#tungsten

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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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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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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Specific Heat and Calorimetry

A 0.2kg tungsten block at 250∘C is dropped into 0.5kg water at 18∘C in a 0.1kg silver calorimeter at 18∘C. What is the f

0.2×134×(250−T) = (0.5×4186+0.1×236)×(T−18). 6700−26.8T = (2093+23.6)×(T−18) = 2116.6T−38098.8. 6700+38098.8 = 2116.6T+26.8T. 44798.8 = 2143.4T⇒T≈20.9∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 20.9°C. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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