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#specific heat capacity

9 public questions tagged with this topic.

Latent Heat and Phase Transitions

Which substance typically has the highest specific heat capacity among common materials?

Water has the highest specific heat capacity (4186J kg−1K−1) among common substances listed (Section 10.6, Table 10.3), making it an effective coolant. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Water. 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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Latent Heat and Phase Transitions

How much heat is required to raise the temperature of 2kg of water from 20∘C to 50∘C? (Specific heat capacity of water =

Given: m = 2kg, ΔT = 50−20 = 30∘C, s = 4186Jkg−1K−1. Heat required: ΔQ = msΔT = 2×4186×30 = 251160J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 251.16 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.

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Latent Heat and Phase Transitions

Why does a calorimeter use a material with low specific heat capacity for its container?

A material with low specific heat capacity absorbs less heat, minimizing its effect on the heat exchange between the substances being measured, ensuring accurate results. As per NCERT, applying relevant law/formula with correct units and sign convention leads to To absorb less heat. 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 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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Temperature Scales and Triple Point

How much heat is required to raise 0.4kg of water from 15∘C to 85∘C and then convert 0.25kg to steam at 100∘C in a 0.2kg

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.

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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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