Chapter 10: Thermal Properties of Matter Case Study Questions | physics Class 11 CBSE

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Chapter 10: Thermal Properties of Matter Case Study Questions | physics Class 11 CBSE | ByteNotes.in

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Case Set 1

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A student performs an experiment using a constant-volume gas thermometer filled with nitrogen at low density. The thermometer is first calibrated at the triple point of water (273.16 K) and then used to measure the temperature of a boiling liquid. The student observes that when the thermometer is dipped in the boiling liquid, the pressure of the gas increases significantly. She also notes that when the same thermometer is placed in a mixture of ice and water, the pressure falls to a value consistent with 273.15 K rather than 273.16 K. She recalls that the Celsius scale and Kelvin scale have the same unit size but different origins, and that absolute zero represents the state of minimum molecular activity for all substances.

Question 1: What is the relationship between Kelvin temperature T and Celsius temperature tC? State the value of absolute zero on the Celsius scale.

The relationship is T = tC + 273.15, meaning the Kelvin and Celsius scales have identical unit sizes but differ in their zero points.
Absolute zero corresponds to –273.15°C on the Celsius scale; it is the temperature at which an ideal gas would theoretically exert zero pressure.
The Kelvin scale is named after Lord Kelvin and uses –273.15°C as its origin (0 K), making all thermodynamic temperatures positive.

Question 2: Why does a gas thermometer give the same temperature reading regardless of which gas is used, unlike liquid-in-glass thermometers?

Liquid-in-glass thermometers depend on the volume expansion of a specific liquid; different liquids have different expansion properties, giving different readings between fixed points.
A gas thermometer at low density uses the fact that all gases at low densities obey the same ideal gas law (PV = μRT) and exhibit identical expansion behaviour.
Therefore, the pressure-temperature relationship is the same for all low-density gases, making the gas thermometer a universal and reproducible standard for temperature measurement.

Question 3: The student finds the ice point reads 273.15 K rather than 273.16 K. Explain the physical reason for this difference. Why is the triple point preferred over the ice point as a fixed reference?

The triple point of water (273.16 K) is the unique temperature at which solid, liquid, and vapour phases of water all coexist in equilibrium; it is a single, precisely reproducible state.
The ice point (melting point of ice at 1 atm) depends on external atmospheric pressure; any variation in pressure shifts the melting point slightly, making it less precise and reproducible than the triple point.
The difference of 0.01 K (273.16 – 273.15) arises because the melting point of ice at 1 atm is very close to but not exactly equal to the triple point temperature.

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Case Set 2

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An engineer is designing railway tracks for a region where temperature swings between 10°C in winter and 50°C in summer. The steel rails are each 10 m long and are laid at 25°C. The engineer must decide how wide the expansion gaps between adjacent rails should be to prevent buckling. He knows the coefficient of linear expansion of steel is 1.2×10⁻⁵ K⁻¹ and its Young's modulus is 2×10¹¹ N m⁻². He also recalls from his physics textbook that if expansion is prevented, the resulting thermal stress can develop forces large enough to bend the rails. The cross-sectional area of each rail is 50 cm².

Question 1: Define the coefficient of linear expansion and write its mathematical expression. What are its SI units?

The coefficient of linear expansion αl is defined as the fractional change in length of a material per unit rise in temperature: αl = (Δl/l)/ΔT.
It is a characteristic property of the material and indicates how much the material expands per degree rise in temperature.
Its SI unit is K⁻¹ (per kelvin); it is also expressed as °C⁻¹ since the size of a kelvin and a degree Celsius are equal.

Question 2: Calculate the minimum gap that must be left between two adjacent 10 m steel rails laid at 25°C to prevent buckling at 50°C.

Maximum temperature rise from laying temperature to summer maximum: ΔT = 50 – 25 = 25°C (the rails were laid at 25°C and must accommodate expansion up to 50°C).
Using Δl = αl × l × ΔT = 1.2×10⁻⁵ × 10 × 25 = 3.0×10⁻³ m = 3.0 mm.
A minimum gap of 3.0 mm must be left between adjacent rails; in practice, a slightly larger gap is used to provide a safety margin for unexpected temperature variations.

Question 3: If the gaps are accidentally not provided and expansion is fully prevented at 50°C, calculate the thermal stress developed and the compressive force on the supports. Comment on the structural consequence.

Compressive strain if expansion is prevented: Δl/l = αl × ΔT = 1.2×10⁻⁵ × 25 = 3.0×10⁻⁴.
Thermal stress = Y × strain = 2×10¹¹ × 3.0×10⁻⁴ = 6.0×10⁷ N m⁻², which is an enormously large compressive stress in the steel rail.
Compressive force = stress × area = 6.0×10⁷ × 50×10⁻⁴ = 3.0×10⁵ N (about 30 tonnes); such a force can easily buckle and bend the rails, causing serious derailment hazards.

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Case Set 3

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During a science laboratory session, students heat 300 g of ice at –10°C in a beaker placed on a constant heat source. They note the temperature every minute and carefully plot a temperature-time graph. They observe that the temperature rises initially, then stays constant for several minutes, then rises again, stays constant again, and finally rises once more. The teacher explains that these flat regions in the graph correspond to phase changes, during which all the heat supplied goes into changing the state of the substance rather than raising its temperature. The teacher also asks them to note the different slopes of the rising portions and relate them to specific heat capacities of different phases.

Question 1: Why does the temperature remain constant during a change of state even though heat is being continuously supplied?

During a change of state, the heat supplied is entirely used to overcome the intermolecular forces holding the molecules in their current arrangement, rather than increasing the kinetic energy of molecules.
Since temperature is a measure of the average kinetic energy of molecules, and kinetic energy does not increase during this process, temperature remains constant.
The heat absorbed during a change of state is called latent heat; it is stored as potential energy of the molecules in the new phase arrangement.

Question 2: The students notice that the slope of the temperature-time graph is steeper for ice than for water during the heating stages. What does this indicate about the specific heat capacities of ice and water?

A steeper slope in the temperature-time graph means the temperature rises more rapidly per unit time, indicating that less heat is needed per unit temperature rise for ice than for water.
This implies that the specific heat capacity of ice (2060 J kg⁻¹ K⁻¹) is significantly less than that of water (4186 J kg⁻¹ K⁻¹); water requires more heat per unit mass per unit temperature change.
The different slopes in the temperature-time graph directly reflect the different specific heat capacities of ice, liquid water, and steam, confirming that specific heat is a state-dependent property.

Question 3: Calculate the total heat required to convert 300 g of ice at –10°C to steam at 100°C. (sice = 2100 J kg⁻¹ K⁻¹, sw = 4186 J kg⁻¹ K⁻¹, Lf = 3.35×10⁵ J kg⁻¹, Lv = 2.256×10⁶ J kg⁻¹)

Q1 (ice –10°C to 0°C) = 0.3 × 2100 × 10 = 6300 J; Q2 (melting ice at 0°C) = 0.3 × 3.35×10⁵ = 100500 J.
Q3 (water 0°C to 100°C) = 0.3 × 4186 × 100 = 125580 J; Q4 (vaporising water at 100°C) = 0.3 × 2.256×10⁶ = 676800 J.
Total Q = 6300 + 100500 + 125580 + 676800 = 909180 J ≈ 9.09×10⁵ J; the dominant contribution is from the latent heat of vaporisation, confirming that Lv >> Lf for water.

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Case Set 4

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A physics teacher demonstrates calorimetry to her class using a copper calorimeter. She heats a 50 g piece of an unknown metal to 100°C in boiling water and quickly transfers it to the calorimeter containing 200 g of water at 20°C. The copper calorimeter has a mass of 100 g. The class observes the thermometer reading rise and stabilise at 24°C. The teacher then challenges students to find the specific heat of the unknown metal and identify it by comparing with values in Table 10.3 of their textbook. She reminds them that the principle is heat lost by the metal equals heat gained by water and calorimeter, and that this works only in a well-insulated system.

Question 1: State the principle of calorimetry and explain why the calorimeter is kept inside a wooden jacket with glass wool insulation.

The principle of calorimetry states that in an isolated system, the heat lost by a hotter body equals the heat gained by a cooler body: heat lost = heat gained.
The wooden jacket and glass wool insulation minimise heat exchange between the calorimeter and the surroundings, making the system approximately isolated.
Without this insulation, heat would escape to or enter from the room, making the heat balance inaccurate and the calculated specific heat unreliable.

Question 2: Calculate the specific heat of the unknown metal. (sw = 4186 J kg⁻¹ K⁻¹, scu = 386 J kg⁻¹ K⁻¹)

Heat gained by water = 0.200 × 4186 × (24–20) = 0.200 × 4186 × 4 = 3348.8 J.
Heat gained by copper calorimeter = 0.100 × 386 × 4 = 154.4 J; total heat gained = 3348.8 + 154.4 = 3503.2 J.
Heat lost by metal = 0.050 × s_metal × (100–24) = 0.050 × s_metal × 76; equating: 0.050 × s_metal × 76 = 3503.2; s_metal = 3503.2 / 3.8 ≈ 922 J kg⁻¹ K⁻¹ ≈ aluminium (900 J kg⁻¹ K⁻¹).

Question 3: If heat losses to the surroundings are not negligible in this experiment, would the calculated value of specific heat be greater or smaller than the actual value? Justify your answer with reasoning.

If heat is lost to surroundings, the total heat gained by water and calorimeter is less than the actual heat lost by the metal; the measured temperature rise is smaller than it should be.
Therefore, the calculated heat gained (= heat lost by metal) is underestimated; dividing this smaller value by the same mass and temperature drop gives a smaller calculated specific heat.
The calculated value would be smaller than the actual specific heat of the metal; to improve accuracy, the experiment should be performed in a better-insulated calorimeter or corrections must be applied for heat losses.

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Case Set 5

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During winter, a student living near a large lake notices that even when the air temperature drops to –5°C, the water in the middle of the lake remains liquid while the edges have frozen. His teacher explains that this is due to the anomalous expansion of water and the unique property of water having maximum density at 4°C. The teacher further explains that this physical property has profound ecological consequences: aquatic organisms survive winter in lakes precisely because of this behaviour of water. The student is asked to trace the sequence of events that leads to ice forming at the surface of a lake rather than at the bottom, and to explain why this is vital for aquatic life.

Question 1: What is meant by the anomalous expansion of water? In what temperature range does it occur?

Anomalous expansion refers to the unusual behaviour of water whereby it contracts (density increases) upon heating between 0°C and 4°C, which is opposite to the normal behaviour of most substances.
Below 4°C, water expands upon further cooling, so its density decreases below 4°C; this means water has its maximum density at 4°C.
This anomalous behaviour occurs in the temperature range 0°C to 4°C; outside this range (above 4°C) water behaves normally and expands on heating.

Question 2: Trace the sequence of events when a lake cools from room temperature to below 0°C in winter. Why does ice form at the top rather than the bottom?

As the lake cools from room temperature to 4°C, surface water becomes denser and sinks; warmer, less dense water from the bottom rises (convective mixing), so the entire lake gradually cools to 4°C.
Once the lake reaches 4°C throughout, further cooling of surface water below 4°C makes it less dense than the water below; it stays at the surface and does not sink.
This less dense surface water then cools to 0°C and freezes; the ice layer remains at the top because ice is less dense than liquid water, while the deeper water stays at ~4°C.

Question 3: If water did not show anomalous expansion and froze from the bottom up, explain the ecological consequences. How does the current behaviour protect aquatic life?

Without anomalous expansion, denser cold water (below 4°C) would continuously sink; the lake would freeze from the bottom upward as the coldest, densest water settles at the bottom.
Freezing from the bottom up would destroy all aquatic plants rooted in the sediment and all bottom-dwelling organisms; fish and other mobile species would be trapped in a progressively shrinking volume of liquid water.
With anomalous expansion, ice forms only at the surface and acts as a thermal insulator; the water below remains liquid at ~4°C all winter, creating a habitable environment where aquatic life can survive even in severely cold conditions.

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Chapter 10: Thermal Properties of Matter | Case Study Questions

Passage

Question 1: What is the relationship between Kelvin temperature T and Celsius temperature tC? State the value of absolute zero on the Celsius scale.

Question 2: Why does a gas thermometer give the same temperature reading regardless of which gas is used, unlike liquid-in-glass thermometers?

Question 3: The student finds the ice point reads 273.15 K rather than 273.16 K. Explain the physical reason for this difference. Why is the triple point preferred over the ice point as a fixed reference?

Passage

Question 1: Define the coefficient of linear expansion and write its mathematical expression. What are its SI units?

Question 2: Calculate the minimum gap that must be left between two adjacent 10 m steel rails laid at 25°C to prevent buckling at 50°C.

Question 3: If the gaps are accidentally not provided and expansion is fully prevented at 50°C, calculate the thermal stress developed and the compressive force on the supports. Comment on the structural consequence.

Passage

Question 1: Why does the temperature remain constant during a change of state even though heat is being continuously supplied?

Question 2: The students notice that the slope of the temperature-time graph is steeper for ice than for water during the heating stages. What does this indicate about the specific heat capacities of ice and water?

Question 3: Calculate the total heat required to convert 300 g of ice at –10°C to steam at 100°C. (sice = 2100 J kg⁻¹ K⁻¹, sw = 4186 J kg⁻¹ K⁻¹, Lf = 3.35×10⁵ J kg⁻¹, Lv = 2.256×10⁶ J kg⁻¹)

Passage

Question 1: State the principle of calorimetry and explain why the calorimeter is kept inside a wooden jacket with glass wool insulation.

Question 2: Calculate the specific heat of the unknown metal. (sw = 4186 J kg⁻¹ K⁻¹, scu = 386 J kg⁻¹ K⁻¹)

Question 3: If heat losses to the surroundings are not negligible in this experiment, would the calculated value of specific heat be greater or smaller than the actual value? Justify your answer with reasoning.

Passage

Question 1: What is meant by the anomalous expansion of water? In what temperature range does it occur?

Question 2: Trace the sequence of events when a lake cools from room temperature to below 0°C in winter. Why does ice form at the top rather than the bottom?

Question 3: If water did not show anomalous expansion and froze from the bottom up, explain the ecological consequences. How does the current behaviour protect aquatic life?

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