This category covers the fundamentals of the Second Law of Thermodynamics, focusing on heat engines and the Kelvin‑Planck statement. It explains how energy conversion limits are defined and why a cyclic engine cannot convert all heat into work.
**Heat engine** absorbs Q_h from hot reservoir T_h, does work W, rejects Q_c to cold T_c, efficiency η = W/Q_h =1 - Q_c/Q_h, second law limits η
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
**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
**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. Intensive variables (e.g., pressure, temperature) do not depend on the system’s size or amount, remaining unchanged when the system is divided. Extensive variables (e.g., volume, internal energy) scale with the system’s size. 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
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
**Heat engine** absorbs Q_h from hot reservoir T_h, does work W, rejects Q_c to cold T_c, efficiency η = W/Q_h =1 - Q_c/Q_h, second law limits η
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
**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
**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. Adiabatic: T₁ V₁^γ-1 = T₂ V₂^γ-1 , V₁ = (μ R T₁)/(P₁) = (0.4 × 8.3 × 340)/(7) ≈ 161.37 L , V₂ = (0.4 × 8.3 × T₂)/(1) = 3.32 T₂ . 340 × 161.37⁰.4 = T₂ × (3.32 T₂)⁰.4 . Approximate: T₂ ≈ 245 K . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
**Heat engine** absorbs Q_h from hot reservoir T_h, does work W, rejects Q_c to cold T_c, efficiency η = W/Q_h =1 - Q_c/Q_h, second law limits η
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
**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
**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 9 × 30^γ = 1 × 90^γ . 9 = ((90)/(30))^γ ⇒ 9 = 3^γ . 3^γ = 3² ⇒ γ = 2 . 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
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
**Heat engine** absorbs Q_h from hot reservoir T_h, does work W, rejects Q_c to cold T_c, efficiency η = W/Q_h =1 - Q_c/Q_h, second law limits η
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
**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
**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. For an ideal gas, the molar specific heat at constant pressure ( C_p ) exceeds that at constant volume ( C_v ) by the gas constant ( R ), due to the work done during expansion at constant pressure: C_p - C_v = R . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
