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#heat calculation

48 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

In an isobaric process, 0.5 moles of gas expand from 300 K to 450 K . What is the heat supplied if C_p = 29.1 J mol⁻¹ K⁻

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. Δ Q = μ C_p Δ T . μ = 0.5 , C_p = 29.1 , Δ T = 450 - 300 = 150 . Δ Q = 0.5 × 29.1 × 150 = 2182.5 J . Using first law ΔU = Q - W, W = ∫ P dV,

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Work Heat Distinction and Energy Transfer Modes

How many calories are equivalent to 1674 J of heat? (1 cal = 4.186 J )

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. Heat in cal = Heat in J4.186 . (1674)/(4.186) ≈ 399.9 ≈ 400 cal . 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 -

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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

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

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

In an isobaric process, 0.9 moles of gas expand from 360 K to 450 K . What is the heat supplied if C_p = 25.5 J mol⁻¹ K⁻

**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. Δ Q = μ C_p Δ T . μ = 0.9 , C_p = 25.5 , Δ T = 450 - 360 = 90 . Δ Q = 0.9 × 25.5 × 90 = 2065.5 J ≈ 2066 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

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

How much heat is required to vaporize 1.2 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**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 L . m = 1.2 , L = 2256 . Δ Q = 1.2 × 2256 = 2707.2 J ≈ 2707 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 2707 J, consistent with thermodynamic laws and energy conservation.

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

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Cyclic Processes and Reversibility Concepts

How much heat is required to raise the temperature of 0.6 kg of silver from 20^circ C to 50^circ C ? (Specific heat of s

**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 = 0.6 , s = 236.1 , Δ T = 50 - 20 = 30 . Δ Q = 0.6 × 236.1 × 30 = 4249.8 J ≈ 4250 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,

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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,

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

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Isochoric Processes and Pressure-Temperature

0.2 moles of an ideal gas at 360 K are compressed isothermally from 8 L to 2 L. What is the heat released? ( R = 8.3 J m

**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. Isothermal: Δ U = 0 , Δ Q = Δ W . W = μ R T ln((V₂)/(V₁)) . μ = 0.2 , T = 360 , V₂ = 2 , V₁ = 8 . W = 0.2 × 8.3 × 360 × ln((2)/(8)) = 597.6 × (-1.386) ≈ -829 J (work by

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

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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

How much heat is required to raise the temperature of 0.4 kg of tungsten from 40^circ C to 70^circ C ? (Specific heat 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. Δ Q = m s Δ T . m = 0.4 , s = 134.4 , Δ T = 70 - 40 = 30 . Δ Q = 0.4 × 134.4 × 30 = 1612.8 J ≈ 1613 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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