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#aluminium

17 public questions tagged with this topic.

Heat Transfer Work Distinction and Internal Energy Change

How much heat is required to raise the temperature of 0.2 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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Zeroth Law Thermal Equilibrium and Quasi-static

How much heat is required to raise the temperature of 0.5 kg of aluminium from 25^circ C to 55^circ C ? (Specific heat o

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. Δ Q = m s Δ T . m = 0.5 , s = 900 , Δ T = 55 - 25 = 30 . Δ Q = 0.5 × 900 × 30 = 13500 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =

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

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First Law of Thermodynamics Applications

How much heat is required to raise the temperature of 0.45 kg of aluminium from 30^circ C to 60^circ C ? (Specific heat

**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. Δ Q = m s Δ T . m = 0.45 , s = 900 , Δ T = 60 - 30 = 30 . Δ Q = 0.45 × 900 × 30 = 12150 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV,

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

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Latent Heat and Phase Transitions

A 0.15kg aluminium block at 280∘C is placed in 0.7kg water at 22∘C in a 0.05kg lead calorimeter at 22∘C. What is the fin

0.15×900×(280−T) = (0.7×4186+0.05×127.7)×(T−22). 37800−135T = (2930.2+6.385)×(T−22) = 2936.585T−64599.87. 37800+64599.87 = 2936.585T+135T. 102399.87 = 3071.585T⇒T≈33.34∘C≈33.3∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 33.3°C. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Specific Heat and Calorimetry

A 0.1kg aluminium block at 80∘C is placed in 0.5kg water at 25∘C. Calculate the final temperature. (Specific heat of alu

Heat lost = Heat gained. 0.1×900×(80−T) = 0.5×4186×(T−25). 7200−90T = 2093T−52325. 7200+52325 = 2093T+90T. 59525 = 2183T⇒T≈27.27∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 27.3°C. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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

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Thermal Expansion of Solids and Liquids

A 0.2kg aluminium block at 120∘C is placed in 0.8kg of water at 20∘C in a 0.1kg copper calorimeter at 20∘C. What is the

Heat lost = Heat gained. 0.2×900×(120−T) = (0.8×4186+0.1×386)×(T−20). 21600−180T = (3348.8+38.6)×(T−20) = 3387.4T−67748. 21600+67748 = 3387.4T+180T. 89348 = 3567.4T⇒T≈25.04∘C≈25∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 25°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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Temperature Scales and Triple Point

A 0.15kg mercury block at 400∘C is placed in 0.5kg water at 20∘C in a 0.05kg aluminium calorimeter at 20∘C. Find the fin

0.15×140×(400−T) = (0.5×4186+0.05×900)×(T−20). 8400−21T = (2093+45)×(T−20) = 2138T−42760. 8400+42760 = 2138T+21T. 51160 = 2159T⇒T≈23.7∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 23.7°C. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

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

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Temperature Scales and Triple Point

How much heat is required to raise 0.8kg of aluminium from 25∘C to 75∘C if its specific heat capacity is 900J kg−1K−1?

Given: m = 0.8kg, ΔT = 75−25 = 50∘C, s = 900Jkg−1K−1. Q = msΔT = 0.8×900×50 = 36000J = 36kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 36 kJ. 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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Elastic Properties and Applications

An aluminium wire of length 1.5m and cross-sectional area 2×10−6m2 is stretched by a force of 140N. If the Young's modul

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 140×1.52×10−6×7×1010 = 2101.4×105 = 1.5×10−3m = 1.5mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.5mm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Bulk Modulus and Compressibility

An aluminium block of dimensions 0.6m×0.4m×0.2m is subjected to a shearing force of 8×104N. If the shear modulus of alum

Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.6×0.4 = 0.24m2, L = 0.2m. Substitute: Δx = 8×104×0.20.24×2.5×1010 = 160006×109≈2.67×10−6m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.67×10−6m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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