5 public questions tagged with this topic.
**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency
Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes
**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. Heat in cal = Heat in J4.186 . (4186)/(4.186) = 1000 cal . 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 1000 cal,
Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature
**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Heat in J = Heat in cal × 4.186 . 250 × 4.186 = 1046.5 J ≈ 1047 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 1047
Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications
**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Heat in cal = Heat in J4.186 . (836)/(4.186) ≈ 199.71 ≈ 200 cal . 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 200 cal, consistent with thermodynamic laws and energy conservation.
Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications
**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. Heat in J = Heat in cal × 4.186 . 300 × 4.186 = 1255.8 J ≈ 1256 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
Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications
