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Physics

Practice questions and tests covering core physics topics like mechanics, thermodynamics, and electromagnetism. Ideal for students preparing for exams or strengthening foundational knowledge.

1587 questions

Ideal Gas Laws and Thermal Properties

How much heat is required to vaporize 0.25kg of ethanol at 78∘C? (Latent heat of vaporization of ethanol = 8.5×105J kg−1

Given: m = 0.25kg, Lv = 8.5×105J kg−1. Q = mLv = 0.25×8.5×105 = 212500J = 212.5kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 212.5 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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Ideal Gas Laws and Thermal Properties

How much heat is required to convert 0.35kg of ice at −28∘C to steam at 115∘C in a 0.1kg brass calorimeter initially at

Q1 = 0.35×2100×28 = 20580J (ice to 0°C). Q2 = 0.35×3.35×105 = 117250J (melting). Q3 = (0.35×4186+0.1×386)×100 = (1465.1+38.6)×100 = 1503.7×100 = 150370J (to 100°C). Q4 = 0.35×2.256×106 = 789600J (vaporization). Q5 = 0.35×4186×15 = 21976.5J (steam to 115°C). Calorimeter cools: 0.1×386×(25−0) = 965J. Total: Q = 20580+117250+150370+789600+21976.5−965 = 1097811.5J = 1097.81kJ.

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 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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Ideal Gas Laws and Thermal Properties

Why does a substance with high specific heat capacity take longer to heat up?

A high specific heat capacity means more heat (ΔQ = msΔT) is required per unit mass to raise the temperature, slowing the heating process (Section 10.6). As per NCERT, applying relevant law/formula with correct units and sign convention leads to It requires more heat per unit temperature rise. 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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Ideal Gas Laws and Thermal Properties

A gas at 1atm and 300K occupies 2L. If it is compressed to 0.5L while temperature rises to 450K, what is the final press

Given: P1 = 1atm, V1 = 2L, T1 = 300K, V2 = 0.5L, T2 = 450K. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 1×20.5×450300 = 1×4×1.5 = 6atm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6 atm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Ideal Gas Laws and Thermal Properties

A 0.5kg silver block at 240∘C is placed in 1.5kg water at 21∘C in a 0.25kg copper calorimeter at 21∘C. What is the final

0.5×236×(240−T) = (1.5×4186+0.25×386)×(T−21). 28320−118T = (6279+96.5)×(T−21) = 6375.5T−133885.5. 28320+133885.5 = 6375.5T+118T. 162205.5 = 6493.5T⇒T≈24.97∘C≈25∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 25°C. 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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