Skip to content
Login Register

#phase change

21 public questions tagged with this topic.

Transverse and Longitudinal Waves

What is the phase change when a longitudinal wave reflects off a rigid boundary?

**Wave classification** depends on particle vibration relative to propagation. Longitudinal waves have particle oscillation parallel to propagation, creating compressions and rarefactions as in sound in air; transverse have perpendicular oscillation. Tuning fork generates longitudinal sound because air cannot sustain shear. For a longitudinal wave, a rigid boundary (fixed end) causes a compression to reflect as a compression, resulting in no phase change (0 radians), unlike transverse waves which invert. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 0 radians, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Transverse and Longitudinal Waves

View Details
Sound Waves, Reflection and Characteristics

What happens to the phase of a transverse wave when it reflects off a free end of a string?

**Reflection at boundaries** follows phase change rules: rigid boundary (fixed end) introduces π phase shift, inverting displacement y → -y, while free boundary reflects without phase change. Reflected wave derived by reversing propagation direction kx → -kx and applying phase shift, preserving k = 2π/λ and ω = 2πf. At a free end, the wave reflects without inversion (phase change of 0 radians) because the boundary allows maximum displacement, aligning the reflected wave with the incident wave. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Remains unchanged, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Sound Waves, Reflection and Characteristics

View Details
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

View Details
CHEMISTRY

Calculate the heat absorbed by 18 g of water to vaporize at 373 K, given Δ_{vap H° = 40.79 kJ/mol (molar mass of water

Given: Calculate the heat absorbed by 18 g of water to vaporize at 373 K, given Δ_{vap H° = 40.79 kJ/mol (molar mass of water = 18 g/mol). These values define the system as per NCERT data. Formula: Moles = 18 / 18 = 1 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Heat absorbed = n × Δ_{vap H° = 1 × 40.79 = 40.79 kJ. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

View Details
CHEMISTRY

The enthalpy change for the reaction Hâ‚‚O(l) -> Hâ‚‚O(g) at 373 K is 40.79 kJ/mol. What is the internal energy change i

Given: The enthalpy change for the reaction H₂O(l) -> H₂O(g) at 373 K is 40.79 kJ/mol. What is the internal energy change if Δ n_g = 1 and R = 8.314 J/mol · K? These values define the system as per NCERT data. Formula: Δ H = Δ U + Δ n_g RT. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given Δ H = 40.79 kJ/mol, Δ n_g = 1, T = 373 K, RT = 8.314 × 373 / 1000 = 3.1 kJ. So, Δ U = 40.79 - 3.1 = 37.69 kJ/mol. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Thermodynamics, Equilibrium and Chemical Kinetics, Topic: Energetics, equilibrium constants and reaction rates.

View Details
PHYSICS

What is the phase change when a longitudinal wave reflects off a rigid boundary?

For a longitudinal wave, a rigid boundary (fixed end) causes a compression to reflect as a compression, resulting in no phase change (0 radians), unlike transverse waves which invert.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

View Details
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.

View Details
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.

View Details
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.

View Details
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.

View Details
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.

View Details