0 °C. Find the heat current and the temperature of the interface.

    Hint

    Thermal resistances R = L/(KA) add in series.

    Answer

    R_wood = 0.02/0.8 = 0.025 K W⁻¹; R_cork = 0.03/0.2 = 0.15 K W⁻¹; H = 30/0.175 = 171 W. Interface: 30 − 171 × 0.025 = 25.7 °C (most of the drop is across the cork).

11. Q11 · 3 marks · Short answerHarder exam level

    A sphere of radius 5 cm and emissivity 0.8 is at 727 °C. Find the power it radiates. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)

    Hint

    T = 1000 K; A = 4πr².

    Answer

    P = 0.8 × 5.67 × 10⁻⁸ × 4π(0.05)² × 10¹² = 1.43 × 10³ W.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) Explain conduction, convection and radiation, giving one everyday example of each. [3]
    (b) Sketch the spectral radiancy of a blackbody at two temperatures and use it to state Wien's displacement law. [2]

    Hint

    Medium needed? Bulk movement?

    Answer

    (a) Conduction: heat passes from particle to particle without bulk movement (a spoon handle in hot tea gets hot). Convection: heated fluid moves, carrying heat (sea breeze; a room heater placed low). Radiation: energy carried by electromagnetic waves, needing no medium (sunlight reaching the Earth). (b) Hump-shaped curves; the hotter one lies higher everywhere with its peak at a shorter wavelength; λ_max T = constant (2.9 × 10⁻³ m K).

13. Q13 · 3 marks · Short answerHOTS / competency

    A metal sphere cools from 70 °C to 60 °C in 5 minutes and from 60 °C to 52 °C in the next 5 minutes. Using Newton's law of cooling, find the temperature of the surroundings.

    Hint

    Write the average form for both intervals and divide.

    Answer

    10/5 = K(65 − T₀) and 8/5 = K(56 − T₀). Dividing: 1.25 = (65 − T₀)/(56 − T₀), so 70 − 1.25T₀ = 65 − T₀ and T₀ = 20 °C.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Relation between α, β and γ for a solid?Show answer

   β = 2α, γ = 3α.

2. 2.At what temperature is water densest?Show answer

   4 °C.

3. 3.Principle of calorimetry?Show answer

   Heat lost by hot bodies = heat gained by cold bodies.

4. 4.Heat current through a rod?Show answer

   H = KA(θ₁ − θ₂)/L.

5. 5.What is the steady state?Show answer

   The state in which the temperature at every point stays constant with time.

6. 6.State Wien's displacement law.Show answer

   λ_max T = constant (≈ 2.9 × 10⁻³ m K).

7. 7.Stefan's constant?Show answer

   5.67 × 10⁻⁸ W m⁻² K⁻⁴.

8. 8.State Newton's law of cooling.Show answer

   For small temperature differences, the rate of cooling is proportional to the excess temperature over the surroundings.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you convert between °C, °F and K?
• ☐Can you explain absolute zero from a gas thermometer?
• ☐Can you find gaps for expansion and thermal stress?
• ☐Can you explain why lakes freeze from the top?
• ☐Can you solve a mixture problem with ice melting?
• ☐Can you describe Searle's experiment and find K?
• ☐Can you use Stefan's and Wien's laws?
• ☐Can you solve a Newton's-law-of-cooling problem with the average method?

Thermal Properties of Matter: frequently asked questions