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#latent heat

30 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

How much heat is required to vaporize 0.9 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). Δ Q = m L . m = 0.9 , L = 2256 . Δ Q = 0.9 × 2256 = 2030.4 J ≈ 2030 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

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Heat Transfer Work Distinction and Internal Energy Change

How much heat is required to vaporize 2 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. Δ Q = m L . m = 2 , L = 2256 . Δ Q = 2 × 2256 = 4512 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

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Cyclic Processes and Reversibility Concepts

How much heat is required to vaporize 1.2 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. Δ Q = m L . m = 1.2 , L = 2256 . Δ Q = 1.2 × 2256 = 2707.2 J ≈ 2707 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields 2707 J, consistent with thermodynamic laws and energy conservation.

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

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Specific Heat Capacity and Latent Heat

How much heat is required to convert 1 g of water from liquid at 100^circ C to vapor at 100^circ C at 1 atm ? (Latent he

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. Heat: Δ Q = m L . m = 1 g , L = 2256 J/g . Δ Q = 1 × 2256 = 2256 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h,

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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First Law of Thermodynamics Applications

How much heat is required to vaporize 0.5 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. Δ Q = m L . m = 0.5 , L = 2256 . Δ Q = 0.5 × 2256 = 1128 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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First Law of Thermodynamics Applications

How much heat is required to vaporize 0.7 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. Δ Q = m L . m = 0.7 , L = 2256 . Δ Q = 0.7 × 2256 = 1579.2 J ≈ 1579 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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Isobaric and Isothermal Processes Work Calculation

How much heat is required to vaporize 1.4 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. Δ Q = m L . m = 1.4 , L = 2256 . Δ Q = 1.4 × 2256 = 3158.4 J ≈ 3158 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields 3158

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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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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