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#latent heat

21 public questions tagged with this topic.

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 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

How much heat is required to raise 0.6kg of water from 25∘C to 75∘C and then convert 0.4kg of it to steam at 100∘C in a

Q1 = (0.6×4186+0.1×386)×(75−25) = (2511.6+38.6)×50 = 2550.2×50 = 127510J (to 75°C). Q2 = (0.6×4186+0.1×386)×(100−75) = 2550.2×25 = 63755J (to 100°C). Q3 = 0.4×2.256×106 = 902400J (vaporization). Total: Q = 127510+63755+902400 = 1093665J = 1093.67kJ.

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

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Latent Heat and Phase Transitions

Why does steam at 100∘C cause more severe burns than water at 100∘C?

Steam releases additional latent heat of vaporization when it condenses on skin, transferring more energy than liquid water at the same temperature. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Steam releases latent heat on condensing. 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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Specific Heat and Calorimetry

How much ice at 0∘C will melt if 16744J of heat is supplied? (Latent heat of fusion of ice = 3.35×105J kg−1)

Given: Q = 16744J, Lf = 3.35×105J kg−1. Q = mLf⇒m = QLf = 167443.35×105≈0.05kg = 50g. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 50 g. 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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Specific Heat and Calorimetry

How much heat is required to convert 0.3kg of ice at −18∘C to water at 40∘C in a 0.1kg copper calorimeter initially at 2

Q1 = 0.3×2100×18 = 11340J (ice to 0°C). Q2 = 0.3×3.35×105 = 100500J (melting). Heat gained by water and calorimeter: (0.3×4186+0.1×386)×(40−0) = (1255.8+38.6)×40 = 1294.4×40 = 51776J. Heat lost by calorimeter: 0.1×386×(25−0) = 965J (assume it cools to 0°C first). Total heat supplied: Q = 11340+100500+51776−965 = 162651J = 162.65kJ.

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

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Thermal Expansion of Solids and Liquids

How much heat is required to vaporize 0.15kg of nitrogen at −196∘C? (Latent heat of vaporization of nitrogen = 2.0×105J

Given: m = 0.15kg, Lv = 2.0×105J kg−1. Q = mLv = 0.15×2.0×105 = 30000J = 30kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 30 kJ. 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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