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Specific Heat and Calorimetry

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25 questions

Specific Heat and Calorimetry

A 0.3kg lead block at 400∘C is placed in 0.7kg water at 15∘C in a 0.1kg aluminium calorimeter at 15∘C. Find the final te

0.3×127.7×(400−T) = (0.7×4186+0.1×900)×(T−15). 15324−38.31T = (2930.2+90)×(T−15) = 3020.2T−45303. 15324+45303 = 3020.2T+38.31T. 60627 = 3058.51T⇒T≈19.82∘C≈19.8∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 19.8°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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Specific Heat and Calorimetry

Which gas law relates pressure and volume when temperature is held constant?

Boyle’s Law describes the inverse relationship between pressure and volume of an ideal gas at constant temperature (PV = constant). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Boyle’s Law. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Specific Heat and Calorimetry

Which law explains why the volume of a gas increases when its temperature rises at constant pressure?

Charles’ Law states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature (V/T = constant, Section 10.4). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Charles’ Law. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Specific Heat and Calorimetry

Which property of a substance determines the amount of heat required to change its temperature by 1∘C per unit mass?

Specific heat capacity (s) is defined as the heat required per unit mass to change the temperature by 1∘C (Section 10.6, s = 1mΔQΔT). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Specific heat capacity. 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 brass rod of length 2.5m at 20∘C is heated to 220∘C. If its cross-sectional area increases by 0.018cm2, what was its o

Given: ΔT = 220−20 = 200∘C, ΔA = 0.018cm2, αl = 1.8×10−5K−1. Area expansion: ΔA = A0×2αlΔT. 0.018 = A0×2×1.8×10−5×200. 0.018 = A0×7.2×10−3⇒A0 = 0.0187.2×10−3 = 2.5cm2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5 cm². 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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Specific Heat and Calorimetry

How much ice at 0∘C will melt if 16744J of heat is supplied? (Latent heat of fusion of ice = 3.35×105J kg−1)

Given: Q = 16744J, Lf = 3.35×105J kg−1. Q = mLf⇒m = QLf = 167443.35×105≈0.05kg = 50g. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 50 g. 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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Specific Heat and Calorimetry

A steel plate has an area of 1.2m2 at 45∘C. What is the decrease in area when cooled to 5∘C? (αl\=1.2×10−5K−1)

Given: A0 = 1.2m2, ΔT = 5−45 = −40∘C, αl = 1.2×10−5K−1. Area expansion: ΔA = A0×2αlΔT = 1.2×2×1.2×10−5×(−40). ΔA = 1.2×2.4×10−5×(−40) = −0.001152m2 (decrease of 0.001152m2). As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.001152 m². This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Specific Heat and Calorimetry

A gas at 300K and 1atm occupies 3L. If it is heated to 450K while the volume is adjusted to 4.5L, what is the final pres

Given: T1 = 300K, P1 = 1atm, V1 = 3L, T2 = 450K, V2 = 4.5L. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 1×34.5×450300 = 1×23×1.5 = 1atm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1 atm. 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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