This category includes questions on different temperature scales, the concept of the triple point, and how these principles apply in thermodynamics and physical systems.
Given: r = 25cm, A0 = πr2 = π×252 = 625πcm2, ΔT = 130−30 = 100∘C, αl = 1.2×10−5K−1. ΔA = A0×2αlΔT = 625π×2×1.2×10−5×100. ΔA = 625π×2.4×10−3 = 1.5πcm2≈4.71cm2 (π≈3.14).
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Q1 = 0.2×2100×15 = 6300J (ice to 0°C). Q2 = 0.2×3.35×105 = 67000J (melting). Q3 = 0.2×4186×100 = 83720J (water to 100°C). Q4 = 0.2×2.256×106 = 451200J (vaporization). Total: Q = 6300+67000+83720+451200 = 608220J = 608.22kJ.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: P1 = 1.9atm, T1 = 67∘C = 340K, V1 = 6.5L, P2 = 2.5atm, T2 = 117∘C = 390K. P1V1T1 = P2V2T2. V2 = V1×P1P2×T2T1 = 6.5×1.92.5×390340 = 6.5×0.76×1.1471≈5.66L.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
The Kelvin scale starts at absolute zero, the theoretical point where all molecular motion ceases, making it an absolute temperature scale. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Kelvin. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
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.
Q1 = (0.4×4186+0.2×236)×(85−15) = (1674.4+47.2)×70 = 1721.6×70 = 120512J (to 85°C). Q2 = (0.4×4186+0.2×236)×(100−85) = 1721.6×15 = 25824J (to 100°C). Q3 = 0.25×2.256×106 = 564000J (vaporization). Total: Q = 120512+25824+564000 = 710336J = 710.34kJ.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: L0 = 80cm, ΔL = −0.0136cm, αl = 1.7×10−5K−1, T1 = 40∘C. ΔL = L0αlΔT⇒−0.0136 = 80×1.7×10−5×ΔT. ΔT = −0.013680×1.7×10−5 = −0.01361.36×10−3 = −10K. T2 = 40−10 = 30∘C.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Q1 = 0.1×140×(39) = 546J (solid from -39°C to 0°C). Q2 = 0.1×0.12×105 = 1200J (melting). Q3 = 0.1×140×357 = 4998J (liquid from 0°C to 357°C). Total: Q = 546+1200+4998 = 6744J = 6.744kJ.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
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.
The heat supplied during melting (latent heat of fusion) is used to change the state from solid to liquid, overcoming intermolecular forces, not to increase temperature (Section 10.8). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Heat is used to change the state, not raise temperature. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Q1 = 0.25×2100×30 = 15750J (ice to 0°C). Q2 = 0.25×3.35×105 = 83750J (melting). Q3 = 0.25×4186×100 = 104650J (water to 100°C). Q4 = 0.25×2.256×106 = 564000J (vaporization). Q5 = 0.25×4186×20 = 20930J (steam to 120°C). Total: Q = 15750+83750+104650+564000+20930 = 789080J = 789.08kJ.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: P1 = 2.4atm, T1 = 87∘C = 360K, V1 = 7L, T2 = 37∘C = 310K, P2 = 3atm. P1V1T1 = P2V2T2. V2 = V1×P1P2×T2T1 = 7×2.43×310360 = 7×0.8×0.8611≈4.82L. Percentage change: V1−V2V1×100 = 7−4.827×100 = 2.187×100≈31.14% (decrease).
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
