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#temperature change

38 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A gas at 11 atm and 90^circ C in a 10 L container is cooled isochorically to 30^circ C . What is the final pressure?

**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 isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 11 atm , T₁ = 90 + 273 = 363 K , T₂ = 30 + 273 = 303 K . (11)/(363) = (P₂)/(303) ⇒ P₂ = (11 × 303)/(363) ≈ 9.18 atm . Using first law ΔU = Q - W, W =

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 monatomic gas undergoes an adiabatic expansion from 720 K to 360 K with 1.5 moles . What is the work done? ( R = 8.3 J

**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 1.5 , R = 8.3 , T₁ = 720 , T₂ = 360 , γ = 1.67 . W = (1.5 × 8.3 × (720 - 360))/(1.67 - 1) = (12.45

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

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,

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

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Cyclic Processes and Reversibility Concepts

A gas at 6 atm in an 8 L container is cooled from 50°C to 10°C at constant volume. What is the final pressure?

**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. Isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 6 atm , T₁ = 50 + 273 = 323 K , T₂ = 10 + 273 = 283 K . (6)/(323) = (P₂)/(283) ⇒ P₂ = (6 × 283)/(323) ≈ 5.26 atm . 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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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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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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Specific Heat Capacity and Latent Heat

A gas at 9 atm and 70^circ C in a 6 L container is cooled isochorically to 10^circ C . What is the final pressure?

**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. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 9 atm , T₁ = 70 + 273 = 343 K , T₂ = 10 + 273 = 283 K . (9)/(343) = (P₂)/(283) ⇒ P₂ = (9 × 283)/(343) ≈ 7.42 atm . Using first law ΔU = Q -

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Specific Heat Capacity and Latent Heat

An ideal gas absorbs 1000 J of heat in an isochoric process, increasing its temperature by 20 K . What is the number of

**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. For isochoric process: Δ Q = μ C_v Δ T . Δ Q = 1000 , C_v = 20 , Δ T = 20 . 1000 = μ × 20 × 20 ⇒ μ = (1000)/(400) = 2.5 moles . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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Zeroth Law Thermal Equilibrium and Quasi-static

A diatomic gas undergoes an adiabatic expansion from 860 K to 430 K with 0.5 moles . What is the work done? ( R = 8.3 J

**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.5 , R = 8.3 , T₁ = 860 , T₂ = 430 , γ = 1.4 . W = (0.5 × 8.3 × (860 - 430))/(1.4 - 1) = (4.15 × 430)/(0.4) = 4467.5 J ≈ 4468 J . Using first law ΔU = Q - W,

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

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the change in internal energy when 1 mole of an ideal gas is heated from 300 K to 350 K at constant volume? ( C_

**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. Δ U = μ C_v Δ T . μ = 1 , C_v = 20.8 , Δ T = 350 - 300 = 50 . Δ U = 1 × 20.8 × 50 = 1040 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

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

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Zeroth Law Thermal Equilibrium and Quasi-static

A gas at 5 atm in a 6 L container is heated from 20°C to 60°C at constant volume. What is the final pressure?

**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. Isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 5 atm , T₁ = 20 + 273 = 293 K , T₂ = 60 + 273 = 333 K . (5)/(293) = (P₂)/(333) ⇒ P₂ = (5 × 333)/(293) ≈ 5.68 atm . 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 =

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

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Second Law Heat Engines and Kelvin-Planck

0.25 kg of a substance at 30°C absorbs 1500 J of heat, reaching 60°C. What is its specific heat capacity?

**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. Specific heat: s = (Δ Q)/(m Δ T) . Δ Q = 1500 J , m = 0.25 kg , Δ T = 60 - 30 = 30 K . s = (1500)/(0.25 × 30) = 200 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

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