#specific heat

Heat Transfer Work Distinction and Internal Energy Change

What is the molar specific heat capacity at constant volume for a diatomic gas if R = 8.3 J mol⁻¹ K⁻¹ ?

**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. For diatomic gas: C_v = (5)/(2) R . C_v = (5)/(2) × 8.3 = 20.75 J mol⁻¹ K⁻¹ . 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

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Heat Transfer Work Distinction and Internal Energy Change

How much heat is required to raise the temperature of 0.25 kg of tungsten from 15^circ C to 45^circ C ? (Specific heat o

**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. Δ Q = m s Δ T . m = 0.25 , s = 134.4 , Δ T = 45 - 15 = 30 . Δ Q = 0.25 × 134.4 × 30 = 1008 J . Using first law ΔU = Q - W, W = ∫ P

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Heat Transfer Work Distinction and Internal Energy Change

0.5 kg of a substance at 15°C absorbs 1800 J of heat at constant volume, reaching 45°C. What is its specific heat capaci

**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. Specific heat: s = (Δ Q)/(m Δ T) . Δ Q = 1800 J , m = 0.5 kg , Δ T = 45 - 15 = 30 K . s = (1800)/(0.5 × 30) = 120 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W,

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Heat Transfer Work Distinction and Internal Energy Change

How much heat is required to raise the temperature of 0.2 kg of aluminium from 45^circ C to 75^circ C ? (Specific heat o

**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. Δ Q = m s Δ T . m = 0.2 , s = 900 , Δ T = 75 - 45 = 30 . Δ Q = 0.2 × 900 × 30 = 5400 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 =

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Cyclic Processes and Reversibility Concepts

What is the molar specific heat capacity at constant pressure for a monatomic gas if R = 8.3 J mol⁻¹ K⁻¹ ?

**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. For monatomic gas: C_v = (3)/(2) R , C_p = C_v + R . C_v = (3)/(2) × 8.3 = 12.45 . C_p = 12.45 + 8.3 = 20.75 J mol⁻¹ K⁻¹ . 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 20.75 J mol⁻¹ K⁻¹, consistent

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Cyclic Processes and Reversibility Concepts

How much heat is required to raise the temperature of 0.4 kg of carbon from 10^circ C to 30^circ C ? (Specific heat of c

**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. Δ Q = m s Δ T . m = 0.4 , s = 600 , Δ T = 30 - 10 = 20 . Δ Q = 0.4 × 600 × 20 = 4800 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 4800

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Cyclic Processes and Reversibility Concepts

What is the molar specific heat capacity at constant volume for a solid predicted by the law of equipartition? ( R = 8.3

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. For a solid: C = 3R (from equipartition, 3 degrees of freedom). C = 3 × 8.3 = 24.9 J mol⁻¹ K⁻¹ . 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 24.9 J mol⁻¹ K⁻¹, consistent with thermodynamic laws and energy conservation.

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Cyclic Processes and Reversibility Concepts

What is the molar specific heat capacity at constant volume for a monatomic gas if R = 8.3 J mol⁻¹ K⁻¹ ?

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. For monatomic gas: C_v = (3)/(2) R . C_v = (3)/(2) × 8.3 = 12.45 J mol⁻¹ K⁻¹ . 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 12.45 J mol⁻¹ K⁻¹, consistent with thermodynamic laws and energy conservation.

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Isochoric Processes and Pressure-Temperature

How much heat is required to raise the temperature of 1 kg of copper from 20^circ C to 50^circ C ? (Specific heat of cop

**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. Heat capacity: Δ Q = m s Δ T . m = 1 kg , s = 386.4 J kg⁻¹ K⁻¹ , Δ T = 50 - 20 = 30 K . Δ Q = 1 × 386.4 × 30 = 11592 J . Using first law ΔU = Q - W, W = ∫ P dV,

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Isochoric Processes and Pressure-Temperature

What is the molar specific heat capacity at constant pressure for a diatomic gas if C_v = 20.75 J mol⁻¹ K⁻¹ and R = 8.3

**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. C_p - C_v = R . C_p = C_v + R = 20.75 + 8.3 = 29.05 J mol⁻¹ K⁻¹ ≈ 29.1 J mol⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P

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Isochoric Processes and Pressure-Temperature

What is the molar specific heat capacity at constant volume for a diatomic gas if R = 8.3 J mol⁻¹ K⁻¹ ?

**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. For diatomic gas: C_v = (5)/(2) R . C_v = (5)/(2) × 8.3 = 20.75 J mol⁻¹ K⁻¹ . 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 20.75 J mol⁻¹ K⁻¹, consistent with thermodynamic laws and energy conservation.

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Specific Heat Capacity and Latent Heat

In an isobaric process, 0.6 moles of gas expand from 400 K to 480 K . What is the heat supplied if C_p = 25.0 J mol⁻¹ K⁻

**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. Δ Q = μ C_p Δ T . μ = 0.6 , C_p = 25.0 , Δ T = 480 - 400 = 80 . Δ Q = 0.6 × 25.0 × 80 = 1200 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic

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