#heat transfer

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

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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 fir

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Heat Transfer Work Distinction and Internal Energy Change

A system in a cyclic process performs 300 J of work and rejects 200 J of heat. What is the heat absorbed?

**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. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . Q_absorb - 200 = 300 ⇒ Q_absorb = 500 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

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Heat Transfer Work Distinction and Internal Energy Change

A gas undergoes an isothermal compression from 8 L to 2 L at 350 K with 0.2 moles . What is the heat released? ( R = 8.3

**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. For isothermal: W = μ R T ln((V₂)/(V₁)) , Δ U = 0 , Q = W . W = 0.2 × 8.3 × 350 × ln((2)/(8)) = 581 × ln(0.25) . ln(0.25) = -ln(4) ≈ -1.386 . W = 581 × (-1.386) ≈ -805 J . Q = -805 J (negative implies heat released). Using first law ΔU = Q - W, W = ∫

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Cyclic Processes and Reversibility Concepts

A system releases 700 J of heat and does 250 J of work. What is 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. First Law: Δ Q = Δ U + Δ W . Δ Q = -700 J (heat released), Δ W = 250 J (work by system). -700 = Δ U + 250 ⇒ Δ U = -700 - 250 = -950 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 =

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

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Cyclic Processes and Reversibility Concepts

Which process is characterized by no heat exchange between the system and its surroundings?

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. An adiabatic process is defined by no heat transfer ( Δ Q = 0 ) between the system and surroundings. This occurs when the system is insulated or the process is too rapid for heat exchange. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV,

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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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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 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

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Zeroth Law Thermal Equilibrium and Quasi-static

A system absorbs 800 J of heat and has 350 J of work done on it. 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 = 800 , Δ W = -350 (work done on system). 800 = Δ U - 350 ⇒ Δ U = 800 + 350 = 1150 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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Zeroth Law Thermal Equilibrium and Quasi-static

A system releases 790 J of heat and has 310 J of work done on it. 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 = -790 (heat released), Δ W = -310 (work on system). -790 = Δ U - 310 ⇒ Δ U = -790 + 310 = -480 J . Using first law ΔU = Q - W, W = ∫ P

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Zeroth Law Thermal Equilibrium and Quasi-static

A system releases 810 J of heat and has 270 J of work done on it. What is the change in internal energy?

**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. First Law: Δ Q = Δ U + Δ W . Δ Q = -810 (heat released), Δ W = -270 (work on system). -810 = Δ U - 270 ⇒ Δ U = -810 + 270 = -540 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

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

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