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

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A blacksmith heats an iron ring before fitting it onto the rim of a wooden wheel of a horse cart. The diameter of the rim is 5.243 m and the iron ring is 5.231 m at 27°C. Explain why the blacksmith heats the ring and calculate the temperature to which it must be heated (αl for iron = 1.20×10⁻⁵ K⁻¹).

When iron is heated, it undergoes linear thermal expansion; the ring's diameter increases proportionally to the rise in temperature, enabling it to fit over the larger wooden rim.
Using the formula: LT2 = LT1[1 + αl(T2 – T1)], substituting LT2 = 5.243 m, LT1 = 5.231 m, T1 = 27°C, and αl = 1.20×10⁻⁵ K⁻¹.
Solving: 5.243 = 5.231[1 + 1.20×10⁻⁵(T2 – 27)], which gives T2 ≈ 218°C; the ring must be heated to about 218°C.
After fitting, the ring cools, contracts, and grips the wooden rim very tightly, acting as a firm mechanical clamp without the need for bolts or adhesive.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A steel rail of length 5 m and cross-sectional area 40 cm² is prevented from expanding while the temperature rises by 10°C. Determine the compressive strain and the force exerted on the supports (αl(steel) = 1.2×10⁻⁵ K⁻¹, Y(steel) = 2×10¹¹ N m⁻²).

When expansion is prevented, the rail develops a compressive strain equal to αlΔT = 1.2×10⁻⁵ × 10 = 1.2×10⁻⁴ (dimensionless).
Thermal stress = Y × strain = 2×10¹¹ × 1.2×10⁻⁴ = 2.4×10⁷ N m⁻², a very large compressive stress in the rail.
The compressive force on supports = stress × area = 2.4×10⁷ × 40×10⁻⁴ = 9.6×10⁴ N ≈ 10⁵ N (about 10 tonnes of force).
Two such rails in contact could bend easily under this force; this is why gaps (expansion joints) are deliberately left between rail sections to allow free expansion.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A sealed metallic lid on a bottle is very tight and difficult to open. Explain why placing the lid in hot water helps to open it. Use the concept of differential thermal expansion.

The metallic lid has a higher coefficient of linear expansion than the glass bottle; when both are heated, the metal expands more than the glass for the same temperature rise.
Hot water raises the temperature of both the lid and the glass neck; since αl(metal) > αl(glass), the metallic lid expands proportionally more than the glass neck.
This differential expansion increases the diameter of the metallic lid relative to the glass neck, reducing the friction and grip between them, making it easier to unscrew.
The principle is the same as thermal expansion used in the horse cart iron ring problem: controlled heating allows tight fits to be loosened by exploiting differences in expansion coefficients.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

A sphere of 0.047 kg aluminium at 100°C is transferred to a copper calorimeter of 0.14 kg containing 0.25 kg water at 20°C. The final steady temperature is 23°C. Calculate the specific heat capacity of aluminium. (sw = 4.18×10³ J kg⁻¹ K⁻¹, scu = 0.386×10³ J kg⁻¹ K⁻¹)

Heat lost by aluminium = m₁ × sAl × ΔT₁ = 0.047 × sAl × (100 – 23) = 0.047 × sAl × 77.
Heat gained by water = 0.25 × 4.18×10³ × 3 = 3135 J; heat gained by calorimeter = 0.14 × 0.386×10³ × 3 = 162.1 J; total heat gained = 3297.1 J.
Applying heat lost = heat gained: 0.047 × sAl × 77 = 3297.1 J, so sAl = 3297.1 / (0.047 × 77) ≈ 911 J kg⁻¹ K⁻¹.
This is the experimental determination of specific heat by calorimetry; the result agrees well with the known value, validating the principle that heat lost equals heat gained in an isolated system.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

When 0.15 kg of ice at 0°C is mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Show how to calculate the latent heat of fusion of ice. (sw = 4186 J kg⁻¹ K⁻¹)

Heat lost by water cooling from 50°C to 6.7°C: Q_lost = 0.30 × 4186 × (50 – 6.7) = 0.30 × 4186 × 43.3 ≈ 54376 J.
Heat needed to melt ice at 0°C: Q₁ = 0.15 × Lf; heat needed to warm melted ice water from 0°C to 6.7°C: Q₂ = 0.15 × 4186 × 6.7 ≈ 4207 J.
Applying heat lost = heat gained: 54376 = 0.15Lf + 4207, so Lf = (54376 – 4207)/0.15 = 50169/0.15 ≈ 3.34×10⁵ J kg⁻¹.
This experimental method determines Lf of ice by calorimetry; the calculated value matches the standard value Lf = 3.33×10⁵ J kg⁻¹, confirming the principle.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

Why is cooking more difficult at high altitudes such as hill stations? Explain using the concept of boiling point and its dependence on pressure.

At high altitudes, atmospheric pressure is lower than at sea level; the boiling point of a liquid decreases with decrease in external pressure.
Since water boils at a temperature below 100°C at high altitudes, the food is cooked at a lower temperature than at sea level.
Lower cooking temperature means slower chemical and physical changes (such as softening of food fibers and starch gelatinization), requiring more time to cook the same food.
A pressure cooker solves this problem by increasing the pressure inside, raising the boiling point above 100°C and thus cooking food faster even at high altitudes.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A pan filled with hot food cools from 94°C to 86°C in 2 minutes when room temperature is 20°C. Using Newton's Law of Cooling, find the time taken to cool from 71°C to 69°C.

Average temperature in first case = (94+86)/2 = 90°C; excess over surroundings = 90 – 20 = 70°C; rate of cooling = 8°C / 2 min = 4°C/min.
Using Newton's law: (change in temp)/time = K × (excess temperature), so 8/2 = K × 70, giving K = 4/70 °C min⁻¹ °C⁻¹.
Average temperature in second case = (71+69)/2 = 70°C; excess over surroundings = 70 – 20 = 50°C; temperature drop = 2°C.
Applying: 2/time = K × 50 = (4/70) × 50; time = 2 × 70/(4 × 50) = 140/200 = 0.7 min = 42 seconds.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.

First interval: average temperature = (80+50)/2 = 65°C; excess over surroundings = 65 – 20 = 45°C; rate of cooling = 30/5 = 6°C/min.
Using Newton's law: 30/5 = K × 45, giving K = 6/45 = 2/15 min⁻¹.
Second interval: average temperature = (60+30)/2 = 45°C; excess over surroundings = 45 – 20 = 25°C; temperature drop = 30°C.
Time = 30/(K × 25) = 30/((2/15)×25) = 30/(50/15) = 30×15/50 = 9 minutes.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 9

Applied Concept

Explain with reference to the chapter why we wear white or light-coloured clothes in summer and dark or black clothes in winter. Relate your answer to the radiation properties of surfaces.

Dark or black surfaces absorb radiant energy more efficiently than light or white surfaces; similarly, black bodies are the best emitters of thermal radiation.
In summer, wearing white or light-coloured clothes minimises absorption of solar radiation (high reflectivity), keeping the body cooler by reducing the amount of radiant heat absorbed.
In winter, wearing dark-coloured clothes maximises absorption of solar radiation, helping to warm the body by absorbing more heat from the Sun.
The same principle applies to the blackening of the bottoms of cooking utensils: a black surface absorbs the maximum heat from the fire and transfers it efficiently to the food being cooked.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

Explain the working of a thermos flask (Dewar flask) with reference to how each of the three modes of heat transfer is minimised in its design.

A Dewar flask has double glass walls with silver coating on both inner surfaces and a vacuum between the walls; the flask is supported on a cork insulator.
Radiation is minimised by the silver coating: the inner silvered wall reflects the radiation from the hot contents back inside; the outer silvered wall reflects external radiation away, preventing it from being absorbed.
Conduction is minimised in two ways: glass is a poor conductor, and the vacuum between the walls eliminates conduction through air (which would otherwise act as a medium for heat transfer).
Convection is eliminated by the vacuum between the walls since convection requires a fluid medium; without any gas between the walls, no convective currents can form to transfer heat.

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Chapter 10: Thermal Properties of Matter | Application Questions

A steel rail of length 5 m and cross-sectional area 40 cm² is prevented from expanding while the temperature rises by 10°C. Determine the compressive strain and the force exerted on the supports (αl(steel) = 1.2×10⁻⁵ K⁻¹, Y(steel) = 2×10¹¹ N m⁻²).

A sealed metallic lid on a bottle is very tight and difficult to open. Explain why placing the lid in hot water helps to open it. Use the concept of differential thermal expansion.

A sphere of 0.047 kg aluminium at 100°C is transferred to a copper calorimeter of 0.14 kg containing 0.25 kg water at 20°C. The final steady temperature is 23°C. Calculate the specific heat capacity of aluminium. (sw = 4.18×10³ J kg⁻¹ K⁻¹, scu = 0.386×10³ J kg⁻¹ K⁻¹)

When 0.15 kg of ice at 0°C is mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Show how to calculate the latent heat of fusion of ice. (sw = 4186 J kg⁻¹ K⁻¹)

Why is cooking more difficult at high altitudes such as hill stations? Explain using the concept of boiling point and its dependence on pressure.

A pan filled with hot food cools from 94°C to 86°C in 2 minutes when room temperature is 20°C. Using Newton's Law of Cooling, find the time taken to cool from 71°C to 69°C.

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.

Explain with reference to the chapter why we wear white or light-coloured clothes in summer and dark or black clothes in winter. Relate your answer to the radiation properties of surfaces.

Explain the working of a thermos flask (Dewar flask) with reference to how each of the three modes of heat transfer is minimised in its design.

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