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#temperature rise

16 public questions tagged with this topic.

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 =

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 0.5 kg of carbon from 25^circ C to 55^circ C ? (Specific heat of c

**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. Δ Q = m s Δ T . m = 0.5 , s = 600 , Δ T = 55 - 25 = 30 . Δ Q = 0.5 × 600 × 30 = 9000 J . Using first law ΔU = Q

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

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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,

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

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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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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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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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Second Law Heat Engines and Kelvin-Planck

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

**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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First Law of Thermodynamics Applications

0.1 kg of a substance absorbs 1200 J of heat, increasing its temperature from 20°C to 50°C. What is its specific heat ca

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. Specific heat: s = (Δ Q)/(m Δ T) . Given Δ Q = 1200 J , m = 0.1 kg , Δ T = 50 - 20 = 30 K . s = (1200)/(0.1 × 30) = 400 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV,

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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First Law of Thermodynamics Applications

How much heat is required to raise the temperature of 0.25 kg of tungsten from 25^circ C to 55^circ C ? (Specific heat o

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Δ Q = m s Δ T . m = 0.25 , s = 134.4 , Δ T = 55 - 25 = 30 . Δ Q = 0.25 × 134.4 × 30 = 1008 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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First Law of Thermodynamics Applications

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

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. Δ Q = m s Δ T . m = 0.3 , s = 236.1 , Δ T = 65 - 35 = 30 . Δ Q = 0.3 × 236.1 × 30 = 2124.9 J ≈ 2125 J . Using first law ΔU = Q - W,

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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