Frank Brothers Solutions for ICSE Class 10 Physics Chapter 5.3 Heat Change Of State

Chapter 5. Heat Change of State

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Question 1. Define specific heat capacity. Is it the same as heat capacity?
Answer: Specific heat capacity is the quantity of heat energy needed to increase the temperature of \( 1\text{ kg} \) of a substance by \( 1^\circ\text{C} \) (or \( 1\text{ K} \)). It is different from heat capacity, which is dependent on the total mass of the object.
In simple words: Specific heat is the heat needed for 1 kg of a material to warm up by 1 degree, and it is not the same as heat capacity.

Exam Tip: Emphasize that specific heat capacity is per unit mass, whereas heat capacity is for the entire mass of the body.

Question 2. State the SI unit of specific heat capacity.
Answer: The SI unit of specific heat capacity is \( \text{J kg}^{-1}\text{ K}^{-1} \).
In simple words: The standard unit for specific heat is Joules per kilogram per Kelvin.

Exam Tip: Make sure you do not forget the negative exponents in the units, which can also be written as \( \text{J / (kg K)} \).

Question 3. Which physical quantity is a characteristic property of a substance and does not depend on its mass?
Answer: Specific heat capacity is the physical property that remains constant for a given substance, regardless of its mass.
In simple words: Specific heat capacity is a fixed property of a material and does not change based on how much of it you have.

Exam Tip: This property is unique to each substance and is often used to identify different materials in thermal experiments.

Question 4. State the principle of calorimetry.
Answer: According to the Principle of Calorimetry, when a hot object is mixed with or placed in contact with a cold object, heat flows from the hotter object to the colder one. This heat transfer continues until they reach thermal equilibrium, such that:
\[ \text{Total heat gained by the colder body} = \text{Total heat lost by the hotter body} \]
This relation is valid provided that no heat energy is lost to the surrounding environment.
In simple words: When hot and cold things touch, the heat lost by the hot one is equal to the heat gained by the cold one, if no heat escapes.

Exam Tip: Don't forget to mention "assuming no loss of heat to the surroundings" as the principle is only valid under adiabatic conditions.

Question 5. Define thermal capacity of a body.
Answer: Thermal capacity (or heat capacity) of an object is the total amount of heat energy needed to raise the temperature of the entire body by \( 1^\circ\text{C} \) (or \( 1\text{ K} \)).
In simple words: Thermal capacity is the heat required to warm up a whole object by 1 degree.

Exam Tip: The formula for thermal capacity is \( C' = m \times c \), where \( m \) is mass and \( c \) is specific heat capacity.

Question 6. What is the mathematical product of mass and specific heat capacity called?
Answer: Multiplying the mass of a body by its specific heat capacity yields its heat capacity (or thermal capacity).
In simple words: If you multiply mass by specific heat, you get the heat capacity.

Exam Tip: This relationship is written as \( C' = m \times c \) and is frequently used to convert between the two quantities in numerical questions.

Question 7. Does the specific heat capacity of a substance depend on its temperature?
Answer: No, specific heat capacity is a constant value for a given substance and does not vary with temperature under standard conditions.
In simple words: No, the specific heat capacity of a material stays the same and does not change when the temperature changes.

Exam Tip: Treat specific heat capacity as a constant value for any given state of matter in your calculations.

Question 8. Why does a copper rod become warmer than an aluminium rod of the same mass when supplied with the same amount of heat?
Answer: A copper rod reaches a higher temperature than an aluminium rod of identical mass because copper has a lower specific heat capacity than aluminium. As a result, copper requires less heat energy to undergo the same temperature rise.
In simple words: Copper heats up faster than aluminium because it has a lower heat capacity, meaning it needs less heat to raise its temperature.

Exam Tip: Explain that a lower specific heat capacity leads to a larger temperature rise for the same heat input and mass, as \( \Delta T \propto \frac{1}{c} \).

Question 9. What is the term used for the quantity of heat needed to raise the temperature of a body by \( 1^\circ\text{C} \)?
Answer: The quantity of heat energy required to increase the temperature of a body by \( 1^\circ\text{C} \) is defined as its heat capacity.
In simple words: The heat needed to raise a body's temperature by 1 degree is called its heat capacity.

Exam Tip: Be sure to use the word "body" to distinguish it from specific heat capacity, which refers to a "unit mass".

Question 10. State a common unit of heat capacity.
Answer: A common unit used to express heat capacity is \( \text{J }^\circ\text{C}^{-1} \) (joules per degree Celsius).
In simple words: Heat capacity is measured in Joules per degree Celsius.

Exam Tip: In the strict SI system, the unit is \( \text{J K}^{-1} \) (joules per Kelvin).

Question 11. What is the specific heat capacity of water?
Answer: The specific heat capacity of water is \( 4200\text{ J kg}^{-1}\text{ K}^{-1} \).
In simple words: The specific heat of water is 4200 Joules per kilogram per Kelvin.

Exam Tip: This is an important constant to remember, as it is very high and explains many natural phenomena.

Question 12. Why does water warm up more slowly than iron when heated?
Answer: Materials with a high specific heat capacity, such as water, require more heat energy to raise their temperature, causing them to warm up slowly. Conversely, substances like iron have a low specific heat capacity and warm up much faster under the same heat input.
In simple words: Water has a high heat capacity, so it takes a lot of heat and time to warm up compared to iron.

Exam Tip: Relate the slow warming rate directly to the "high specific heat capacity" value of water (\( 4200\text{ J kg}^{-1}\text{ K}^{-1} \)) compared to metals.

Question 13. Define latent heat.
Answer: Latent heat is the quantity of heat energy absorbed or released by a substance while it undergoes a change in its physical state (such as ice melting to water, or water boiling to steam) without any change in its temperature.
In simple words: Latent heat is the hidden heat used to change a substance's state (like ice to water) while the temperature stays the same.

Exam Tip: The term "at constant temperature" is highly critical in this definition, so do not omit it.

Question 14. What is the SI unit of specific latent heat?
Answer: The standard SI unit for expressing specific latent heat is \( \text{J kg}^{-1} \) (joules per kilogram).
In simple words: Specific latent heat is measured in Joules per kilogram.

Exam Tip: Remember that specific latent heat is defined per unit mass, which is why the unit contains "per kilogram".

Question 15. What is meant by the statement 'the specific latent heat of fusion of ice is \( 336\text{ J g}^{-1} \)'?
Answer: This statement means that \( 1\text{ g} \) of ice at \( 0^\circ\text{C} \) requires exactly \( 336\text{ J} \) of heat energy to completely melt into water at the same temperature of \( 0^\circ\text{C} \).
In simple words: It means you need 336 Joules of heat to melt 1 gram of ice at 0°C into water without raising its temperature.

Exam Tip: Ensure you specify both the mass (1 g) and the constant temperature (0°C) to make your explanation complete.

Question 16. What changes in volume occur when a liquid solidifies? Explain with the example of water.
Answer: When a liquid transitions into a solid, it can experience either an expansion or a contraction in volume. For example, when water freezes to become ice, it expands, resulting in a volume increase of approximately 10%.
In simple words: Liquids can get bigger or smaller when they freeze. Water is unique because it expands and gets about 10% bigger when it turns to ice.

Exam Tip: This expansion of water on freezing is anomalous and is a key concept in physics that explains why pipes burst in winter.

Question 17. How does the addition of impurities affect the melting point of ice?
Answer: Introducing impurities (such as salt) into ice lowers its melting point.
In simple words: Adding things like salt to ice makes it melt at a lower temperature.

Exam Tip: This property is why salt is sprinkled on snowy roads to melt ice quickly at sub-zero temperatures.

Question 18. How does an increase in pressure affect the melting point of substances like ice that contract upon melting?
Answer: For substances that contract when they melt, such as ice, an increase in applied pressure results in a decrease in their melting point.
In simple words: Squeezing ice with pressure makes it melt at a lower temperature than 0°C.

Exam Tip: Remember that the melting point of substances that expand on melting (like wax) increases with pressure, which is the opposite of ice.

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Question 19. Define the phenomenon of regelation.
Answer: Regelation is the process where ice melts under high pressure and solidifies (freezes) back into ice once that pressure is released.
In simple words: Regelation is when ice turns to water under pressure and then turns back to ice when the pressure is gone.

Exam Tip: This phenomenon can be demonstrated by pressing two ice cubes together; they melt at the contact surface and fuse when released.

Question 20. Define specific latent heat of vaporization.
Answer: The quantity of heat energy required to convert a unit mass of a liquid at its boiling point into vapour without changing its temperature is known as its specific latent heat of vaporization.
In simple words: Latent heat of vaporization is the heat needed to turn boiling liquid into steam at the same temperature.

Exam Tip: Make sure to mention "at constant temperature" or "at its boiling point" to secure full marks.

Question 21. How does pressure affect the boiling point of a liquid?
Answer: The boiling point of a liquid rises when pressure is increased, and drops when pressure is reduced.
In simple words: Higher pressure raises the boiling point, while lower pressure lowers it.

Exam Tip: This principle explains why food cooks faster in a pressure cooker where the pressure is high.

Question 22. What is meant by the latent heat of fusion of ice?
Answer: The latent heat of fusion of ice is the total heat energy required to transform ice at \( 0^\circ\text{C} \) into liquid water without any change in temperature.
In simple words: It is the heat needed to melt ice at 0°C into water at 0°C.

Exam Tip: Specify that this transition happens at \( 0^\circ\text{C} \) to keep the definition precise.

Question 23. Define the latent heat of vaporization of steam.
Answer: The latent heat of vaporization of steam is the quantity of heat energy needed to convert liquid water at \( 100^\circ\text{C} \) into steam at the identical temperature.
In simple words: It is the heat required to turn water at 100°C into steam at 100°C.

Exam Tip: Since this refers to steam, the standard reference boiling point of \( 100^\circ\text{C} \) must be stated clearly.

Question 24. Which physical quantity remains constant during a change of state?
Answer: The temperature of a body stays completely constant while it is undergoing a change of state.
In simple words: The temperature of an object does not change while it is melting or boiling.

Exam Tip: Even though heat is continuously supplied or removed, the temperature remains unchanged because the energy goes into changing molecular bonds.

Question 25. Why does ice at \( 0^\circ\text{C} \) appear much colder than water at the same temperature?
Answer: Every kilogram of ice at \( 0^\circ\text{C} \) must absorb \( 336,000\text{ J} \) of latent heat energy to melt into water at \( 0^\circ\text{C} \). Consequently, a kilogram of water at \( 0^\circ\text{C} \) contains \( 336,000\text{ J} \) more heat energy than ice at the same temperature. Since ice absorbs this extra heat from our hands to melt, it feels significantly colder than water.
In simple words: Ice feels colder because it absorbs a large amount of heat (336,000 Joules per kg) from your hand just to melt into water, even though both are at 0°C.

Exam Tip: Use the term "specific latent heat of fusion" and state its value (\( 336,000\text{ J kg}^{-1} \)) to write a high-scoring answer.

Question 26. Why do burns caused by steam at \( 100^\circ\text{C} \) tend to be more severe than those caused by boiling water at the same temperature?
Answer: Steam at \( 100^\circ\text{C} \) causes more severe burns than boiling water at \( 100^\circ\text{C} \) because every gram of steam releases an additional \( 2260\text{ J} \) of latent heat of vaporization while condensing into liquid on the skin. This extra thermal energy is not present in boiling water, resulting in more intense tissue damage.
In simple words: Steam carries an extra 2260 Joules of hidden heat per gram compared to boiling water. When it touches your skin, it releases this extra heat as it turns back into water, causing deeper burns.

Exam Tip: Mention the latent heat of vaporization of steam (\( 2260\text{ J g}^{-1} \)) and the process of condensation on the skin to score full marks.

Question 27. State the unit of heat capacity in the CGS system.
Answer: In the CGS system, the unit for measuring heat capacity is \( \text{cal }^\circ\text{C}^{-1} \) (calories per degree Celsius).
In simple words: In the CGS system, heat capacity is measured in calories per degree Celsius.

Exam Tip: Be sure to write the unit clearly, and avoid confusing CGS units with SI units.

Question 28. Convert \( 1\text{ cal g}^{-1}\text{ }^\circ\text{C}^{-1} \) into the SI unit of specific heat capacity.
Answer: To convert the CGS unit of specific heat capacity to the SI unit, we use the relations \( 1\text{ cal} \approx 4.2\text{ J} \) and \( 1\text{ g} = 10^{-3}\text{ kg} \):
\[ 1\text{ cal g}^{-1}\text{ }^\circ\text{C}^{-1} = 4200\text{ J kg}^{-1}\text{ K}^{-1} \]
In simple words: One calorie per gram per degree Celsius is equal to 4200 Joules per kilogram per Kelvin.

Exam Tip: This conversion is extremely important for solving numerical problems where units are mixed.

Question 29. What is meant by the statement 'the specific heat capacity of a substance is \( 0.2\text{ cal g}^{-1}\text{ }^\circ\text{C}^{-1} \)'?
Answer: This statement means that \( 0.2\text{ calories} \) of heat energy is required to increase the temperature of \( 1\text{ g} \) of the given substance by \( 1^\circ\text{C} \).
In simple words: It means you need 0.2 calories of heat to warm up 1 gram of this material by 1 degree Celsius.

Exam Tip: Ensure you specify both the unit of mass (1 g) and the temperature rise (1°C) in your explanation.

Question 30. A copper block of mass \( 100\text{ g} \) has a specific heat capacity of \( 0.04\text{ cal g}^{-1}\text{ }^\circ\text{C}^{-1} \). Calculate its heat capacity.
Answer: Given:
- Mass, \( m = 100\text{ g} \)
- Specific heat capacity, \( C = 0.04\text{ cal g}^{-1}\text{ }^\circ\text{C}^{-1} \)

Using the heat capacity formula:
\[ \text{Heat capacity} = m \times C \ ]
\[ \text{Heat capacity} = 100 \times 0.04 = 4\text{ cal }^\circ\text{C}^{-1} \]
Thus, the heat capacity of the block is \( 4\text{ cal }^\circ\text{C}^{-1} \).
In simple words: By multiplying the mass of 100 grams by the specific heat of 0.04, we find the heat capacity is 4 calories per degree Celsius.

Exam Tip: Always state the formula used clearly before substituting numerical values to secure partial marks in case of calculation errors.

Question 31. State the relation between heat capacity and specific heat capacity of a substance.
Answer: The relationship is given by the formula:
\[ \text{Heat capacity} = \text{mass} \times \text{specific heat capacity} \]
In simple words: Heat capacity is found by multiplying the mass of the object by its specific heat capacity.

Exam Tip: This formula is written symbolically as \( C' = m \times c \), where \( C' \) is the heat capacity.

Question 32. Does the specific heat capacity of a substance depend on its mass?
Answer: No, specific heat capacity is an intensive property and is completely independent of the mass of the substance.
In simple words: No, the specific heat of a substance does not change when the amount of substance changes.

Exam Tip: Remember that while heat capacity depends on mass, specific heat capacity is constant for a given substance.

Question 33. What is the specific heat capacity of water in SI units?
Answer: In SI units, the specific heat capacity of water is \( 4200\text{ J kg}^{-1}\text{ K}^{-1} \).
In simple words: The specific heat capacity of water is 4200 Joules per kilogram per Kelvin.

Exam Tip: This high specific heat capacity makes water a very useful coolant and a major climate regulator.

Question 34. Which common liquid has an exceptionally high specific heat capacity, even higher than water?
Answer: Among common chemical compounds, liquid ammonia possesses the highest specific heat capacity.
In simple words: Liquid ammonia has the highest specific heat capacity of any common substance.

Exam Tip: Water has the highest specific heat capacity among common everyday liquids, but ammonia exceeds it.

Question 35. On what factors does the amount of heat energy absorbed or released by a body during heating or cooling depend?
Answer: The quantity of heat gained or lost by a body is determined by:
1. The total mass of the substance (\( m \)).
2. The specific heat capacity of the material (\( c \)).
3. The change in temperature (\( \Delta T \)).
In simple words: The amount of heat gained or lost depends on how heavy the object is, what it is made of, and how much its temperature changes.

Exam Tip: This is represented by the formula \( Q = m \times c \times \Delta T \), which shows the direct dependency on mass, material nature, and temperature difference.

Question 36. Explain why oceans are regarded as vast reservoirs or storehouses of heat energy.
Answer: Since oceans cover over 70% of the Earth's surface, they act as massive collectors of solar radiation. The surface water absorbs and stores immense amounts of solar thermal energy, while the deeper water remains cooler. This massive absorption, combined with water's exceptionally high specific heat capacity, makes the oceans the largest storehouses of heat energy on the planet.
In simple words: Oceans are huge heat storehouses because they cover most of the Earth and absorb massive amounts of sun's heat, keeping it stored due to water's high heat capacity.

Exam Tip: Relate this storing capacity directly to the massive surface area of the oceans and the high specific heat capacity of water.

Question 37. Why is water used as a cooling agent in the radiators of automobiles?
Answer: Water is used as an effective coolant in car radiators due to its high specific heat capacity. This allows it to absorb a large quantity of heat energy from the engine while experiencing only a small increase in its own temperature, protecting the engine from overheating.
In simple words: Water is used in radiators because it can absorb a lot of heat from the engine without getting too hot itself.

Exam Tip: Highlighting the high specific heat capacity (\( 4200\text{ J kg}^{-1}\text{ K}^{-1} \)) is the core scientific reason for this application.

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Question 38. What is meant by a change of state? Give an example.
Answer: A change of state refers to the physical transition of a substance from one state of matter to another (such as solid to liquid, liquid to gas, or solid to gas) under constant temperature. An everyday example is ice melting into water.
In simple words: Change of state is when a material changes form, like solid ice melting into liquid water.

Exam Tip: Emphasize that a change of state is a physical change, meaning the chemical properties of the substance remain identical.

Question 39. Is heat energy absorbed or released when ice melts?
Answer: During the melting of ice, heat energy is absorbed from the surroundings to break the intermolecular bonds.
In simple words: Ice absorbs heat from its surroundings in order to melt.

Exam Tip: This absorbed heat is the latent heat of fusion, which does not cause a temperature rise.

Question 40. Is heat energy absorbed or released when water freezes to form ice?
Answer: During the freezing process of water, heat energy is released into the surroundings.
In simple words: Water releases heat into the environment when it turns into ice.

Exam Tip: This released heat is the latent heat of solidification, which matches the latent heat of fusion.

Question 41. How does the temperature change when ice melts at \( 0^\circ\text{C} \)?
Answer: The temperature remains completely constant at \( 0^\circ\text{C} \) until all the ice has fully transitioned into water.
In simple words: The temperature stays exactly at 0°C while ice is melting.

Exam Tip: Be sure to mention that temperature only starts rising after the change of state is complete.

Question 42. Explain why the temperature of a solid remains constant at its melting point while it is melting.
Answer: In a solid, molecules are bound together by strong intermolecular forces. To melt the solid, these bonds must be broken. At the melting point, all the supplied thermal energy is used exclusively to do work against these molecular bonds rather than increasing the kinetic energy of the molecules. Consequently, the average kinetic energy remains unchanged, which explains why the temperature stays constant during the entire melting process.
In simple words: While a solid is melting, the heat supplied is used solely to break the bonds holding the molecules together, so the temperature does not rise.

Exam Tip: Explain that temperature is a measure of average kinetic energy, which does not increase since the heat is used only to break intermolecular bonds (potential energy).

Question 43. Explain why the temperature of a boiling liquid remains constant until all of it has converted to vapour.
Answer: During a phase change like boiling, the absorbed heat energy is utilized solely to overcome the intermolecular attractive forces between liquid molecules to convert them into gas. Because this energy is consumed in breaking these molecular interactions, the temperature stays constant. Once the entire liquid has evaporated, any further heat input will begin raising the temperature of the vapour.
In simple words: During boiling, all the heat goes into turning the liquid molecules into gas. The temperature only starts rising again after everything has turned to steam.

Exam Tip: Emphasize that the supplied heat changes the potential energy of the molecules rather than their kinetic energy.

Question 44. Why does the weather become warm during snowfall, but extremely cold when the snow starts to melt?
Answer: During a snowfall, water vapor in the atmosphere freezes to form snow, releasing \( 336,000\text{ J} \) of latent heat of fusion per kilogram into the air. This release of heat makes the weather feel warm. Conversely, when the snow begins to melt, each kilogram of ice absorbs \( 336,000\text{ J} \) of heat from the surrounding atmosphere, causing the temperature of the air to drop drastically.
In simple words: When snow forms, it releases heat into the air, making it warmer. When snow melts, it absorbs heat from the air, making the weather feel freezing cold.

Exam Tip: Clearly explain both processes: the release of latent heat during freezing (warming) and its absorption during melting (cooling).

Question 45. Calculate the amount of heat energy required to completely convert \( 2\text{ g} \) of ice at \( 0^\circ\text{C} \) into water at the same temperature. (Take the specific latent heat of fusion of ice as \( 336\text{ J g}^{-1} \)).
Answer: Given:
- Mass of ice, \( m = 2\text{ g} \)
- Specific latent heat of fusion of ice, \( L = 336\text{ J g}^{-1} \)

Using the latent heat formula:
\[ Q = m \times L \]
\[ Q = 2 \times 336 = 672\text{ J} \]
Thus, the total heat energy required is \( 672\text{ J} \).
In simple words: We multiply the mass of 2 grams by the latent heat of 336 Joules per gram to get the total heat needed, which is 672 Joules.

Exam Tip: Since the temperature is constant at 0°C, we only use the latent heat formula \( Q = mL \) and do not include any temperature change terms.

Question 46. Calculate the amount of heat energy required to convert \( 100\text{ g} \) of water at \( 100^\circ\text{C} \) into steam at \( 100^\circ\text{C} \). (Take specific latent heat of vaporization of water as \( 540\text{ cal g}^{-1} \)).
Answer: Given:
- Mass of water, \( m = 100\text{ g} \)
- Specific latent heat of vaporization of water, \( L = 540\text{ cal g}^{-1} \)

Using the latent heat formula:
\[ Q = m \times L \]
\[ Q = 100 \times 540 = 54000\text{ cal} = 54\text{ kcal} \]
Thus, the total heat energy required is \( 54\text{ kcal} \).
In simple words: We multiply the mass of 100 grams by the latent heat of 540 calories per gram to get the total heat needed, which is 54 kilocalories.

Exam Tip: Always double-check your units (calories vs Joules) to ensure your final answer has the correct corresponding unit (kcal or kJ).

Question 47. Explain why it is possible to skate on ice, but not on a sheet of glass.
Answer: When the sharp steel blades of ice skates press down on ice, they exert high pressure. This increased pressure lowers the melting point of the ice, causing it to melt beneath the blades. The resulting thin film of water acts as an excellent lubricant, drastically reducing friction and allowing the skates to slide smoothly. On the other hand, applying pressure to glass does not cause it to melt or form a lubricating layer, making skating on glass impossible.
In simple words: Skates apply high pressure that melts the ice into a thin layer of water. This water acts like oil to let you slide smoothly, which does not happen on glass.

Exam Tip: Mention the lowering of the melting point under pressure (regelation) and how the water acts as a lubricant to write a complete answer.

Question 48. Why are bottled drinks cooled more effectively by surrounding them with lumps of ice at \( 0^\circ\text{C} \) than with cold water at the same temperature?
Answer: Lumps of ice at \( 0^\circ\text{C} \) cool drinks much more effectively than water at \( 0^\circ\text{C} \) because every gram of ice absorbs an additional \( 336\text{ J} \) of heat (specific latent heat of fusion) from the drinks to melt into water. Water at \( 0^\circ\text{C} \) only absorbs heat through a temperature rise, thereby taking away much less heat from the bottles initially.
In simple words: Ice at 0°C needs to absorb a lot of extra heat (336 Joules per gram) just to melt into water, which cools the drinks much faster than water that is already melted.

Exam Tip: Always focus on "latent heat of fusion" when comparing the cooling effect of ice and water at the same temperature.

Question 49. With reference to a heating curve of ice, which parts of the curve correspond to the substance existing in two states simultaneously?
Answer: The horizontal flat sections of the heating curve, which represent the phase transition periods (melting and boiling), correspond to the substance existing in two states simultaneously.
In simple words: During melting (solid and liquid) and boiling (liquid and gas), the substance exists in two states at the same time.

Exam Tip: Identify that the temperature remains constant during these phase transition segments.

Question 50. What do the first and second horizontal flat regions of a temperature-time graph of a heated solid represent?
Answer: The first flat region where the temperature remains constant indicates the phase change from solid to liquid (melting). The second flat region of constant temperature represents the phase change from liquid to vapour (boiling).
In simple words: The first flat line shows the solid turning into liquid, and the second flat line shows the liquid turning into steam.

Exam Tip: Label these flat regions as "latent heat of fusion" and "latent heat of vaporization" respectively to show deep conceptual understanding.

Question 51. A cooling curve of a substance is plotted.
(a) What is its boiling point?
(b) What does the region DE represent?
(c) What is the melting point of the substance?
Answer:
(a) The boiling point of the substance is \( 150^\circ\text{C} \), as the flat region BC represents condensation where vapour transitions into liquid at a constant temperature.
(b) The region DE represents the freezing of the substance, where the liquid transitions into a solid state at a constant temperature of \( 100^\circ\text{C} \).
(c) The melting point of the substance is \( 100^\circ\text{C} \), which corresponds to the constant temperature of the freezing region DE.
In simple words: (a) The substance condenses at 150°C. (b) Part DE shows the liquid freezing into a solid. (c) The melting (and freezing) point of this substance is 100°C.

Exam Tip: Remember that the melting point of a pure substance is identical to its freezing point under the same pressure.

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Question 52. Draw a labeled temperature-time graph (heating curve) showing the change of state of ice at \( -10^\circ\text{C} \) to steam at \( 100^\circ\text{C} \) or above.
Answer: The temperature-time heating curve of ice is shown below:

In simple words: The graph has two flat lines showing when ice melts at 0°C and when water boils at 100°C because the temperature stays constant during these phase changes.

Exam Tip: When drawing this heating curve, remember to show two distinct flat regions at \( 0^\circ\text{C} \) and \( 100^\circ\text{C} \).

Question 53. Based on the heating curve of ice, explain:
(i) What does part AB represent?
(ii) What does part CD represent?
(iii) Describe the temperature change when ice at \( -10^\circ\text{C} \) is heated until it melts completely.
Answer:
(i) Part AB corresponds to the phase change from solid to liquid, indicating that ice is melting at \( 0^\circ\text{C} \).
(ii) Part CD corresponds to the phase change from liquid to gas, indicating that water is boiling at \( 100^\circ\text{C} \).
(iii) Initially, the ice is in a solid state at \( -10^\circ\text{C} \). Upon heating, its temperature rises steadily to \( 0^\circ\text{C} \). At this temperature, it absorbs latent heat of fusion to completely melt into water at \( 0^\circ\text{C} \).
In simple words: (i) AB is ice melting at 0°C. (ii) CD is water boiling at 100°C. (iii) Ice starts at -10°C, warms to 0°C, and then melts into water at 0°C.

Exam Tip: Clearly link each line segment (AB and CD) to its corresponding physical phase change to present a well-structured answer.