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#final temperature

10 public questions tagged with this topic.

Heat Transfer Work Distinction and Internal Energy Change

0.3 moles of an ideal gas at 400 K expand adiabatically from 6 atm to 2 atm. If gamma = 1.5 , what is the final temperat

**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. Adiabatic: T₁ V₁^γ-1 = T₂ V₂^γ-1 , P V = μ R T ⇒ V₁ = (μ R T₁)/(P₁) = (0.3 × 8.3 × 400)/(6) = 166 L , V₂ = (0.3 × 8.3 × T₂)/(2) = 1.245 T₂ . 400 × 166⁰.5 = T₂ × (1.245 T₂)⁰.5 .

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

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Adiabatic Processes and Gamma Determination

In an adiabatic process, 1 mole of gas at 600 K does 2000 J of work. What is the final temperature? ( gamma = 1.4 , R =

**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. W = (μ R (T₁ - T₂))/(γ - 1) . 2000 = (1 × 8.3 × (600 - T₂))/(1.4 - 1) . 2000 = (8.3 × (600 - T₂))/(0.4) . 2000 × 0.4 = 8.3 × (600 - T₂) ⇒ 800 = 8.3 × (600 - T₂) . 600 - T₂ = (800)/(8.3) ≈ 96.39 . T₂ = 600

Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination

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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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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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Temperature Scales and Triple Point

A 0.05kg gold block at 1500∘C is placed in 0.6kg water at 25∘C in a 0.1kg copper calorimeter at 25∘C. Find the final tem

0.05×134×(1500−T) = (0.6×4186+0.1×386)×(T−25). 10050−6.7T = (2511.6+38.6)×(T−25) = 2550.2T−63755. 10050+63755 = 2550.2T+6.7T. 73805 = 2556.9T⇒T≈28.87∘C≈28.9∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 28.9°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

A 0.15kg copper block at 200∘C is dropped into 0.9kg water at 15∘C in a calorimeter of mass 0.1kg (specific heat = 386J

Heat lost = Heat gained. 0.15×386×(200−T) = (0.9×4186+0.1×386)×(T−15). 11580−57.9T = (3767.4+38.6)×(T−15) = 3806T−57090. 11580+57090 = 3806T+57.9T. 68670 = 3863.9T⇒T≈17.77∘C≈17.8∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 17.8°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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