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#first law

19 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A gas expands adiabatically, doing 360 J of work. What is the change in its internal energy?

**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

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Work Heat Distinction and Energy Transfer Modes

A system in a cyclic process absorbs 940 J of heat and rejects 360 J . What is the net work done?

**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). For cyclic: Δ U = 0 , Q_net = W . Q_net = Q_absorb - Q_reject = 940 - 360 = 580 J . W = 580 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

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Heat Transfer Work Distinction and Internal Energy Change

A system absorbs 850 J of heat and has 300 J of work done on it. What is the change in internal energy?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. First Law: Δ Q = Δ U + Δ W . Δ Q = 850 J , Δ W = -300 J (work on system). 850 = Δ U - 300 ⇒ Δ U = 850 + 300 = 1150 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

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Heat Transfer Work Distinction and Internal Energy Change

A gas is compressed adiabatically, doing 300 J of work on the system. What is the change in internal energy?

**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 adiabatic ( Δ Q = 0 ), First Law: Δ U = -Δ W . Work on system: Δ W = -300 J (negative by convention). Δ U = -(-300) = 300 J . Using first law ΔU = Q - W,

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

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Cyclic Processes and Reversibility Concepts

In a cyclic process, what is true about the change in internal energy?

**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. In a cyclic process, the system returns to its initial state. Since internal energy ( U ) is a state variable, its change ( Δ U ) is zero over a complete cycle, regardless of the path taken. 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 > Cyclic Processes and Reversibility Concepts

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Isochoric Processes and Pressure-Temperature

A system releases 670 J of heat and has 230 J of work done on it. What is the change in internal energy?

**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

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Isochoric Processes and Pressure-Temperature

A system releases 760 J of heat and performs 240 J of work. What is the change in internal energy?

**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

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Specific Heat Capacity and Latent Heat

Which of the following correctly describes the First Law of Thermodynamics?

**Specific heat capacity** c = Q/(m ΔT) (J/kg·K), molar C = Q/(n ΔT), heat required to raise temperature, Q = m c ΔT, for water c=4186 J/kg·K, latent heat L = Q/m for phase change at constant temperature, fusion L_f and vaporization L_v, Q = m L, e.g., ice melting L_f=3.34×10⁵ J/kg, water vaporization 2.26×10⁶ J/kg. The First Law ( Δ Q = Δ U + Δ W ) states that heat added equals the increase in internal energy plus work done by the system, a form of energy conservation. Option B is correct. Using first law ΔU = Q - W, W = ∫

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

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the primary implication of the First Law of Thermodynamics?

**Quasi-static process** infinitely slow, system always near equilibrium, reversible, can be represented as continuous path on P-V diagram, non-quasi-static rapid process non-equilibrium, work W = ∫ P_ext dV, for quasi-static P_ext = P_system, work = ∫ P dV, zeroth law ensures temperature defined throughout quasi-static. The First Law of Thermodynamics is a statement of energy conservation: Δ Q = Δ U + Δ W . It implies that the total energy supplied to a system (as heat) equals the increase in internal energy plus the work done by the system. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

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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Zeroth Law Thermal Equilibrium and Quasi-static

A system releases 550 J of heat and performs 200 J of work. What is the change in internal energy?

**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

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Second Law Heat Engines and Kelvin-Planck

A system releases 500 J of heat and has 200 J of work done on it. What is the change in internal energy?

**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

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