#water

Wave Speed, Energy and Power

The bulk modulus of water is 2.2 × 10⁹ Pa and its density is 1000 kg/m³. What is the speed of sound in water?

**Transverse wave velocity** depends on medium not frequency alone. For string under tension, v ∝ √(T/μ), calculation requires μ from mass and length, then square root evaluation, giving v in m/s, then f = v/λ for given wavelength. Speed: v = √((B/rho)) = √((2.2 × 10⁹/1000)) = √(2.2 × 10⁶) ≈ 1483 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1480 m/s, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Wave Speed, Energy and Power

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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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Second Law Heat Engines and Kelvin-Planck

How much heat is required to vaporize 1.5 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**Second law Kelvin-Planck statement** no process possible whose sole result is absorption of heat from reservoir and complete conversion to work, heat engine must have at least two reservoirs hot and cold, efficiency η = W/Q_h =1 - Q_c/Q_h

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

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Second Law Heat Engines and Kelvin-Planck

How much heat is required to vaporize 0.8 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**Second law Kelvin-Planck statement** no process possible whose sole result is absorption of heat from reservoir and complete conversion to work, heat engine must have at least two reservoirs hot and cold, efficiency η = W/Q_h =1 - Q_c/Q_h

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

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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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Latent Heat and Phase Transitions

Which substance typically has the highest specific heat capacity among common materials?

Water has the highest specific heat capacity (4186J kg−1K−1) among common substances listed (Section 10.6, Table 10.3), making it an effective coolant. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Water. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.