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#phase change

12 public questions tagged with this topic.

Ideal Gas Laws and Thermal Properties

How much heat is required to convert 0.2kg of ice at −22∘C to steam at 110∘C in a 0.05kg aluminium calorimeter initially

Q1 = 0.2×2100×22 = 9240J (ice to 0°C). Q2 = 0.2×3.35×105 = 67000J (melting). Q3 = (0.2×4186+0.05×900)×100 = (837.2+45)×100 = 88220J (to 100°C). Q4 = 0.2×2.256×106 = 451200J (vaporization). Q5 = 0.2×4186×10 = 8372J (steam to 110°C). Calorimeter cools: Q6 = 0.05×900×(30−0) = 1350J (assume it cools to 0°C). Total: Q = 9240+67000+88220+451200+8372−1350 = 622682J = 622.68kJ.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Ideal Gas Laws and Thermal Properties

How much heat is required to convert 1kg of ice at −10∘C to water at 0∘C? (Specific heat of ice = 2100J kg−1K−1, Latent

Q1 = msΔT = 1×2100×10 = 21000J. Q2 = mLf = 1×3.35×105 = 335000J. Total heat: Q = Q1+Q2 = 21000+335000 = 356000J = 356kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 356 kJ. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Latent Heat and Phase Transitions

What process occurs when dry ice changes directly from solid to vapor?

Sublimation is the direct transition from solid to vapor without passing through the liquid state, as seen with dry ice (Section 10.8). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Sublimation. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Fluid Pressure, Pascal's Law and Manometers

Which of the following is true about latent heat during a phase change?

Latent heat is the heat absorbed or released per unit mass during a phase change without a temperature change (Section 10.8.1, Q = mL). As per NCERT, applying relevant law/formula with correct units and sign convention leads to It changes the state without changing temperature. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Thermal Expansion of Solids and Liquids

How much heat is needed to convert 0.2kg of water at 100∘C to steam at 100∘C? (Latent heat of vaporization = 2.256×106J

Given: m = 0.2kg, Lv = 2.256×106J kg−1. Q = mLv = 0.2×2.256×106 = 451200J = 451.2kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 451.2 kJ. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Thermal Expansion of Solids and Liquids

Which process describes the direct transition from solid to gas observed in iodine?

Sublimation is the process where a substance transitions directly from solid to gas, as observed with iodine under certain conditions. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Sublimation. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Temperature Scales and Triple Point

During the melting of ice at 0∘C, why does the temperature remain constant despite heat being supplied?

The heat supplied during melting (latent heat of fusion) is used to change the state from solid to liquid, overcoming intermolecular forces, not to increase temperature (Section 10.8). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Heat is used to change the state, not raise temperature. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Temperature Scales and Triple Point

How much heat is required to convert 0.25kg of ice at −30∘C to steam at 120∘C? (Specific heat of ice = 2100J kg−1K−1, la

Q1 = 0.25×2100×30 = 15750J (ice to 0°C). Q2 = 0.25×3.35×105 = 83750J (melting). Q3 = 0.25×4186×100 = 104650J (water to 100°C). Q4 = 0.25×2.256×106 = 564000J (vaporization). Q5 = 0.25×4186×20 = 20930J (steam to 120°C). Total: Q = 15750+83750+104650+564000+20930 = 789080J = 789.08kJ.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Temperature Scales and Triple Point

During vaporization, what does the supplied heat primarily do to the liquid?

During vaporization, the supplied heat (latent heat of vaporization) breaks intermolecular bonds to convert the liquid into vapor, without changing its temperature. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Converts it to vapor without temperature change. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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