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5 public questions tagged with this topic.

Second Law Heat Engines and Kelvin-Planck

How much heat is required to raise the temperature of 0.3 kg of lead from 35^circ C to 65^circ C ? (Specific heat of lea

**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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Isobaric and Isothermal Processes Work Calculation

What is the change in internal energy when 0.1 kg of lead is heated from 20^circ C to 40^circ C ? (Specific heat of lead

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. Δ U = m s Δ T (no work in constant volume or solid). m = 0.1 , s = 127.7 , Δ T = 40 - 20 = 20 . Δ U = 0.1 × 127.7 × 20 = 255.4 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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Isobaric and Isothermal Processes Work Calculation

How much heat is required to raise the temperature of 0.2 kg of lead from 50^circ C to 80^circ C ? (Specific heat of lea

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. Δ Q = m s Δ T . m = 0.2 , s = 127.7 , Δ T = 80 - 50 = 30 . Δ Q = 0.2 × 127.7 × 30 = 766.2 J ≈ 766 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

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

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