#calorimeter

Ideal Gas Laws and Thermal Properties

How much heat is required to raise 0.6kg of water from 25∘C to 75∘C and then convert 0.4kg of it to steam at 100∘C in a

Q1 = (0.6×4186+0.1×386)×(75−25) = (2511.6+38.6)×50 = 2550.2×50 = 127510J (to 75°C). Q2 = (0.6×4186+0.1×386)×(100−75) = 2550.2×25 = 63755J (to 100°C). Q3 = 0.4×2.256×106 = 902400J (vaporization). Total: Q = 127510+63755+902400 = 1093665J = 1093.67kJ.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Ideal Gas Laws and Thermal Properties

What determines the final temperature when two objects of different temperatures are mixed in a calorimeter?

The final temperature is determined by the principle of conservation of energy, where heat lost by the hotter object equals heat gained by the colder one, reaching thermal equilibrium. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Conservation of heat energy. 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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Latent Heat and Phase Transitions

A 0.15kg aluminium block at 280∘C is placed in 0.7kg water at 22∘C in a 0.05kg lead calorimeter at 22∘C. What is the fin

0.15×900×(280−T) = (0.7×4186+0.05×127.7)×(T−22). 37800−135T = (2930.2+6.385)×(T−22) = 2936.585T−64599.87. 37800+64599.87 = 2936.585T+135T. 102399.87 = 3071.585T⇒T≈33.34∘C≈33.3∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 33.3°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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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

A 0.45kg copper block at 200∘C is placed in 1.1kg water at 23∘C in a 0.2kg brass calorimeter at 23∘C. What is the final

Heat lost = Heat gained. 0.45×386×(200−T) = (1.1×4186+0.2×386)×(T−23). 34740−173.7T = (4604.6+77.2)×(T−23) = 4681.8T−107678.4. 34740+107678.4 = 4681.8T+173.7T. 142418.4 = 4855.5T⇒T≈29.33∘C≈29.3∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 29.3°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 0.2kg 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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Thermal Expansion of Solids and Liquids

A 0.2kg aluminium block at 120∘C is placed in 0.8kg of water at 20∘C in a 0.1kg copper calorimeter at 20∘C. What is the

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.

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Temperature Scales and Triple Point

A 0.15kg mercury block at 400∘C is placed in 0.5kg water at 20∘C in a 0.05kg aluminium calorimeter at 20∘C. Find the fin

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.

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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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