Questions on how solids and liquids expand when heated, including linear and volume expansion coefficients. Tests understanding of thermal effects on materials and practical implications.
Given: m = 0.15kg, Lv = 2.0×105J kg−1. Q = mLv = 0.15×2.0×105 = 30000J = 30kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 30 kJ. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: L0 = 5cm, L = 5.009cm, αl = 1.8×10−5K−1, T1 = 30∘C. ΔL = 5.009−5 = 0.009cm. ΔL = L0αlΔT⇒0.009 = 5×1.8×10−5×ΔT. ΔT = 0.0095×1.8×10−5 = 100K. T2 = 30+100 = 130∘C.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Q1 = 0.5×2100×20 = 21000J (ice from -20°C to 0°C). Q2 = 0.5×3.35×105 = 167500J (melting). Q3 = 0.5×4186×10 = 20930J (water from 0°C to 10°C). Total: Q = 21000+167500+20930 = 208430J = 208.43kJ.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: m = 0.3kg, Lf = 3.33×105J kg−1. Q = mLf = 0.3×3.33×105 = 99900J = 99.9kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 99.9 kJ. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
Given: r = 8cm, V0 = 43π×83 = 20483πcm3, ΔT = 20−70 = −50∘C, αl = 1.9×10−5K−1. αv = 3αl = 3×1.9×10−5 = 5.7×10−5K−1. ΔV = V0αvΔT = 20483π×5.7×10−5×(−50). ΔV = 20483π×(−2.85×10−3)≈−6.12π≈−19.22cm3 (decrease of 19.22cm3).
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
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.
At the boiling point of water under standard pressure, the Celsius scale reads 100∘C, while the Fahrenheit scale reads 212∘F, reflecting different interval sizes and zero points. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Celsius is 100∘, Fahrenheit is 212∘. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Relation: T = tC+273.15. T = −23+273.15 = 250.15K≈250K. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 250 K. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
The Kelvin scale is offset from the Celsius scale by 273.15, so T = tC+273.15 (Section 10.4), where T is Kelvin and tC is Celsius. As per NCERT, applying relevant law/formula with correct units and sign convention leads to T\=tC+273.15. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Given: m = 0.2kg, Lv = 2.256×106J kg−1. Q = mLv = 0.2×2.256×106 = 451200J = 451.2kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 451.2 kJ. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
0.4×900×(160−T) = (1.2×4186+0.2×127.7)×(T−28). 57600−360T = (5023.2+25.54)×(T−28) = 5048.74T−141364.72. 57600+141364.72 = 5048.74T+360T. 198964.72 = 5408.74T⇒T≈36.78∘C≈36.8∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 36.8°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.
Heat lost = Heat gained. 0.25×236×(150−T) = (1×4186+0.2×900)×(T−25). 8850−59T = (4186+180)×(T−25) = 4366T−109150. 8850+109150 = 4366T+59T. 118000 = 4425T⇒T≈26.67∘C≈26.7∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 26.7°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.
