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#vaporization

12 public questions tagged with this topic.

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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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 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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Temperature Scales and Triple Point

How much heat is required to raise 0.4kg of water from 15∘C to 85∘C and then convert 0.25kg to steam at 100∘C in a 0.2kg

Q1 = (0.4×4186+0.2×236)×(85−15) = (1674.4+47.2)×70 = 1721.6×70 = 120512J (to 85°C). Q2 = (0.4×4186+0.2×236)×(100−85) = 1721.6×15 = 25824J (to 100°C). Q3 = 0.25×2.256×106 = 564000J (vaporization). Total: Q = 120512+25824+564000 = 710336J = 710.34kJ.

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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