19 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
**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
**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. For an ideal gas, internal energy ( U ) depends only on temperature. In an isothermal process, temperature remains constant ( Δ T = 0 ), so Δ U = 0 . Using first law ΔU = Q - W, W = ∫
Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change
**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. An isothermal process is characterized by constant temperature ( T = constant ). For an ideal gas, this implies P V = constant , with heat exchange balancing work done. 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
**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. First Law: Δ Q = Δ U + Δ W . Δ Q = -670 (heat released), Δ W = -230 (work on system). -670 = Δ U - 230 ⇒ Δ U = -670 + 230 = -440 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R
Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature
**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. First Law: Δ Q = Δ U + Δ W . Δ Q = -760 (heat released), Δ W = 240 (work by system). -760 = Δ U + 240 ⇒ Δ U = -760 - 240 = -1000 J . Using first law ΔU = Q - W, W = ∫ P
Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature
**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. First Law: Δ Q = Δ U + Δ W . Δ Q = -870 (heat released), Δ W = 330 (work by system). -870 = Δ U + 330 ⇒ Δ U = -870 - 330 = -1200 J . Using first law ΔU = Q - W, W = ∫ P
Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature
**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. First Law: Δ Q = Δ U + Δ W . Δ Q = -550 (heat released), Δ W = 200 (work by system). -550 = Δ U + 200 ⇒ Δ U = -550 - 200 = -750 J . Using first law ΔU = Q - W, W = ∫ P
Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static
**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. First Law: Δ Q = Δ U + Δ W . Δ Q = -500 J (heat released), Δ W = -200 J (work on system). -500 = Δ U - 200 ⇒ Δ U = -500 + 200 = -300 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
**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. First Law: Δ Q = Δ U + Δ W . Δ Q = 690 , Δ W = -210 (work on system). 690 = Δ U - 210 ⇒ Δ U = 690 + 210 = 900 J . Using first law ΔU = Q - W,
Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications
