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#work done

45 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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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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Isochoric Processes and Pressure-Temperature

A system absorbs 600 J of heat and does 150 J of work. What is the change in internal energy?

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. First Law: Δ Q = Δ U + Δ W . Given Δ Q = 600 J , Δ W = 150 J (work by system). 600 = Δ U + 150 ⇒ Δ U = 600 - 150 = 450 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Specific Heat Capacity and Latent Heat

A system releases 600 J of heat and performs 250 J of work. What is the change in internal energy?

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. First Law: Δ Q = Δ U + Δ W . Δ Q = -600 J (heat released), Δ W = 250 J (work by system). -600 = Δ U + 250 ⇒ Δ U = -600 - 250 = -850 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

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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Second Law Heat Engines and Kelvin-Planck

In an isobaric process, 1 mole of an ideal gas expands from 8 L to 16 L at 360 K . What is the work done by the gas? ( R

**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

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First Law of Thermodynamics Applications

A gas undergoes an isothermal expansion at 300 K from a volume of 2 L to 6 L . If the number of moles of the gas is 0.1

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. For an isothermal process, W = μ R T ln((V₂)/(V₁)) .Substitute: μ = 0.1 , R = 8.3 , T = 300 , V₂ = 6 , V₁ = 2 . W = 0.1 × 8.3 × 300 × ln((6)/(2)) = 249 × ln(3) . ln(3) ≈ 1.0986 , so W ≈ 249 × 1.0986

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

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First Law of Thermodynamics Applications

A system absorbs 500 J of heat and does 200 J of work. What is the change in its internal energy?

**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 .Given: Δ Q = 500 J , Δ W = 200 J . 500 = Δ U + 200 . Δ U = 500 - 200 = 300 J . Using first law ΔU = Q - W,

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

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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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Isobaric and Isothermal Processes Work Calculation

A system absorbs 700 J of heat while 300 J of work is done on it. What is the change in internal energy?

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. First Law: Δ Q = Δ U + Δ W . Δ Q = 700 , Δ W = -300 (work done on system, negative work by system). 700 = Δ U - 300 ⇒ Δ U = 700 + 300 = 1000 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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Orbital Velocity and Time Period

A 5kg mass is moved from RE to 2RE from Earth’s center. What is the work done? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N

W = ΔV = −GMEm(1r2−1r1). r1 = 6.4×106m, r2 = 1.28×107m. W = −6.67×10−11×6×1024×5(11.28×107−16.4×106). W = −2.001×1015(−7.8125×10−8)≈1.56×108J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.6 × 10⁸ J. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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