#energy balance

Work Heat Distinction and Energy Transfer Modes

In an adiabatic process, what compensates for the absence of heat transfer?

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). In an adiabatic process ( Δ Q = 0 ), the change in internal energy ( Δ U ) is entirely due to work done ( Δ U = -Δ W ). Work done by or on the system adjusts the internal energy, as no heat is exchanged. Using first law ΔU

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Work Heat Distinction and Energy Transfer Modes

A system absorbs 730 J of heat and performs 190 J of work. What is the change in internal energy?

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). First Law: Δ Q = Δ U + Δ W . Δ Q = 730 , Δ W = 190 (work by system). 730 = Δ U + 190 ⇒ Δ U = 730 - 190 = 540 J . Using first law ΔU = Q - W, W = ∫ P

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Isochoric Processes and Pressure-Temperature

A system in a cyclic process absorbs 900 J of heat and rejects 400 J . What is the net work done?

**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. For cyclic process: Δ U = 0 , W = Q_net . Q_net = Q_absorb - Q_reject = 900 - 400 = 500 J . W = 500 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

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Specific Heat Capacity and Latent Heat

A system in a cyclic process absorbs 650 J of heat and performs 200 J of work. What is the heat rejected?

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . 650 - Q_reject = 200 ⇒ Q_reject = 650 - 200 = 450 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

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Zeroth Law Thermal Equilibrium and Quasi-static

A system in a cyclic process absorbs 980 J of heat and performs 420 J of work. What is the heat rejected?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . 980 - Q_reject = 420 ⇒ Q_reject = 980 - 420 = 560 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

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Zeroth Law Thermal Equilibrium and Quasi-static

A system in a cyclic process absorbs 860 J of heat and performs 340 J of work. What is the heat rejected?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . 860 - Q_reject = 340 ⇒ Q_reject = 860 - 340 = 520 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

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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First Law of Thermodynamics Applications

A system absorbs 450 J of heat and performs 150 J of work. What is the change in internal energy?

**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. First Law: Δ Q = Δ U + Δ W . Δ Q = 450 , Δ W = 150 . 450 = Δ U + 150 ⇒ Δ U = 450 - 150 = 300 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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