This MCQ module is based on: Temperature Thermal Expansion
TOPIC 9 OF 33
Temperature Thermal Expansion
🎓 Class 11
Physics
CBSE
Theory
Ch 10 – Thermal Properties of Matter
⏱ ~14 min
🌐 Language: [gtranslate]
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Temperature Thermal Expansion
10.1 Introduction: Heat & Temperature in Daily Life
A kettle of boiling water feels much "hotter" than a glass of iced water. Common sense links this hotness with a quantity called temperature. In physics we sharpen this idea: temperature decides the direction in which heat flows. Heat is energy in transit; temperature is the property that controls that transit.
This first part of Chapter 10 covers (i) what temperature really means, (ii) how thermometers measure it, (iii) the different thermometric scales (Celsius, Fahrenheit, Kelvin), (iv) the constant-volume gas thermometer that leads us to absolute zero, and (v) thermal expansion — how solids, liquids and gases enlarge when heated.
10.2 Temperature and Heat
Heat is the form of energy transferred between two (or more) bodies (or system and surroundings) by virtue of temperature difference. The SI unit of heat is the joule (J). The cgs unit is the calorie (cal); 1 cal = 4.186 J.
Temperature vs. Heat: Temperature is an intensive property (does not depend on amount of matter). Heat is an extensive energy in transit. A spoonful and a bucket of boiling water have the same temperature (100 °C) but the bucket carries far more heat.
10.3 Measurement of Temperature
Any physical property that varies smoothly and reproducibly with temperature can serve as a thermometric property. Examples are: the volume of a liquid (mercury, alcohol thermometer), the pressure of a fixed mass of gas at constant volume, the resistance of a platinum wire (resistance thermometer), or the colour of a hot body (pyrometer).
10.3.1 The Celsius and Fahrenheit Scales
Most thermometers used in everyday life are liquid-in-glass thermometers, calibrated against two reference points:
- Ice point (lower fixed point): melting point of pure ice at 1 atm.
- Steam point (upper fixed point): boiling point of pure water at 1 atm.
On the Celsius scale these are 0 °C and 100 °C, divided into 100 equal parts. On the Fahrenheit scale they are 32 °F and 212 °F, divided into 180 parts. The conversion is:
\[\boxed{\,t_F = \tfrac{9}{5}t_C + 32\,}\qquad\text{or}\qquad t_C = \tfrac{5}{9}(t_F - 32)\]
10.3.2 The Constant-Volume Gas Thermometer
Real-gas thermometers are far more reproducible than mercury thermometers. The constant-volume gas thermometer (Fig 10.2) keeps a fixed amount of gas in a bulb at constant volume; the gas pressure \(P\) becomes the thermometric property. Different gases (H₂, He, N₂, O₂) all give straight-line P–T graphs that, when extrapolated backwards, meet the temperature axis at the same intercept.
10.3.3 Absolute (Kelvin) Scale
The intercept −273.15 °C is therefore a natural lower limit of temperature: absolute zero. Lord Kelvin defined a new scale starting there, with the same step size as Celsius. The kelvin (K) is now the SI unit of temperature.
\[\boxed{\,T(\text{K}) = t(°\text{C}) + 273.15\,}\]
The triple point of water — at which ice, water and water-vapour coexist — is fixed at 273.16 K. Absolute zero (0 K) is the temperature at which an ideal gas would exert zero pressure.
10.4 Ideal Gas Equation and Absolute Temperature
For a fixed mass of an ideal gas the experimental laws of Boyle, Charles and Avogadro combine into the single relation:
\[\boxed{\,PV = \mu R T\,}\]
where \(\mu\) is the number of moles, \(R = 8.314\) J mol\(^{-1}\) K\(^{-1}\) is the universal gas constant, and \(T\) is the absolute (kelvin) temperature.
For a constant amount of gas: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\). At low pressures real gases obey this relation closely; at high pressures and low temperatures deviations appear (covered in Chapter 12 — Kinetic Theory).
10.5 Thermal Expansion
Most substances expand on heating and contract on cooling. The microscopic reason: at higher temperature the atoms vibrate with larger amplitude about their mean positions, and because the inter-atomic potential is asymmetric, the mean separation between atoms increases.
10.5.1 Linear Expansion
Consider a rod of length \(L\) at temperature \(T\). When heated by \(\Delta T\) its length increases by \(\Delta L\). Experiment shows \(\Delta L \propto L\) and \(\Delta L \propto \Delta T\):
\[\Delta L = \alpha_L\, L\, \Delta T \qquad \text{or} \qquad \alpha_L = \frac{1}{L}\frac{\Delta L}{\Delta T}\]
\(\alpha_L\) is the coefficient of linear expansion with SI unit K\(^{-1}\) (or °C\(^{-1}\), since they have the same step).
| Material | αL (×10⁻⁶ K⁻¹) | Material | αL (×10⁻⁶ K⁻¹) |
|---|---|---|---|
| Aluminium | 23 | Steel | 11 |
| Copper | 17 | Glass (ordinary) | 9 |
| Brass | 19 | Glass (Pyrex) | 3.2 |
| Iron | 12 | Invar (Fe-Ni alloy) | 0.9 |
10.5.2 Area (Superficial) Expansion
For an isotropic plate of area \(A\) heated by \(\Delta T\):
\[\Delta A = \alpha_A\, A\, \Delta T,\qquad \alpha_A = 2\alpha_L\]
Derivation: a square plate of side \(L\) has area \(A = L^2\). After heating, side becomes \(L(1+\alpha_L\Delta T)\). New area \(A' = L^2(1+\alpha_L\Delta T)^2 \approx L^2(1 + 2\alpha_L \Delta T)\) for small \(\alpha_L \Delta T\). So \(\Delta A/A = 2\alpha_L\Delta T\).
10.5.3 Volume (Cubical) Expansion
For an isotropic solid of volume \(V\):
\[\Delta V = \alpha_V\, V\, \Delta T,\qquad \alpha_V = 3\alpha_L\]
For liquids and gases there is only volume expansion (no fixed shape). The coefficient \(\alpha_V\) of liquids is about 10× that of solids; gases expand even more — for an ideal gas at constant pressure, \(\alpha_V = 1/T \approx 3.66\times 10^{-3}\) K\(^{-1}\) at 273 K.
10.5.4 Anomalous Expansion of Water
Most substances expand monotonically. Water is an oddity: between 0 °C and 4 °C its volume decreases with rising temperature; above 4 °C it expands normally. So water is densest at 4 °C. This is why ice floats and why ponds freeze top-down, leaving aquatic life alive in the warmer water below.
Worked Examples
Example 10.1: Convert temperatures between scales
(a) Express 37 °C (normal body temperature) in Fahrenheit and Kelvin. (b) Express −40 °C in Fahrenheit (note the surprise!).
(a) \(t_F = \tfrac{9}{5}\times 37 + 32 = 66.6 + 32 = 98.6\) °F.
\(T = 37 + 273.15 = 310.15\) K.
(b) \(t_F = \tfrac{9}{5}(-40) + 32 = -72 + 32 = -40\) °F. The Celsius and Fahrenheit scales coincide at \(\boxed{-40°}\) — useful trick to remember!
\(T = 37 + 273.15 = 310.15\) K.
(b) \(t_F = \tfrac{9}{5}(-40) + 32 = -72 + 32 = -40\) °F. The Celsius and Fahrenheit scales coincide at \(\boxed{-40°}\) — useful trick to remember!
Example 10.2: Steel railway track expansion (NCERT-style)
A steel railway line is 1.0 km long when the temperature is 5 °C. By how much does its length increase if the temperature rises to 50 °C? (\(\alpha_L\) for steel = 1.2 × 10⁻⁵ K⁻¹). What practical engineering arrangement avoids buckling?
\(\Delta T = 50 - 5 = 45\) K (or °C, same).
\(\Delta L = \alpha_L L \Delta T = (1.2\times 10^{-5}) \times 1000 \times 45 = 0.54\) m = 54 cm.
Engineers leave small expansion gaps between rail sections, or use continuously welded rails clamped tightly so that compressive stress, not buckling, absorbs the expansion.
\(\Delta L = \alpha_L L \Delta T = (1.2\times 10^{-5}) \times 1000 \times 45 = 0.54\) m = 54 cm.
Engineers leave small expansion gaps between rail sections, or use continuously welded rails clamped tightly so that compressive stress, not buckling, absorbs the expansion.
Example 10.3: Hot iron ring on a wooden wheel (the blacksmith trick)
A blacksmith fits an iron ring of inner diameter 5.231 m onto a wooden wheel of outer diameter 5.243 m. To what temperature must the iron ring be heated so that it just fits over the wheel? Initial temperature = 27 °C; \(\alpha_L\) (iron) = 1.20 × 10⁻⁵ K⁻¹.
The diameter of the ring must increase from 5.231 m to 5.243 m, i.e. \(\Delta L = 0.012\) m, with \(L = 5.231\) m.
\(\Delta T = \dfrac{\Delta L}{\alpha_L L} = \dfrac{0.012}{(1.20\times 10^{-5})(5.231)} = 191\) K.
So the ring must be heated to \(T = 27 + 191 = \boxed{218\,°\text{C}}\). On cooling, the ring contracts and grips the wheel firmly — this is exactly how cartwheels were rimmed traditionally.
\(\Delta T = \dfrac{\Delta L}{\alpha_L L} = \dfrac{0.012}{(1.20\times 10^{-5})(5.231)} = 191\) K.
So the ring must be heated to \(T = 27 + 191 = \boxed{218\,°\text{C}}\). On cooling, the ring contracts and grips the wheel firmly — this is exactly how cartwheels were rimmed traditionally.
Example 10.4: Volume expansion of mercury
A glass flask contains 1000 cm³ of mercury at 20 °C. The temperature rises to 80 °C. Find the apparent change in mercury level if the glass flask itself expands. \(\alpha_V\) (Hg) = 1.82 × 10⁻⁴ K⁻¹; \(\alpha_L\) (glass) = 9 × 10⁻⁶ K⁻¹.
\(\Delta T = 60\) K.
Volume coeff. of glass = \(3\alpha_L = 27\times 10^{-6}\) K⁻¹.
Apparent expansion = \((\alpha_V^{Hg} - \alpha_V^{glass})V\Delta T\)
= \((1.82\times 10^{-4} - 0.27\times 10^{-4})\times 1000 \times 60\)
= \((1.55\times 10^{-4})\times 60000 = 9.30\) cm³.
So mercury appears to rise by 9.30 cm³ — much less than the absolute mercury expansion (10.92 cm³) because the flask also expands.
Volume coeff. of glass = \(3\alpha_L = 27\times 10^{-6}\) K⁻¹.
Apparent expansion = \((\alpha_V^{Hg} - \alpha_V^{glass})V\Delta T\)
= \((1.82\times 10^{-4} - 0.27\times 10^{-4})\times 1000 \times 60\)
= \((1.55\times 10^{-4})\times 60000 = 9.30\) cm³.
So mercury appears to rise by 9.30 cm³ — much less than the absolute mercury expansion (10.92 cm³) because the flask also expands.
Interactive 1: Temperature Scale Converter L3 Apply
Type a value in any one box; the others update live.
Identical reading on all three scales.
Interactive 2: Linear Expansion Calculator L3 Apply
Choose material and starting length; vary temperature change.
Activity — Bimetallic Strip in Action
L4 Analyse
Predict: Two thin strips, one of brass and one of iron, are riveted together to form a single "bimetallic strip". Predict its shape change when heated.
- Take a long, thin bimetallic strip (or simulate by riveting brass and iron strips together).
- At room temperature note that the strip is straight.
- Hold one end and heat the other end gently with a candle flame.
- Observe the bending and note which metal forms the convex (outer) side.
Observation: The strip bends with brass on the convex (outer) side and iron on the concave (inner) side, because brass has a larger coefficient of linear expansion (19 × 10⁻⁶ K⁻¹ vs 12 × 10⁻⁶ K⁻¹).
Application: Bimetallic strips act as thermal switches in electric irons, fire alarms and refrigerator thermostats — when the temperature exceeds a set value, the bend breaks (or makes) the circuit.
Application: Bimetallic strips act as thermal switches in electric irons, fire alarms and refrigerator thermostats — when the temperature exceeds a set value, the bend breaks (or makes) the circuit.
Competency-Based Questions
A 30 m long aluminium electric cable strung between two pylons is taut at 20 °C. During a heat-wave the temperature reaches 50 °C, while in winter it drops to −10 °C. \(\alpha_L\) (Al) = 2.3 × 10⁻⁵ K⁻¹.
Q1. L1 Remember Define the coefficient of linear expansion of a substance.
It is the fractional change in length per unit rise in temperature: \(\alpha_L = \dfrac{1}{L}\dfrac{\Delta L}{\Delta T}\). SI unit K⁻¹.
Q2. L3 Apply By how much does the cable expand on a 50 °C summer day?
\(\Delta L = (2.3\times 10^{-5})\times 30 \times 30 = 0.0207\) m = \(\boxed{2.07\text{ cm}}\). The cable visibly sags more in summer — a familiar sight on hot days.
Q3. L3 Apply By how much does it shorten in winter (from 20 °C to −10 °C)?
\(\Delta T = -30\) K, so \(\Delta L = (2.3\times 10^{-5})(30)(-30) = -2.07\times 10^{-2}\) m = −2.07 cm. The cable contracts and pulls tightly on the pylons.
Q4. L4 Analyse If the cable were not allowed to contract (rigidly fixed at both ends) the resulting tensile stress is \(\sigma = Y\alpha_L\Delta T\). For Y(Al) = 7 × 10¹⁰ Pa, find σ at −10 °C.
\(\sigma = (7\times 10^{10})(2.3\times 10^{-5})(30) = 4.83\times 10^{7}\) Pa ≈ 48 MPa. This is well within Al's tensile strength (~90 MPa), so the cable survives but is under heavy stress.
Q5. L5 Evaluate Why are pendulums for accurate clocks made of Invar rather than steel?
Invar (an Fe-Ni alloy) has \(\alpha_L\) ≈ 0.9 × 10⁻⁶ K⁻¹ — over ten times smaller than steel. The pendulum length, and therefore its time period \(T = 2\pi\sqrt{L/g}\), is almost insensitive to temperature changes, keeping the clock accurate across seasons.
Assertion-Reason Questions
Assertion (A): A hollow brass sphere expands the same as a solid brass sphere of the same outer radius when heated.
Reason (R): Thermal expansion depends only on the coefficient α and the linear dimensions, not on whether the body is hollow or solid.
Answer: A. Every linear dimension scales by (1 + αΔT), so the cavity (if any) also enlarges by the same factor. The hollow and solid spheres expand identically.
Assertion (A): Water at 4 °C is at its maximum density.
Reason (R): Water expands on warming above 4 °C and also expands on cooling below 4 °C.
Answer: A. Both statements are true and R is exactly why ρ is maximum at 4 °C — this is the celebrated anomalous expansion of water.
Assertion (A): The Kelvin and Celsius scales have the same step size.
Reason (R): The Kelvin scale was constructed by shifting the zero of the Celsius scale by 273.15 units, keeping the divisions the same.
Answer: A. A change of 1 °C is exactly a change of 1 K. Only the zero is shifted.
Frequently Asked Questions - Temperature Thermal Expansion
What is the main concept covered in Temperature Thermal Expansion?
In NCERT Class 11 Physics Chapter 10 (Thermal Properties of Matter), "Temperature Thermal Expansion" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Temperature Thermal Expansion useful in real-life applications?
Real-life applications of Temperature Thermal Expansion from NCERT Class 11 Physics Chapter 10 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Temperature Thermal Expansion?
Key formulas in Temperature Thermal Expansion (NCERT Class 11 Physics Chapter 10 Thermal Properties of Matter) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 10?
NCERT Class 11 Physics Chapter 10 (Thermal Properties of Matter) is structured so each part builds on the previous one. Temperature Thermal Expansion connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Temperature Thermal Expansion?
CBSE board questions from Temperature Thermal Expansion typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Temperature Thermal Expansion lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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