10th Class Physics Notes 2026 — Chapter 10
Class 10 Physics — Chapter 10: Thermal Properties of Matter
Punjab Board — Exercise Solutions
A. Multiple Choice Questions (MCQs)
10.1 When the temperature of a rod of copper is increased, its length:
(a) Remains the same
(b) Increases
(c) Decreases
(d) Becomes double
Correct Answer: (b) Increases
10.2 The amount by which the unit length of a material increases when the temperature is raised by 1°C is called the coefficient of:
(a) Cubic expansion
(b) Volume expansion
(c) Linear expansion
(d) None of these
Correct Answer: (c) Linear expansion
10.3 What does a liquid turn into once it has evaporated?
(a) Gas
(b) Water vapour
(c) Air
(d) Both (a) and (b)
Correct Answer: (b) Water vapour
10.4 Which property determines how much heat a solid can absorb before its temperature changes significantly?
(a) Density
(b) Specific heat capacity
(c) Colour
(d) Hardness
Correct Answer: (b) Specific heat capacity
10.5 High-temperature superconductors typically involve which type of material?
(a) Pure metals
(b) Organic compounds
(c) Ceramics
(d) Liquid crystals
Correct Answer: (c) Ceramics
B. Short Answer Questions
10.1 What factors influence the thermal expansion of solids?
Answer:
The thermal expansion of solids depends on three main factors:
Original Dimensions: The initial length, area, or volume of the solid. A greater initial length or volume generally leads to greater expansion.
Change in Temperature (\(\Delta T\)): A greater temperature change results in greater expansion.
Nature of the Material: Different materials have different coefficients of expansion (\(\alpha\) or \(\beta\)) because of differences in their atomic bonding structures.
10.2 What is latent heat of fusion?
Answer:
Latent heat of fusion (\(H_f\)) is the amount of heat energy required to change a unit mass (1 kg) of a substance from the solid state to the liquid state at its melting point without any change in temperature.
Formula:
Or,
Where:
\(H_f\) = Latent heat of fusion
\(\Delta Q_f\) = Heat energy supplied
\(m\) = Mass of the substance
SI Unit: Joule per kilogram (\(\text{J kg}^{-1}\))
10.3 How do different materials exhibit varying thermal expansion behaviours?
Answer:
Different materials exhibit different thermal expansion behaviours because of variations in their interatomic bonding forces and atomic packing.
Solids with strong chemical bonds, such as steel or quartz, generally expand less because their atoms are tightly bound.
Materials such as brass and aluminium generally expand more because of their different bonding structures and coefficients of thermal expansion.
10.4 How does evaporation contribute to cooling processes in everyday life?
Answer:
During evaporation, high-kinetic-energy molecules escape from the surface of a liquid into the air. The remaining molecules have lower average kinetic energy, causing the temperature of the liquid to decrease.
Everyday examples:
Sweating: Sweat evaporates from the skin, absorbing heat from the body and helping maintain body temperature.
Earthen Pots (Matkas): Water seeps through the pores of an earthen pot and evaporates, keeping the stored water cool.
10.5 What is latent heat of vaporization, and how is it calculated?
Answer:
Latent heat of vaporization (\(H_v\)) is the amount of heat energy required to convert a unit mass (1 kg) of a liquid completely into gas or vapour at its boiling point without any change in temperature.
Formula:
Or,
Where:
\(H_v\) = Latent heat of vaporization
\(\Delta Q_v\) = Heat supplied
\(m\) = Mass of the liquid
SI Unit: Joule per kilogram (\(\text{J kg}^{-1}\))
C. Constructed Response Questions
10.1 Why do bridges need expansion joints but glass jars with metal lids do not, even though both experience temperature changes? Discuss in terms of the nature of thermal expansion.
Answer:
Bridges:
Bridges are long structures made of concrete and steel. Linear thermal expansion is given by:
Since expansion is directly proportional to the original length (\(L_0\)), even a small temperature change can cause considerable expansion in a long bridge. Without expansion joints, this expansion may cause cracking or structural damage.
Glass Jars with Metal Lids:
Metal generally has a higher coefficient of thermal expansion than glass. When a metal lid is heated, for example, by pouring hot water over it, the lid expands more than the glass neck, making it easier to open. The smaller dimensions of a jar also mean that its total expansion is much less than that of a large bridge.
10.2 If two liquids are heated under the same conditions, but one of them heats up faster, what does this indicate about specific heat capacity?
Answer:
The heat equation is:
Rearranging:
This shows that, for the same amount of heat supplied to equal masses, the temperature change is inversely proportional to specific heat capacity (\(c\)).
The liquid that heats up faster has a lower specific heat capacity.
The liquid that heats up more slowly has a higher specific heat capacity.
10.3 Why does a liquid evaporate faster when the air is dry?
Answer:
Dry air contains less water vapour, meaning its humidity is low. This allows more liquid molecules to escape into the surrounding air without as much vapour returning to the liquid through condensation. Therefore, evaporation occurs faster in dry air.
10.4 When a piece of ice melts, the temperature does not increase even though heat is supplied. Where does the energy go, and why is this important in phase change?
Answer:
Where the Energy Goes:
The absorbed thermal energy is used as latent heat of fusion to overcome the intermolecular forces holding the solid crystal structure together. Instead of increasing the average kinetic energy of the molecules, the energy helps change the state of the substance.
Importance in Phase Change:
Temperature measures the average kinetic energy of particles. During melting, the temperature remains constant at the melting point because the supplied energy is used for the phase change. For ice at normal atmospheric pressure, this temperature is approximately \(0^\circ\text{C}\).
10.5 Why do electric transmission wires sag in summer and tighten in winter?
Answer:
In Summer: Higher temperatures cause metallic transmission wires to expand. Their increased length makes them slacken and sag between supporting poles.
In Winter: Lower temperatures cause the wires to contract, reducing their length and making them tighter.
D. Comprehensive Questions
10.1 How does thermal expansion differ between solids, liquids, and gases? Provide examples to illustrate these differences.
Answer:
Thermal expansion differs among solids, liquids, and gases because of differences in particle arrangement and intermolecular forces.
1. Solids
Solids have fixed shapes and tightly packed particles. They generally expand less than liquids and gases when heated.
Example: A bimetallic strip bends when heated because its two metals expand by different amounts.
2. Liquids
Liquids have no fixed shape and generally expand more than solids for the same temperature rise.
Example: Mercury expands and rises inside a glass thermometer when heated.
3. Gases
Gases have widely spaced particles and weak intermolecular forces. At constant pressure, they generally expand more than solids and liquids when heated.
Example: Heated air expands and becomes less dense, helping a hot-air balloon rise.
10.2 What are some practical consequences of thermal expansion in everyday life, such as in construction materials or household appliances?
Answer:
Thermal expansion has several practical applications in construction and household appliances.
Construction Materials
Railway Tracks: Gaps or suitable expansion provisions are provided to accommodate thermal expansion and help prevent buckling.
Bridges: Expansion joints and rollers allow bridges to expand and contract without suffering structural damage.
Pipelines: Loops and bends are incorporated into long pipelines to accommodate changes in length caused by temperature variations.
Household Appliances
Thermostats (Bimetallic Strips): These consist of two different metals joined together. When heated, they bend because the metals expand by different amounts. This action can make or break electrical circuits in appliances such as irons and heaters.
Pyrex/Glassware: Borosilicate glass has a low coefficient of thermal expansion, making it more resistant to cracking caused by sudden temperature changes.
10.3 What is latent heat, and why is it important in phase transitions like melting and vaporization?
Answer:
Latent heat is the heat energy absorbed or released by a substance during a change of state at a constant temperature.
Importance of Latent Heat
Phase Changes: It supplies the energy needed to overcome intermolecular forces during melting or vaporization without increasing the temperature.
Natural Systems: When lakes freeze, latent heat is released into the surroundings, slowing the cooling process and helping protect aquatic life.
Weather and Climate: Latent heat of vaporization plays an important role in atmospheric convection and weather patterns by transferring energy through evaporation and condensation.
10.4 Why do solids, liquids, and gases expand when heated? Explain with examples.
Answer:
When a substance is heated, its particles gain thermal energy. Their motion or vibrations increase, generally increasing the average separation between particles and causing expansion.
Solids:
Atoms vibrate with greater amplitude around their fixed mean positions. This can increase the average distance between neighbouring atoms.
Example: Overhead power lines become longer in summer and may sag.
Liquids:
Liquid molecules gain kinetic energy and move more rapidly. Their average separation generally increases, causing the liquid to expand.
Example: The liquid level rises in a thermometer when heated.
Gases:
Gas particles move faster and collide more energetically. At constant pressure, a gas expands when heated.
Example: Air inside a balloon expands when heated, provided the balloon can stretch.
10.5 If we have an equal amount of water in a wide, shallow dish and a tall, narrow container kept at the same temperature, in which container will evaporation occur faster? Explain the reason scientifically.
Answer:
Evaporation will occur faster in the wide, shallow dish.
Scientific Reason:
Evaporation is a surface phenomenon. Its rate generally increases with the surface area exposed to the atmosphere. The wide dish exposes a larger surface area, allowing more high-kinetic-energy water molecules to escape into the air simultaneously than in the narrow container.
E. Numerical Problems
10.1 A metal rod of length 1 m expands by 0.02 m when heated from \(20^\circ\text{C}\) to \(120^\circ\text{C}\). Calculate its coefficient of linear expansion.
Given Data:
Original length, \(L_0=1\,\text{m}\)
Change in length, \(\Delta L=0.02\,\text{m}\)
Initial temperature, \(T_1=20^\circ\text{C}\)
Final temperature, \(T_2=120^\circ\text{C}\)
To Find:
Coefficient of linear expansion, \(\alpha=?\)
Formula:
Therefore,
Calculation:
Step 1: Calculate the temperature change.
Step 2: Calculate the coefficient of linear expansion.
Result: The coefficient of linear expansion of the metal rod is \(2.0\times10^{-4}\,\text{K}^{-1}\).
10.2 A container holds 1 litre of water at \(20^\circ\text{C}\). What will its volume be at \(80^\circ\text{C}\), assuming water’s coefficient of volume expansion is \(2.1\times10^{-4}\) per °C?
Given Data:
Initial volume, \(V_0=1\,\text{litre}=1000\,\text{cm}^3\)
Initial temperature, \(T_1=20^\circ\text{C}\)
Final temperature, \(T_2=80^\circ\text{C}\)
Coefficient of volume expansion, \(\beta=2.1\times10^{-4}\,^\circ\text{C}^{-1}\)
To Find:
Final volume, \(V=?\)
Formula:
Calculation:
Step 1: Calculate the temperature change.
Step 2: Calculate the final volume.
Result: The volume of water at \(80^\circ\text{C}\) is \(1012.6\,\text{cm}^3\).
10.3 A steel rod initially measures 2 m at \(20^\circ\text{C}\). If its coefficient of linear expansion is \(1.2\times10^{-5}\,^\circ\text{C}^{-1}\), what will its length be at \(100^\circ\text{C}\)?
Given Data:
Initial length, \(L_0=2\,\text{m}\)
Initial temperature, \(T_1=20^\circ\text{C}\)
Final temperature, \(T_2=100^\circ\text{C}\)
Coefficient of linear expansion, \(\alpha=1.2\times10^{-5}\,^\circ\text{C}^{-1}\)
To Find:
Final length, \(L=?\)
Formula:
Calculation:
Step 1: Calculate the temperature change.
Step 2: Calculate the final length.
Rounding to three decimal places:
Result: The length of the steel rod at \(100^\circ\text{C}\) will be approximately \(2.002\,\text{m}\).
10.4 A steel bridge expands by 5 cm on a hot summer day. If the bridge originally spanned 100 m, what is the temperature change?
Given Data:
Expansion, \(\Delta L=5\,\text{cm}=0.05\,\text{m}\)
Original length, \(L_0=100\,\text{m}\)
Coefficient of linear expansion for steel, \(\alpha=1.2\times10^{-5}\,\text{K}^{-1}\)
To Find:
Temperature change, \(\Delta T=?\)
Formula:
Therefore,
Calculation:
Result: The temperature change is approximately \(41.67^\circ\text{C}\).
10.5 How much heat is required to raise the temperature of a 2 kg iron bar from \(20^\circ\text{C}\) to \(100^\circ\text{C}\), given that the specific heat capacity of iron is \(450\,\text{J kg}^{-1}\text{K}^{-1}\)?
Given Data:
Mass, \(m=2\,\text{kg}\)
Initial temperature, \(T_1=20^\circ\text{C}\)
Final temperature, \(T_2=100^\circ\text{C}\)
Specific heat capacity, \(c=450\,\text{J kg}^{-1}\text{K}^{-1}\)
To Find:
Heat required, \(\Delta Q=?\)
Formula:
Calculation:
Step 1: Calculate the temperature change.
Step 2: Calculate the heat required.
Converting joules into kilojoules:
Result: The total heat required is \(72\,\text{kJ}\).
10.6 How much heat is required to melt 500 g of ice at \(0^\circ\text{C}\) into water at \(0^\circ\text{C}\)? (Latent heat of fusion of ice = \(3.36\times10^5\,\text{J kg}^{-1}\))
Given Data:
Mass of ice, \(m=500\,\text{g}=0.5\,\text{kg}\)
Temperature, \(T=0^\circ\text{C}\)
Latent heat of fusion, \(H_f=3.36\times10^5\,\text{J kg}^{-1}\)
To Find:
Heat required, \(\Delta Q_f=?\)
Formula:
Calculation:
Converting into kilojoules:
Result: The heat required to melt the ice is \(168\,\text{kJ}\).
10.7 Calculate the heat required to completely vaporize 1 kg of water at \(100^\circ\text{C}\). (Latent heat of vaporization of water = \(2.26\times10^6\,\text{J kg}^{-1}\))
Given Data:
Mass of water, \(m=1\,\text{kg}\)
Temperature, \(T=100^\circ\text{C}\)
Latent heat of vaporization, \(H_v=2.26\times10^6\,\text{J kg}^{-1}\)
To Find:
Heat required, \(\Delta Q_v=?\)
Formula:
Calculation:
Converting into megajoules:
Result: The heat required to vaporize the water is \(2.26\,\text{MJ}\).
Notes by Sir Dilawar
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