Maharashtra Board Class 11 Physics Chapter 7 Thermal Properties of Matter Solutions

1. Choose The Correct Option.

Question 1. Range of temperature in a clinical thermometer, which measures the temperature of human body, is
(A) 70 °C to 100 °C
(B) 34 °C to 42 °C
(C) 0 °F to 100 °F
(D) 34 °F to 80 °F
Answer: (B) 34 °C to 42 °C
In simple words: Clinical thermometers are designed for human body temperature, which typically ranges from 34°C to 42°C. This specific range is vital for accurate medical readings.

🎯 Exam Tip: Remember the specific temperature range for clinical thermometers as it's a common fact-based question.

Question 2. A glass bottle completely filled with water is kept in the freezer. Why does it crack?
(A) Bottle gets contracted
(B) Bottle is expanded
(C) Water expands on freezing
(D) Water contracts on freezing
Answer: (C) Water expands on freezing
In simple words: Unlike most liquids, water expands when it freezes, increasing its volume. If a bottle is completely full, this expansion exerts immense pressure, causing the glass to crack.

🎯 Exam Tip: Understanding the anomalous expansion of water upon freezing is key to solving this type of problem.

Question 3. If two temperatures differ by 25 °C on Celsius scale, the difference in temperature on Fahrenheit scale is
(A) 65°
(B) 45°
(C) 38°
(D) 25°
Answer: (B) 45°
In simple words: A temperature difference of 25°C is equivalent to a difference of 45°F because the Fahrenheit scale has 1.8 times more divisions than the Celsius scale for the same temperature range.

🎯 Exam Tip: To convert a temperature difference from Celsius to Fahrenheit, multiply the Celsius difference by 9/5 (or 1.8).

Question 4. If \( \alpha \), \( \beta \) and \( \gamma \) are coefficients of linear, area I and volume expansion of a solid then
(A) \( \alpha:\beta:\gamma \) 1:3:2
(B) \( \alpha:\beta:\gamma \) 1:2:3
(C) \( \alpha:\beta:\gamma \) 2:3:1
(D) \( \alpha:\beta:\gamma \) 3:1:2
Answer: (B) \( \alpha:\beta:\gamma \) 1:2:3
In simple words: For an isotropic solid, the coefficients of linear, area, and volume expansion are in the ratio 1:2:3 respectively, meaning \( \beta = 2\alpha \) and \( \gamma = 3\alpha \).

🎯 Exam Tip: Remember the fundamental relationships between the coefficients of thermal expansion: \( \beta = 2\alpha \) and \( \gamma = 3\alpha \) are crucial for many problems.

Question 5. Consider the following statements-
(I) The coefficient of linear expansion has dimension K^-1
(II) The coefficient of volume expansion has dimension K^-1
(A) I and II are both correct
(B) I is correct but II is wrong
(C) II is correct but I is wrong
(D) I and II are both wrong
Answer: (A) I and II are both correct
In simple words: Both linear and volume expansion coefficients represent a fractional change per unit temperature, so their dimensions are inverse of temperature, typically K^-1 or °C^-1.

🎯 Exam Tip: The dimension of any thermal expansion coefficient is always the inverse of temperature, irrespective of whether it's linear, area, or volume.

Question 6. Water falls from a height of 200 m. What is the difference in temperature between the water at the top and bottom of a water fall given that specific heat of water is 4200 J kg^-1 °C^-1?
(A) 0.96 °C
(B) 1.02 °C
(C) 0.46 °C
(D) 1.16 °C
Answer: (C) 0.46 °C
In simple words: As water falls, its potential energy converts into thermal energy, causing a slight increase in its temperature at the bottom of the waterfall. This temperature rise can be calculated using the principles of energy conservation and specific heat capacity.

🎯 Exam Tip: For problems involving energy conversion to heat, use the formula \( mgh = mc\Delta T \), where potential energy becomes thermal energy.

2. Answer The Following Questions.

Question 1. Clearly state the difference between heat and temperature?
Answer:

In simple words: Heat is the energy flowing between objects due to a temperature difference, while temperature is a measure of the average kinetic energy of particles within an object, indicating its hotness or coldness.

🎯 Exam Tip: Clearly distinguishing between heat as energy transfer and temperature as a state property is crucial for conceptual clarity and scoring well.

Question 2. How a thermometer is calibrated?
Answer: 1. For the calibration of a thermometer, a standard temperature interval is selected between two easily reproducible fixed temperatures. 2. The fact that substances change state from solid to liquid to gas at fixed temperatures is used to define reference temperature called fixed point. 3. The two fixed temperatures selected for this purpose are the melting point of ice or freezing point of water and the boiling point of water. 4. This standard temperature interval is divided into sub-intervals by utilizing some physical property that changes with temperature. 5. Each sub-interval is called as a degree of temperature. Thus, an empirical scale for temperature is set up. In simple words: A thermometer is calibrated by using two fixed points, like the freezing and boiling points of water, and then dividing the interval between them into uniform smaller units called degrees.

🎯 Exam Tip: Highlight the use of reproducible fixed points (ice point and steam point) and the division of the interval into degrees as key steps in thermometer calibration.

Question 3. What are different scales of temperature? What is the relation between them?
Answer: 1. Celsius scale: • The ice point (melting point of pure ice) is marked as 0 °C (lower point) and steam point (boiling point of water) is marked as 100 °C (higher point). • Both are taken at one atmospheric pressure. • The interval between these points is divided into two equal parts. Each of these parts is called as one degree celsius and it is 'written as 1 °C. 2. Fahrenheit scale: • The ice point (melting point of pure ice) is marked as 32 °F and steam point (boiling point of water) is marked as 212 °F. • The interval between these two reference points is divided into 180 equal parts. Each part is called as degree fahrenheit and is written as 1 °F. 3. Kelvin scale: • The temperature scale that has its zero at -273.15 °C and temperature intervals are same as that on the Celsius scale is called as kelvin scale or absolute scale. • The absolute temperature, T and celsius temperature, T_c are related as, \( T = T_c + 273.15 \)
eg.: when \( T_c = 27 \) °C,
\( T = 27+273.15 \) K = 300.15 K

Relation between different scales of temperature: \[ \frac{T_F-32}{180} = \frac{T_C-0}{100} = \frac{T_K-273.15}{100} \] where,
\( T_F \) = temperature in fahrenheit scale,
\( T_C \) = temperature in celsius scale,
\( T_K \) = temperature in kelvin scale,
[Note: At zero of the kelvin scale, every substance in nature has the least possible activity.] In simple words: The main temperature scales are Celsius, Fahrenheit, and Kelvin, each defined by different reference points. They are related by specific conversion formulas that allow transformation between them.

🎯 Exam Tip: Memorize the conversion formula relating Celsius, Fahrenheit, and Kelvin scales, and understand the definition of each scale's fixed points.

Question 4. What is absolute zero?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गैसों के लिए दाब-ताप और आयतन-ताप ग्राफ़ दर्शाता है। चित्र (a) में, नियत आयतन पर तीन आदर्श गैसों (A, B, C) के लिए दाब (P) को तापमान (T) के विरुद्ध प्लॉट किया गया है, जिसमें सीधी रेखाएँ दिखाई देती हैं जो -273.15°C पर तापमान अक्ष को काटती हैं। चित्र (b) में, नियत दाब पर तीन आदर्श गैसों (A, B, C) के लिए आयतन (V) को तापमान (T) के विरुद्ध प्लॉट किया गया है, जो भी सीधी रेखाएँ हैं और -273.15°C पर तापमान अक्ष पर प्रतिच्छेद करती हैं। 1. When the graph of pressure (P) against temperature T (°C) at constant volume for three ideal gases A, B and C is plotted, in each case, P -T graph is straight line indicating direct proportion between them. The slopes of these graphs are different. 2. The individual straight lines intersect the pressure axis at different values of pressure at 0 °C. but each line intersects the temperature axis at the same point, i.e., at absolute temperature (-273.15 °C). 3. Similarly graph at constant pressure for three different ideal gases A, B and C extrapolate to the same temperature intercept -273.15 °C i.e., absolute zero temperature. 4. It is seen that all the lines for different gases Cut the temperature axis at the same point at -273.15 °C. 5. This point is termed as the absolute zero of temperature. 6. It is not possible to attain a temperature lower than this value. Even to achieve absolute zero temperature is not possible in practice. [Note: The point of zero pressure or zero volume does not depend on am specific gas.] In simple words: Absolute zero is the lowest possible temperature (-273.15 °C or 0 Kelvin) at which all molecular motion ceases, and a substance would ideally have zero pressure and volume according to gas laws.

🎯 Exam Tip: Clearly define absolute zero in terms of temperature and its implications for molecular activity, referencing the extrapolation of gas laws.

Question 5. Derive the relation between three coefficients of thermal expansion.
Answer: Consider a square plate of side l₀ at 0 °C and l_T at T °C. 1. \( l_T = l_0 (1 + \alpha T) \) If area of plate at 0 °C is A₀, \( A_0 = l_0^2 \) If area of plate at T °C is A_T,
\( A_T = l_T^2 = l_0^2 (1 + \alpha T)^2 \)
or \( A_T = A_0 (1 + \alpha T)^2 ............... (1) \) Also,
\( A_T = A_0(1 + \beta T) ................ (2) \)
\[ \therefore \beta = \frac{A_T-A_0}{A_0(T-T_0)} \] 2. Using Equations (1) and (2),
\( A_0 (1 + \alpha T)^2 = A_0(1 + \beta T) \)
\( \therefore 1 + 2\alpha T + \alpha^2 T^2 = 1 + \beta T \) 3. Since the values of \( \alpha \) are very small, the term \( \alpha^2 T^2 \) is very small and may be neglected,
\( \therefore \beta = 2\alpha \) 4. The result is general because any solid can be regarded as a collection of small squares.

Relation between coefficient of linear expansion (\( \alpha \)) and coefficient of cubical expansion (\( \gamma \)). 1. Consider a cube of side l₀ at 0 °C and l_T at T °C.
\( \therefore l_T = l_0(1 + \alpha T) \) If volume of the cube at 0 °C is V₀, \( V_0 = l_0^3 \) If volume of the cube at T °C is
V_T, \( V_T = l_T^3 = l_0^3 (1 + \alpha T)^3 \)
\( V_T = V_0 (1 + \alpha T)^3 ...... (1) \) Also,
\( V_T = V_0(1 + \gamma T) ............. (2) \)
\[ \therefore \gamma = \frac{V_T-V_0}{V_0(T-T_0)} \] 2. Using Equations (1) and (2),
\( V_0(1 + \alpha T)^3 = V_0(1 + \gamma T) \)
\( \therefore 1 + 3\alpha T + 3\alpha^2 T^2 + \alpha^3 T^3 = 1 + \gamma T \) 3. Since the values of \( \alpha \) are very small, the terms with higher powers of \( \alpha \) may be neglected,
\( \therefore \gamma = 3\alpha \) 4. The result is general because any solid can be regarded as a collection of small cubes.

Relation between \( \alpha \), \( \beta \) and \( \gamma \) is given by,
\[ \alpha = \frac{\beta}{2} = \frac{\gamma}{3} \] where, \( \alpha \) = coefficient of linear expansion.
\( \beta \) = coefficient of superficial expansion,
\( \gamma \) = coefficient of cubical expansion. In simple words: The coefficients of linear, area, and volume expansion for a solid are related as \( \alpha : \beta : \gamma = 1 : 2 : 3 \), meaning \( \beta = 2\alpha \) and \( \gamma = 3\alpha \), which are derived by considering how a small change in length affects area and volume.

🎯 Exam Tip: The derivations for \( \beta = 2\alpha \) and \( \gamma = 3\alpha \) are important; practice simplifying the terms with higher powers of \( \alpha \) by neglecting them due to their small magnitude.

Question 6. State applications of thermal expansion.
Answer: Applications of thermal expansion: • The steel wheel is heated to expand. This expanded wheel can easily fit over axle. The wheel is then cooled quickly. Upon cooling, wheel contracts and fits tightly upon the axle. • An electric light bulb gets hot quickly when in use. The wire leads to the filament are sealed into the glass. If the glass of the bulbs has significantly different thermal expansivity from the wire leads, the glass and the wire would separate, breaking down the vacuum. To prevent this, wires are made of platinum or some suitable alloy with the same expansivity as ordinary glass. In simple words: Thermal expansion is used in practical applications like fitting steel wheels onto axles (by heating the wheel to expand it, then cooling it to contract onto the axle) and in designing materials for light bulbs to ensure wire leads and glass expand similarly, preventing damage.

🎯 Exam Tip: Provide at least two distinct practical examples of thermal expansion, ensuring the explanation for each is clear and concise.

Question 7. Why do we generally consider two specific heats for a gas?
Answer: • A slight change in temperature causes considerable change in pressure as well as volume of the gas. • Therefore, two principal specific heats are defined for a gas viz., specific heat capacity at constant volume (S_v) and specific heat capacity at constant pressure (S_p). In simple words: We consider two specific heats for a gas (at constant volume and constant pressure) because a gas's volume and pressure change significantly with temperature, unlike solids or liquids, requiring different amounts of heat to raise its temperature under these varying conditions.

🎯 Exam Tip: Emphasize that the large changes in volume and pressure of a gas distinguish it from solids/liquids, necessitating two specific heat capacities.

Question 8. Are freezing point and melting point same with respect to change of state ? Comment.
Answer: Though freezing point and melting point mark same temperature (0°C or 32° F), state of change is different for the two points. At freezing point liquid gets converted into solid, whereas at melting point solid gets converted into its liquid state. In simple words: Freezing and melting points occur at the same temperature for a given substance, but they represent opposite phase changes: freezing is liquid to solid, while melting is solid to liquid.

🎯 Exam Tip: While the temperature is the same, clearly state the distinct direction of phase transition (liquid to solid vs. solid to liquid) for freezing and melting points.

Question 9. Define
(i) Sublimation
(ii) Triple point.
Answer: 1. The change from solid state to vapour stale without passing through the liquid state is called sublimation and the substance is said to sublime. Examples: Dry ice (solid CO₂) and iodine. 2. The triple point of water is that point where water in a solid, liquid and gas state co-exists in equilibrium and this occurs only at a unique temperature and a pressure. In simple words: Sublimation is the direct transition from solid to gas, bypassing the liquid phase (like dry ice). The triple point is a specific temperature and pressure where a substance can exist simultaneously as a solid, liquid, and gas in equilibrium.

🎯 Exam Tip: For definitions, provide a clear, concise statement and, where applicable, relevant examples (for sublimation) or conditions (for triple point).

Question 10. Explain the term 'steady state'.
Answer: 1. When one end of a metal rod is heated, the heat flows by conduction from hot end to the cold end.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक धातु की छड़ को दर्शाता है जिसमें ऊष्मा गर्म सिरे (T1) से ठंडे सिरे (T2) की ओर प्रवाहित हो रही है, जिससे छड़ के भीतर विभिन्न अनुभागों में एक स्थिर तापमान प्रवणता (steady state temperature gradient) स्थापित होती है। छड़ के अनुप्रस्थ काट पर एक 'X' चिह्न दिखाया गया है। 2. As a result, the temperature of every section of the rod starts increasing. 3. Under this condition, the rod is said to be in a variable temperature state. 4. After some time, the temperature at each section of the rod becomes steady i.e., does not change. 5. Temperature of each cross-section of the rod now becomes constant though not the same. This is called steady state condition. In simple words: In a steady state for heat conduction, the temperature at every point within a material remains constant over time, even though there might be a temperature difference across the material, causing a continuous flow of heat.

🎯 Exam Tip: Emphasize that in a steady state, temperatures at points are constant, but not necessarily equal across the body, and there is a continuous heat flow.

Question 11. Define coefficient of thermal conductivity. Derive its expression.
Answer: Coefficient of thermal conductivity of a material is defined as the quantity of heat that flows in one second between the opposite faces of a cube of side 1 m, the faces being kept at a temperature difference of 1°C (or 1 K).

Expression for coefficient of thermal conductivity: 1. Under steady state condition, the quantity of heat 'Q' that flows from the hot face at temperature T₁ to the cold face at temperature T₂ of a cube with side x and area of cross- section A is • directly proportional to the cross-sectional area A of the face. i.e.. \( Q \propto A \) • directly proportional to the temperature difference between the two faces i.e., \( Q \propto (T_1 - T_2) \) • directly proportional to time t (in seconds) for which heat flows i.e.. \( Q \propto t \) • inversely proportional to the perpendicular distance x between hot and cold faces i.e., \( Q \propto 1/x \) 2. Combining the above four factors, we have the quantity of heat
\[ Q \propto \frac{A(T_1-T_2)t}{x} \] \[ \therefore Q = \frac{kA(T_1-T_2)t}{x} \] where k is a constant of proportionality and is called coefficient of thermal conductivity. Its value depends upon the nature of the material. In simple words: Thermal conductivity (k) measures a material's ability to conduct heat, defined as the amount of heat flowing through a unit area per second for a unit temperature gradient. It's derived by combining the proportionalities of heat flow with area, temperature difference, time, and inversely with thickness.

🎯 Exam Tip: Clearly state the definition and ensure all four proportionalities (A, \( \Delta T \), t, \( 1/x \)) are correctly combined to derive the expression for 'k'.

Question 12. Give any four applications of thermal conductivity in every day life.
Answer: Applications of thermal conductivity: 1. Thick walls are used in the construction of cold storage rooms. • Brick being a bad conductor of heat is used to reduce the flow of heat from the surroundings to the rooms. • Better heat insulation is obtained by using hollow bricks. • Air being a poorer conductor than a brick, it further avoids the conduction of heat from outside. 2. Street vendors keep ice blocks packed in saw dust to prevent them from melting rapidly. 3. The handle of a cooking utensil is made of a bad conductor of heat, such as ebonite, to protect our hand from the hot utensil. 4. Two bedsheets used together to cover the body help retain body heat better than a single bedsheet of double the thickness. Trapped air being a bad conductor of heat, the layer of air between the two sheets reduces thermal conduction better than a sheet of double the thickness. Similarly, a blanket coupled with a bedsheet is a cheaper alternative to using two blankets. In simple words: Thermal conductivity applications include using thick or hollow walls and sawdust for insulation (poor conductors), making utensil handles from bad conductors like ebonite, and using multiple thin blankets for better warmth (trapped air).

🎯 Exam Tip: Provide diverse examples that clearly illustrate how both good and bad thermal conductors are utilized in everyday life.

Question 13. Explain the term thermal resistance. State its Sl unit and dimensions.
Answer: 1. Consider expression for conduction rate,
\[ P_{cond} = kA \frac{(T_1-T_2)}{X} \] \[ \implies \frac{P_{cond}}{T_1-T_2} = \frac{kA}{X} ............... (1) \] 2. Ratio \( \frac{T_1-T_2}{P_{cond}} \) is called as thermal resistance (R_T) of material.

The Sl unit of thermal resistance is °C s/kcal or °C s/J and its dimensional formula is \( [L^{-2}M^{-1}T^3K^1] \). In simple words: Thermal resistance (R_T) is a measure of how effectively a material opposes the flow of heat, defined as the temperature difference divided by the rate of heat flow. Its SI unit is °C s/J and its dimension is \( [L^{-2}M^{-1}T^3K^1] \).

🎯 Exam Tip: Define thermal resistance clearly as the reciprocal of thermal conductance and provide its correct SI unit and dimensional formula for full marks.

Question 14. How heat transfer occurs through radiation in absence of a medium?
Answer: 1. All objects possess thermal energy due to their temperature T (T > 0 K). 2. The rapidly moving molecules of a hot body emit EM waves travelling with the velocity of light. These are called thermal radiations. 3. These carry energy with them and transfer it to the low-speed molecules of a cold body on which they fall. 4. This results in an increase in the molecular motion of the cold body and its temperature rises. 5. Thus transfer of heat by radiation is a two fold process-the conversion of thermal energy into waves and reconversion of waves into thermal energy by the body on which they fall. In simple words: Heat transfer by radiation occurs through electromagnetic waves emitted by hot objects, which travel through space (even a vacuum) at the speed of light and are absorbed by colder objects, increasing their thermal energy.

🎯 Exam Tip: Highlight that radiation involves electromagnetic waves, does not require a medium, and is a two-step process of emission and absorption for heat transfer.

Question 15. State Newton's law of cooling and explain how it can be experimentally verified.
Answer: The rate of loss of heat \( dT/dt \) of the both' is directly proportional to the difference of temperature \( (T – T_0) \) of the body and the surroundings provided the difference in temperatures is small.

Mathematically, Newton's law of cooling can be expressed as:
\[ \frac{dT}{dt} \propto (T - T_0) \]
\[ \therefore \frac{dT}{dt} = C(T - T_0) \] where, C is constant of proportionality. Experimental verification of Newton's law of cooling: 1. Fill a calorimeter upto \( \frac{2}{3} \) of its capacity with a boiling water. Cover it with lid with a hole for passing the thermometer. 2. Insert the thermometer through the hole and adjust it so that the bulb of the thermometer is fully immersed in hot water. 3. Keep calorimeter vessel in constant temperature enclosure or just in open air since room temperature will not change much in the experiment. 4. Note down the temperature (T) on the thermometer at every one minute interval until the temperature of water decreases by about 25 °C. 5. Plot a graph of temperature (T) on Y-axis against time (t) on X-axis. This graph is called cooling curve as shown in figure (a).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र तापमान (Y-अक्ष) बनाम समय (X-अक्ष) का एक शीतलन वक्र (cooling curve) दर्शाता है, जिसमें गर्म वस्तु का तापमान समय के साथ धीरे-धीरे घटता जाता है। वक्र पर एक बिंदु पर ढलान \( \Delta T / \Delta t \) के रूप में दिखाया गया है, जो शीतलन की दर को दर्शाता है। 6. Draw tangents to the curve at suitable points on the curve. The slope of each tangent is \( \lim_{\Delta t \to 0} \frac{\Delta T}{\Delta t} \) and gives the rate of fall of temperature at that temperature (T). 7. Now the graph of \( \frac{dT}{dt} \) on Y-axis against \( (T – T_0) \) on X-axis is plotted with (0, 0) origin. The graph is straight line and passes through origin as shown in figure (b), which verities Newton's law of cooling. In simple words: Newton's Law of Cooling states that an object's cooling rate is proportional to the temperature difference between it and its surroundings. Experimentally, this is verified by plotting a cooling curve and then plotting the rate of cooling against the temperature difference, which should yield a straight line.

🎯 Exam Tip: State the law clearly with its mathematical form. For verification, describe the two graphs (Temperature vs. time, and Rate of cooling vs. Temperature difference) and their expected shapes.

Question 16. What is thermal stress? Give an example of disadvantages of thermal stress in practical use?
Answer: 1. Consider a metallic rod of length l₀ fixed between two rigid supports at T °C. If the temperature of rod is increased by \( \Delta T \), length of rod would become,
\( l = l_0(1 + \alpha \Delta T) \) Where, \( \alpha \) is the coefficient of linear expansion of material of the rod. But the supports prevent expansion of rod. As a result, rod exerts stress on the supports. Such stress is termed as thermal stress. 2. Disadvantage: Thermal stress can lead to fracture or deformation in substance under certain conditions. 3. Railway tracks are made up of metals which expand upon heating. If no gap is kept between tracks, in hot weather, expansion of metal tracks may exert thermal stress on track. This may lead to bending of tracks which would be dangerous. Hence, railway track is not a continuous piece but is made up of segments separated by gaps. In simple words: Thermal stress is the internal stress developed in a material when its expansion or contraction due to temperature change is constrained. A disadvantage is that in railway tracks, if no gaps are left for thermal expansion, the stress can cause tracks to buckle, leading to dangerous deformations.

🎯 Exam Tip: Define thermal stress by explaining the constraint on expansion/contraction. Use the railway track example to clearly illustrate a practical disadvantage.

Question 17. Which materials can be used as thermal insulators and why?
Answer: 1. Substances such as glass, wood, rubber, plastic, etc. can be used as thermal insulators. 2. These substances do not have free electrons to conduct heat freely throughout the body. Hence, they are poor conductors of heat. In simple words: Materials like glass, wood, rubber, and plastic are good thermal insulators because they lack free electrons, which are necessary for efficient heat conduction, thus impeding heat transfer.

🎯 Exam Tip: Name common thermal insulators and concisely explain *why* they are effective (lack of free electrons).

3. Solve The Following Problems.

Question 1. A glass flask has volume \( 1 \times 10^{-4} \) m³. It is filled with a liquid at 30 °C. If the temperature of the system is raised to 100 °C, how much of the liquid will overflow. (Coefficient of volume expansion of glass is \( 1.2 \times 10^{-5} \) (°C)^-1 while that of the liquid is \( 75 \times 10^{-5} \) °C^-1)
Solution: Given: \( V_1 = 1 \times 10^{-4} \) m³ \( = 10^{-4} \) m³, \( T_1 = 30 \)°C,
\( T_2 = 100 \) °C
To find: Volume of liquid that overflows
Formula: \( \gamma = \frac{V_2-V_1}{V_1(T_2-T_1)} \)
Calculation: From formula,
Increase in volume \( = V_2 - V_1 \)
\( = \gamma V_1(T_2 – T_1) \)
increase in volume of beaker
\( = \gamma_{glass} \times V_1 (T_2 – T_1) \)
\( = 1.2 \times 10^{-5} \times 10^{-4} \times (100 – 30) \)
\( = 1.2 \times 70 \times 10^{-9} \)
\( = 84 \times 10^{-9} \) m³
\( \therefore \) Increase in volume of beaker
\( = 84 \times 10^{-9} \) m³
Increase in volume of liquid
\( = \gamma_{liquid} \times V_1 (T_2-T_1) \)
\( = 75 \times 10^{-5} \times 10^{-4} \times (100 – 30) \)
\( = 75 \times 70 \times 10^{-9} \)
\( = 5250 \times 10^{-9} \) m³
\( \therefore \) Increase in volume of liquid \( = 5250 \times 10^{-9} \) m³
\( \therefore \) Volume of liquid which overflows
\( = (5250 – 84) \times 10^{-9} \) m³
\( = 5166 \times 10^{-9} \) m³
\( = 0.5166 \times 10^{-7} \) m³
Volume of liquid that overflows is \( 0.5166 \times 10^{-7} \) m³.
[Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]
In simple words: The overflow volume is the difference between the liquid's expansion and the flask's expansion when heated. Both expand, but since the liquid's coefficient of expansion is much higher, a net overflow occurs.

🎯 Exam Tip: Remember to calculate the volume expansion for *both* the liquid and the container, then find the difference to determine the overflow.

Question 2. Which will require more energy, heating a 2.0 kg block of lead by 30 K or heating a 4.0 kg block of copper by 5 K? (S_lead \( = 128 \) J kg^-1 K^-1, S_copper \( = 387 \) J kg^-1 K^-1)
Solution: Given: m_lead \( = 2 \) kg, \( \Delta T_{lead} = 30 \) K,
S_lead \( = 128 \) J/kg K,
m_Cu \( = 4 \) kg, \( \Delta T_{Cu} = 5 \) K,
S_Cu \( = 387 \) J/kg K
To find: Substance requiring more heat energy.
Formula: \( Q = ms \Delta T \)
Calculation: From formula,
For lead, \( Q_{lead} = 2 \times 128 \times 30 = 7680 \)J
For Copper, \( Q_{Cu} = 4 \times 387 \times 5 = 7740 \) J
\( Q_{Cu} > Q_{lead} \), copper will require more heat energy.
Copper will require more heat energy.
In simple words: By calculating the heat required for each metal using their mass, specific heat, and temperature change, we find that the copper block needs slightly more energy (7740 J) compared to the lead block (7680 J).

🎯 Exam Tip: Apply the formula \( Q = mc\Delta T \) for each substance and compare the calculated heat values carefully. Ensure units are consistent.

**Question 3.** Specific latent heat of vaporization of water is 2.26 × 10\(^6\) J/kg. Calculate the energy needed to change 5.0 g of water into steam at 100 °C.
Answer: Solution: Given: \( L_{vap} \) = 2.26 × 10\(^6\) J/kg m = 5g = 5 × 10\(^{-3}\) kg In this case, no temperature change takes place only change of state occurs. To find: Heat required to convert water into steam. Formula: Heat required = \( mL_{vap} \) Calculation: From formula, Heat required = 5 × 10\(^{-3}\) × 2.26 × 10\(^6\) = 11300J = 1.13 × 10\(^4\) J Heat required to convert water into steam is 1.13 × 10\(^4\) J [Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]
In simple words: To convert water into steam at its boiling point, we need to provide enough energy to overcome the latent heat of vaporization. This is calculated by multiplying the mass of water by its specific latent heat of vaporization, resulting in the total heat energy required for the phase change.

🎯 Exam Tip: Remember to convert mass to kilograms (kg) if the latent heat is given in J/kg. Pay attention to significant figures and standard form when presenting the final answer in numerical problems.

**Question 4.** A metal sphere cools at the rate of 0.05 °C/s when its temperature is 70°C and at the rate of 0.025 °C/s when its temperature is 50 °C. Determine the temperature of the surroundings and find the rate of cooling when the temperature of the metal sphere is 40 °C.
Answer: Solution: Given: \( T_1 \) = 70 °C, \( \left(\frac{dT}{dt}\right)_1 \) = 0.05 °C/s \( T_2 \) = 50 °C, \( \left(\frac{dT}{dt}\right)_2 \) = 0.025 °C/s \( T_3 \) = 40 °C. To find: (i) Temperature of surrounding (\( T_0 \)) (ii) Rate of cooling \( \left(\frac{dT}{dt}\right)_3 \) Formula: \( \frac{dT}{dt} \) = C (\( T - T_0 \)) Calculation: From formula, \( \left(\frac{dT}{dt}\right)_1 \) = C (\( T_1 - T_0 \)) and \( \left(\frac{dT}{dt}\right)_2 \) = C (\( T_2 - T_0 \)). \( \frac{0.05}{0.025} \) = \( \frac{C (70 - T_0)}{C (50 - T_0)} \)
\( \implies \) 2(50 - \( T_0 \)) = 70 - \( T_0 \)
\( \implies \) \( T_0 \) = 30 °C Substituting value of \( T_0 \). 0.05 = C (70 - 30)
\( \implies \) C = \( \frac{0.05}{40} \) = 0.00125/s. For \( T_3 \) = 40 °C \( \left(\frac{dT}{dt}\right)_3 \) = C(\( T_3 - T_0 \)) = 0.00125 (40 - 30) = 0.00125 × 10 = 0.0125°C/s. (i) Temperature of surrounding is 30 °C. (ii) Rate of cooling at 40 °C is 0.0125 °C/s.
In simple words: Using Newton's Law of Cooling, which states that the rate of cooling is proportional to the temperature difference between an object and its surroundings, we can calculate the unknown ambient temperature and the cooling rate at a specific object temperature by setting up simultaneous equations from the given cooling data.

🎯 Exam Tip: Newton's Law of Cooling problems often involve solving for an unknown constant of proportionality (C) and the surrounding temperature (\(T_0\)) using two given cooling rates and temperatures. Ensure consistent units throughout the calculation.

**Question 5.** The volume of a gas varied linearly with absolute temperature if its pressure is held constant. Suppose the gas does not liquefy even at very low temperatures, at what temperature the volume of the gas will be ideally zero?
Answer: At temperature of -273.15 °C, the volume of the gas will be ideally zero.
In simple words: The theoretical temperature at which the volume of an ideal gas would become zero, assuming it doesn't liquefy, is -273.15 °C, also known as absolute zero.

🎯 Exam Tip: This question refers to Charles's Law and the concept of absolute zero. Understanding the theoretical limit of temperature and its relation to gas volume is crucial for conceptual questions.

**Question 6.** In olden days, while laying the rails for trains, small gaps used to be left between the rail sections to allow for thermal expansion. Suppose the rails are laid at room temperature 27 °C. If maximum temperature in the region is 45 °C and the length of each rail section is 10 m, what should be the gap left given that \( \alpha \) = 1.2 × 10\(^{-5}\) K\(^{-1}\) for the material of the rail section?
Answer: Solution: Given: \( T_1 \) = 27 °C, \( T_2 \) = 45 °C, \( L_1 \) = 10m. \( \alpha \) = 1.2 × 10\(^{-5}\) K\(^{-1}\) To find: Gap that should be left (\( L_2 - L_1 \)) Formula: \( L_2 - L_1 \) = \( L_1 \alpha(T_2 - T_1) \) Calculation: From formula, \( L_2 - L_1 \) = 10 × 1.2 × 10\(^{-5}\) × (45 - 27) = 2.16 × 10\(^{-3}\) m = 2.16 mm The gap that should be left between rail sections is 2.16 mm.
In simple words: To prevent railway tracks from bending due to thermal expansion in hot weather, small gaps are intentionally left between sections. This required gap is calculated by considering the initial length of the rail, the material's coefficient of linear expansion, and the maximum expected temperature increase.

🎯 Exam Tip: Thermal expansion problems require careful attention to units (e.g., °C vs. K for temperature difference, meters vs. millimeters for length). The coefficient of linear expansion (α) relates directly to the change in length for a given temperature change.

**Question 7.** A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the wooden rim and the iron ring are 1.5 m and 1.47 m respectively at room temperature of 27 °C. To what temperature the iron ring should be heated so that it can fit the rim of the wheel (\( \alpha_{iron} \) = 1.2 × 10\(^{-5}\) K\(^{-1}\)).
Answer: Solution: Given: \( d_w \) = 1.5 m, \( d_i \) = 1.47 m, \( T_1 \) = 27 °C. \( \alpha_i \) = 1.2 × 10\(^{-5}\) / K To find: Temperature (\( T_2 \)) Formula: \( \alpha \) = \( \frac{d_w - d_i}{d_i(T_2 - T_1)} \) Calculation: From formula, \( T_2 \) = \( \frac{d_w - d_i}{d_i \alpha_i} \) + \( T_1 \) = \( \frac{1.5 - 1.47}{1.47 \times 1.2 \times 10^{-5}} \) + 27 = 1700.7 + 27 = 1727.7 °C Iron ring should be heated to temperature of 1727.7 °C.
In simple words: To fit a slightly smaller iron ring onto a wooden wheel, the iron ring must be heated to expand its diameter. We calculate the target temperature by using the linear expansion formula, considering the initial diameters, the desired final diameter, the initial temperature, and the iron's coefficient of linear expansion.

🎯 Exam Tip: In thermal expansion problems involving diameters or circumferences, the linear expansion formula applies directly to these dimensions. Ensure to correctly identify the initial and final dimensions and the temperature change. Units must be consistent.

**Question 8.** In a random temperature scale X, water boils at 200 °X and freezes at 20 °X. Find the boiling point of a liquid in this scale if it boils at 62 °C.
Answer: Solution: Here thermometric property P is temperature at some random scale X. Using equation, \[ T = \frac{100(P_T - P_1)}{(P_2 - P_1)} \] For \( P_1 \) = 20 °X, \( P_2 \) = 200 °X, T = 62°C
\( \implies \) 62 = \( \frac{100(P_T - 20)}{(200 - 20)} \)
\( \implies \) \( P_T \) = \( \frac{62 \times (200 - 20)}{100} \) + 20 = 111.6 + 20 = 131.6 °X The boiling point of a liquid in this scale is 131.6 °X.
In simple words: To convert a temperature from Celsius to a custom scale (X), we use a linear conversion formula that relates the known freezing and boiling points of water on both scales. By substituting the boiling point in Celsius into this formula, we can find its equivalent value on the X scale.

🎯 Exam Tip: Temperature scale conversion problems rely on the principle of linear proportionality. Remember the general formula: (Reading - Lower Fixed Point) / (Upper Fixed Point - Lower Fixed Point) = constant for all scales. Be careful with algebra during rearrangement.

**Question 9.** A gas at 900°C is cooled until both its pressure and volume are halved. Calculate its final temperature.
Answer: Solution: Given: \( T_1 \) = 900 °C = 900 + 273.15 = 1173.15 K \( V_2 \) = \( \frac{V_1}{2} \), \( P_2 \) = \( \frac{P_1}{2} \) To find: Final temperature (\( T_2 \)) Formula: \( \frac{P_1 V_1}{T_1} \) = \( \frac{P_2 V_2}{T_2} \) Calculation: From formula. \( \frac{P_1 V_1}{1173.15} \) = \( \frac{(P_1/2) (V_1/2)}{T_2} \)
\( \implies \) \( \frac{P_1 V_1}{1173.15} \) = \( \frac{P_1 V_1}{4 T_2} \)
\( \implies \) \( T_2 \) = \( \frac{1173.15}{4} \) = 293.29 K Final temperature of gas is 293.29 K.
In simple words: When a gas's pressure and volume are both halved, its final absolute temperature will be a quarter of its initial absolute temperature, according to the combined gas law.

🎯 Exam Tip: Always convert temperatures to Kelvin when using gas laws (e.g., ideal gas law, combined gas law), as these laws are based on absolute temperature. Misusing Celsius will lead to incorrect results.

**Question 10.** An aluminium rod and iron rod show 1.5 m difference in their lengths when heated at all temperature. What are their lengths at 0 °C if coefficient of linear expansion for aluminium is 24.5 × 10\(^{-6}\) /°C and for iron is 11.9 × 10\(^{-6}\) /°C
Answer: Solution: Given: (\( L_T \))\(_i\) - (\( L_T \))\(_\text{al}\) = 1.5 m, \( T_0 \) = 0 °C \( \alpha_\text{al} \) = 24.5 × 10\(^{-6}\)/°C \( \alpha_i \) = 11.9 × 10\(^{-6}\) /°C To find: Lengths of aluminium and iron rod (\( L_0 \))\(_\text{al}\) and (\( L_0 \))\(_i\) Formula: \( L_T \) = \( L_0 \)[(1 + \( \alpha(T - T_0) \))] Calculation: For \( T_0 \) = 0 °C From formula, \( L_T \) = \( L_0 \)(1 + \( \alpha T \)) For aluminium, (\( L_T \))\(_\text{al}\) = (\( L_0 \))\(_\text{al}\)(1 + \( \alpha_\text{al} T \)) ............... (1) For iron, (\( L_T \))\(_i\) = (\( L_0 \))\(_i\)(1 + \( \alpha_i T \)) ............. (2) Subtracting equation (2) by (1), (\( L_T \))\(_i\) - (\( L_T \))\(_\text{al}\) = [(\( L_0 \))\(_i\) + (\( L_0 \))\(_i\) \( \alpha_i T \)] - [(\( L_0 \))\(_\text{al}\) + (\( L_0 \))\(_\text{al}\) \( \alpha_\text{al} T \)] = (\( L_0 \))\(_i\) - (\( L_0 \))\(_\text{al}\) + [(\( L_0 \))\(_i\) \( \alpha_i \) - (\( L_0 \))\(_\text{al}\) \( \alpha_\text{al} \)]T
\( \implies \) 1.5 = 1.5 + [(\( L_0 \))\(_i\) \( \alpha_i \) - (\( L_0 \))\(_\text{al}\) \( \alpha_\text{al} \)]T
\( \implies \) [(\( L_0 \))\(_i\) \( \alpha_i \) - (\( L_0 \))\(_\text{al}\) \( \alpha_\text{al} \)] T = 0
\( \implies \) (\( L_0 \))\(_\text{al}\) \( \alpha_\text{al} \) = (\( L_0 \))\(_i\) \( \alpha_i \)
\( \implies \) (\( L_0 \))\(_\text{al}\) = (\( L_0 \))\(_i\) \( \frac{\alpha_i}{\alpha_\text{al}} \) = (\( L_0 \))\(_i\) x \( \frac{11.9 \times 10^{-6}}{24.5 \times 10^{-6}} \) = (\( L_0 \))\(_i\) x \( \frac{17}{35} \) We also know that the difference between their lengths at all temperatures is 1.5m. So, (\( L_0 \))\(_i\) - (\( L_0 \))\(_\text{al}\) = 1.5 m. Substitute (\( L_0 \))\(_\text{al}\) = (\( L_0 \))\(_i\) x \( \frac{17}{35} \): (\( L_0 \))\(_i\) - (\( L_0 \))\(_i\) x \( \frac{17}{35} \) = 1.5 (\( L_0 \))\(_i\) (1 - \( \frac{17}{35} \)) = 1.5 (\( L_0 \))\(_i\) (\( \frac{35 - 17}{35} \)) = 1.5 (\( L_0 \))\(_i\) (\( \frac{18}{35} \)) = 1.5 (\( L_0 \))\(_i\) = 1.5 x \( \frac{35}{18} \) = \( \frac{52.5}{18} \) = 2.917 m Now, (\( L_0 \))\(_\text{al}\) = (\( L_0 \))\(_i\) - 1.5 = 2.917 - 1.5 = 1.417 m Length of aluminium rod at 0 °C is 1.417 m and that of iron rod is 2.917 m.
In simple words: When two rods of different materials maintain a constant length difference despite temperature changes, their initial lengths at 0°C can be found using their coefficients of linear expansion. This implies that the expansion of one rod perfectly compensates for the expansion of the other, maintaining the difference.

🎯 Exam Tip: For problems where length difference remains constant, the product of initial length and coefficient of expansion must be equal for both materials, i.e., \(L_{01}\alpha_1 = L_{02}\alpha_2\). This simplifies the calculation for their initial lengths. Ensure to set up the equations correctly from the problem statement.

**Question 11.** What is the specific heat of a metal if 50 cal of heat is needed to raise 6 kg of the metal from 20°C to 62 °C ?
Answer: Solution: Given: Q = 50 cal, m =6 kg, \( \Delta T \) = 62 – 20 = 42 °C To find: Specific heat (s) Formula: Q = ms \( \Delta T \) Calculation: From formula, S = \( \frac{Q}{m \Delta T} \) = \( \frac{50}{6 \times 42} \) = 0.198 cal/kg °C Specific heat of metal is copper 0.198 cal/kg °C.
In simple words: The specific heat capacity of a substance tells us how much heat energy is needed to raise the temperature of a unit mass of that substance by one degree Celsius. We calculate it by dividing the total heat supplied by the product of the mass and the temperature change.

🎯 Exam Tip: Be mindful of units in specific heat calculations. If heat is in calories, mass in kg, and temperature in °C, the specific heat unit will be cal/kg°C. Always ensure consistency or convert units as required by the problem.

**Question 12.** The rate of flow of heat through a copper rod with temperature difference 30 °C is 1500 cal/s. Find the thermal resistance of copper rod.
Answer: Solution: Given: \( \Delta T \) = 30 °C, \( P_\text{cond} \) = 1500 cal/s To find: Thermal resistance (\( R_T \)) Formula: \( R_T \) = \( \frac{\Delta T}{P_\text{cond}} \) Calculation: From formula, \( R_T \) = \( \frac{30}{1500} \) = 0.02 °C s/cal. Thermal resistance of copper rod is 0.02 °C s/cal.
In simple words: Thermal resistance measures how effectively a material opposes the flow of heat. It is calculated by dividing the temperature difference across the material by the rate of heat flow through it.

🎯 Exam Tip: Thermal resistance is analogous to electrical resistance (R = V/I). Ensure that the units for temperature difference and heat flow rate are consistent to obtain the correct unit for thermal resistance.

**Question 13.** An electric kettle takes 20 minutes to heat a certain quantity of water from 0°C to its boiling point. It requires 90 minutes to turn all the water at 100°C into steam. Find the latent heat of vaporisation. (Specific heat of water = 1cal/g°C)
Answer: Solution: Let heat supplied by kettle in 20 minutes be \( Q_1 \) and that in 90 min. be \( Q_2 \). Using heat temperature of water is raised from 0 °C to 100 °C. If mass of water in the kettle is 'm' then. \( Q_1 \) = \( mS_\text{water} \Delta T \) m × 1 × (100 – 0) = 100 m ................. (i) (:: \( S_\text{water} \) = 1 cal/g °C) Similarly using heat \( Q_2 \) water is converted from liquid to gas,
\( \implies \) \( Q_2 \) = \( mL_{vap} \) ................. (ii) Given that heat \( Q_1 \), \( Q_2 \) are supplied to water in 20 min. (\( t_1 \)) and 90 min (\( t_2 \)) respectively. Kettle being same its conduction rate (\( P_\text{cond} \)) is same. Using \( P_\text{cond} \) = \( \frac{Q_1}{t_1} \) = \( \frac{Q_2}{t_2} \) ................. (iii) From (i), (ii) and (iii), \( \frac{100 \text{ m}}{20} \) = \( \frac{mL_{vap}}{90} \)
\( \implies \) \( L_{vap} \) = 5 × 90 = 450 cal/g Latent heat of vaporisation for water is 450 cal/g.
In simple words: By comparing the time taken for an electric kettle to heat water to its boiling point with the time taken to convert that boiling water into steam, we can determine the latent heat of vaporization. This is because the kettle provides heat at a constant rate, allowing us to equate the heat supplied per unit time for both processes.

🎯 Exam Tip: This problem utilizes the concept of constant power/heat rate from the kettle. Equating \(P = Q/t\) for both heating and phase change allows you to solve for unknown quantities like latent heat. Ensure temperatures are appropriate for specific heat and phase change calculations (e.g., 0°C to 100°C for heating, 100°C for vaporization).

**Question 14.** Find the temperature difference between two sides of a steel plate 4 cm thick, when heat is transmitted through the plate at the rate of 400 k cal per minute per square metre at steady state. Thermal conductivity of steel is 0.026 kcal/m s K.
Answer: Solution: Given: \( \frac{Q}{At} \) = 400 kcal/min m\(^2\) = \( \frac{400}{60} \) kcal/s m\(^2\) x = 4 cm = 4 x 10\(^{-2}\) m, k = 0.026 kcal/m s K To find: Temperature difference (\( T_1 - T_2 \)). Formula: Q = \( \frac{kA (T_1 - T_2) t}{x} \) Calculation: From formula, \( T_1 - T_2 \) = \( \frac{Q}{At} \) \( \frac{x}{k} \) \( T_1 - T_2 \) = \( \frac{400}{60} \) × \( \frac{4 \times 10^{-2}}{0.026} \) = 10.26 K Temperature difference between two sides is 10.26 K. [Note: Above answer is expressed in K ('kelvin considering that thermal conductivity is expressed in units of kcal / ms K, and not as kcal / m s °C. As 1 °C equivalent to 1 K. conceptually temperature difference of 10.26 K will correspond to 10.26 t]
In simple words: The temperature difference across a material, required to maintain a specific rate of heat flow, can be found using Fourier's Law of Heat Conduction. This calculation considers the material's thermal conductivity, thickness, and the heat flow rate per unit area.

🎯 Exam Tip: In thermal conduction problems, ensure all units are consistent (e.g., convert minutes to seconds, cm to meters). Pay close attention to the units of thermal conductivity to determine the final unit of temperature difference, often K or °C for differences.

**Question 15.** A metal sphere cools from 80 °C to 60 °C in 6 min. How much time with it take to cool from 60 °C to 40 °C if the room temperature is 30°C?
Answer: Solution: Given: \( T_1 \) = 80 °C, \( T_2 \) = 60 °C, \( T_3 \) = 40 °C, \( T_0 \) = 30 °C, \( (dt)_1 \) = 6 min. To find: Time taken in cooling \( (dt)_2 \) Formula: \( \frac{dT}{dt} \) = C (\( T - T_0 \)) Calculation: From formula, Average temperature for first interval = \( \frac{80 + 60}{2} \) = 70 °C For the first cooling interval: \( \frac{(80 - 60)}{6} \) = C (70 - 30)
\( \implies \) \( \frac{20}{6} \) = C (40)
\( \implies \) C = \( \frac{20}{6 \times 40} \) = \( \frac{1}{12} \) min\(^{-1}\). Now, for the second cooling interval: Average temperature = \( \frac{60 + 40}{2} \) = 50 °C \( \frac{(60 - 40)}{(dt)_2} \) = C (\( T_{\text{avg}} - T_0 \)) \( \frac{20}{(dt)_2} \) = \( \frac{1}{12} \) (50 - 30) \( \frac{20}{(dt)_2} \) = \( \frac{1}{12} \) (20)
\( \implies \) \( (dt)_2 \) = 12 min Time taken in cooling is 12 min.
In simple words: Using Newton's Law of Cooling, which links the cooling rate to the temperature difference with the surroundings, we can calculate the time needed for a metal sphere to cool through a second temperature range by first determining the cooling constant from the initial cooling data and then applying it to the new conditions.

🎯 Exam Tip: When applying Newton's Law of Cooling for discrete time intervals, use the average temperature of the object during that interval for \(T\) in the formula \(\frac{dT}{dt} = C(T - T_0)\). Calculate the cooling constant (C) from the first set of data, then use it to find the unknown time or temperature in the second scenario.

Can You Tell? (Textbook Page No. 125)

**Question 1.** (i) Why the metal wires for electrical transmission lines sag?
(ii) Why a railway track is not a continuous piece but is made up of segments separated by gaps?
(iii) How a steel wheel is mounted on an axle to fit exactly?
Answer: 1. In hot weather, metal wires get heated due to increased temperature of surrounding. As a result, they expand increasing the slack between transmission line structure, causing them to sag. 2. Railway tracks are made up of metals which expand upon heating. If no gap is kept between tracks, in hot weather, expansion of metal tracks may exert thermal stress on track. This may lead to bending of tracks which would be dangerous. Hence, railway track is not a continuous piece but is made up of segments separated by gaps. 3. The steel wheel is heated to expand. This expanded wheel can easily fit over axle. The wheel is then cooled quickly. Upon cooling, wheel contracts and fits tightly upon the axle.
In simple words: Thermal expansion explains why power lines sag (they expand in heat), why railway tracks have gaps (to prevent buckling from expansion), and how steel wheels are fitted onto axles (by heating the wheel to expand it, then cooling it to shrink-fit onto the axle).

🎯 Exam Tip: These are classic examples illustrating the practical implications of thermal expansion. Be prepared to explain how expansion and contraction affect engineering designs and everyday phenomena.

Intext Question. (Textbook Page No 124)

**Question 1.** Can you now tell why the balloon bursts sometimes when you try to fill air in it?
Answer: 1. When balloon is blown, air that is blown inside makes the balloon expand. 2. A given size of balloon can expand upto certain limit. 3. Once that limit is reached and air is still blown inside the balloon, balloon cannot expand further. 4. As a result, air causes additional pressure on inner surface of balloon. 5. Since, pressure inside balloon is now greater than pressure outside balloon, balloon bursts equalizing the two pressures.
In simple words: A balloon bursts when overinflated because the inward tension from its stretched material can no longer contain the increasing outward pressure exerted by the air inside, leading to a rupture to equalize pressure.

🎯 Exam Tip: This question relates to the concepts of pressure, elasticity, and the mechanical limits of materials. Focus on explaining the balance (or imbalance) of forces and pressures involved.

Can You Tell? (Textbook Page No. 125)

**Question 1.** Why lakes freeze first at the surface?
Answer: 1. In cold climate, temperature of water in ponds and lakes starts falling. 2. On getting colder, water contracts. As a result, density of water increases and it goes down. To replace it, warmer water from below rises up. This process continues till temperature of water at the bottom of pond becomes 4 °C. 3. Water, due to its anomalous behaviour possesses maximum density at 4 °C. 4. If the temperature lowers further, ice is formed at the surface of pond with water below it. 5. Ice being poor conductor of heat blocks the further heat exchange between atmosphere and water in the pond and maintains water below surface in liquid state.
In simple words: Lakes freeze from the surface downwards because water has an anomalous expansion: it is densest at 4°C. As surface water cools below 4°C, it becomes less dense and stays at the top, eventually freezing, while warmer, denser water remains below, protecting aquatic life.

🎯 Exam Tip: The anomalous expansion of water (maximum density at 4°C) is a crucial concept. Understand how this property affects the freezing patterns of bodies of water and its ecological significance.

Activity (Textbook Page No. 129)

**Question 1.** To understand the process of change of state: Take some cubes of ice in a beaker. Note the temperature of ice (0 °C). Start heating it slowly on a constant heat source. Note the temperature after every minute. Continuously stir the mixture of water and ice. Observe the change in temperature. Continue heating even after the whole of ice gets converted into water. Observe the change in temperature as before till vapours start coming out. Plot the graph of temperature (along Y-axis) versus time (along X-axis). Obtain a graph of temperature versus time.
Answer: [Students are expected to attempt the activity on their own.]
In simple words: This activity involves observing and plotting the temperature change of ice as it melts into water and then boils into steam, demonstrating how temperature remains constant during phase changes despite continuous heat input.

🎯 Exam Tip: While direct answers to activities might be "Do it yourself," understanding the expected outcome (a temperature-time graph with plateaus during phase changes) is important for conceptual understanding of latent heat.

Can You Tell? (Textbook Page No. 130)

**Question 1.** What is observed after point D in graph? Can steam be hotter than 100 °C?
Answer: Beyond point D, thermometer again shows rise in temperature. This means, steam can be hotter than 100 °C and is termed as superheated steam.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक पदार्थ के तापमान में समय के साथ परिवर्तन को दर्शाता है जब उसे लगातार गर्म किया जाता है। रेखाखंड A-B बर्फ को गर्म करना दर्शाता है, B-C बर्फ के पानी में पिघलने (अवस्था परिवर्तन) के दौरान तापमान स्थिर रहता है, C-D पानी को गर्म करना दर्शाता है, और D बिंदु पर पानी उबलना शुरू होता है। बिंदु D के बाद, भाप का तापमान बढ़ सकता है, जिसे सुपरहीटेड भाप कहा जाता है।
In simple words: After water reaches its boiling point and turns into steam (point D), further heating can increase the steam's temperature above 100°C, creating superheated steam.

🎯 Exam Tip: Remember that phase changes occur at constant temperature (e.g., boiling at 100°C for water), but once a phase change is complete, the new phase (like steam) can continue to absorb heat and increase its temperature (superheating).

**Question 2.** Why steam at 100 °C causes more harm to our skin than water at 100 °C?
Answer: 1. Though steam and boiling water have same temperature, the heat contained in steam is more than that in boiling water. 2. Steam is formed when boiling water absorbs specific latent heat of vaporisation i.e.. 22.6 × 10\(^5\) J/kg. 3. As a result, when steam comes in contact with the skin of a person, it gives off additional 22.6 × 10\(^5\) joule per kilogram causing severe (more serious) burns. Hence, burns caused from steam are more serious than those caused from boiling water at same temperature.
In simple words: Steam at 100°C causes more severe burns than water at 100°C because steam contains additional latent heat of vaporization, which is released upon condensation on the skin, transferring a significantly larger amount of energy.

🎯 Exam Tip: The concept of latent heat is key here. Be able to explain how the phase change (condensation of steam) releases a large amount of energy, making steam burns more dangerous than hot water burns at the same temperature.

Activity (Textbook Page No. 130)

Activity to understand the dependence of boiling point on pressure: Take a round bottom flask, more than half filled with water. Keep it over a burner and fix a thermometer and steam outlet through the cork of the flask as shown in figure. As water in the flask gets heated, note that first the air, which was dissolved in the water comes out as small bubbles. Later bubbles of steam form at the bottom but as they rise to the cooler water near the top, they condense and disappear. Finally, as the temperature of the entire mass of the water reaches 100 °C, bubbles of steam reach the surface and boiling is said to occur. The steam in the flask may not be visible but as it comes out of the flask, It condenses as tiny droplets of water giving a foggy appearance.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक गोल पेंदे वाले फ्लास्क में पानी को गर्म करके उसके क्वथनांक (boiling point) पर दबाव के प्रभाव को दर्शाने वाले प्रायोगिक सेटअप को दिखाता है। फ्लास्क में पानी, एक थर्मामीटर और भाप निकालने के लिए एक नली लगी है, जो क्वथनांक और दबाव के बीच संबंध को समझने के लिए उपयोग किया जाता है। If now the steam outlet is closed for a few seconds to increase the pressure in the flask, you will notice that boiling stops. More heat would be required to raise the temperature (depending on the increase in pressure) before boiling starts again. Thus, boiling point increases with increase in pressure. Let us now remove the burner. Allow water to cool to about 80°C. Remove the thermometers and steam outlet. Close the flask with a air tight cork. Keep the flask turned upside down on a stand. Pour icecold water on the flask. Water vapours in the flask condense reducing the pressure on the water surface inside the flask. Water begins to boil again, now at a lower temperature. Thus boiling point decreases with decrease in pressure and increases with increase in pressure.
In simple words: This activity demonstrates that the boiling point of water is directly dependent on pressure. Increasing the pressure raises the boiling point, while decreasing the pressure lowers it, showcasing how external conditions influence phase transitions.

🎯 Exam Tip: The relationship between boiling point and pressure is fundamental to phase diagrams. Higher pressure raises the boiling point, and lower pressure lowers it. This principle is applied in pressure cookers (higher boiling point) and during cooking at high altitudes (lower boiling point).

Can You Tell? (Textbook Page No. 131)

**Question 1.** (i) Why is cooking difficult at high altitude?
(ii) Why is cooking faster in pressure cooker?
Answer: 1. At high altitude density of air is low which causes reduction in atmospheric pressure.
• As pressure is less, boiling point of water lowers.
• Water, at high altitude, starts boiling below 100 °C.
• As food is cooked mostly through the water boiling, cooking of food becomes difficult. 2. Pressure cooker operates by expelling air within the cooker and trapping steam produced from the liquid. (mostly water) boiling inside.
• Due to high internal pressure, boiling point of liquid increases and liquid boils at temperature higher than its boiling point.
• The increased boiling point allows more absorption of heat by liquid and steam formed is superheated.
• As a result, food gets cooked quickly.
In simple words: Cooking is difficult at high altitudes because reduced atmospheric pressure lowers water's boiling point, meaning food cooks at a lower temperature. In contrast, pressure cookers increase internal pressure, raising the water's boiling point and allowing food to cook faster at higher temperatures.

🎯 Exam Tip: This question directly relates to the pressure dependence of boiling point. For high altitudes, lower pressure means lower boiling point; for pressure cookers, higher pressure means higher boiling point. Be able to explain the implications for cooking efficiency.