#heat exchange

Ideal Gas Laws and Thermal Properties

A 0.5kg silver block at 240∘C is placed in 1.5kg water at 21∘C in a 0.25kg copper calorimeter at 21∘C. What is the final

0.5×236×(240−T) = (1.5×4186+0.25×386)×(T−21). 28320−118T = (6279+96.5)×(T−21) = 6375.5T−133885.5. 28320+133885.5 = 6375.5T+118T. 162205.5 = 6493.5T⇒T≈24.97∘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 B is scientifically correct.

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

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Ideal Gas Laws and Thermal Properties

A 0.1kg iron block at 200∘C is dropped into 0.4kg water at 30∘C in a 0.05kg copper calorimeter at 30∘C. Find the final t

0.1×450×(200−T) = (0.4×4186+0.05×386)×(T−30). 9000−45T = (1674.4+19.3)×(T−30) = 1693.7T−50811. 9000+50811 = 1693.7T+45T. 59811 = 1738.7T⇒T≈34.4∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 34.4°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.

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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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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 0.05kg gold block at 1200∘C is placed in 0.5kg water at 20∘C. What is the final temperature? (Specific heat of gold =

0.05×134×(1200−T) = 0.5×4186×(T−20). 8040−6.7T = 2093T−41860. 8040+41860 = 2093T+6.7T. 49900 = 2099.7T⇒T≈23.77∘C≈23.8∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 23.8°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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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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Thermal Expansion of Solids and Liquids

Why does water in a calorimeter reach a steady temperature when mixed with a hot object?

In calorimetry (Section 10.7), heat lost by the hot object equals heat gained by the water and calorimeter at thermal equilibrium, where no further heat transfer occurs. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Heat lost equals heat gained. 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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Thermodynamics

Which process is characterized by no heat exchange?

Adiabatic correctly identifies the characteristic feature or property described in this question. In Thermodynamics, specific characteristics define and distinguish biological molecules, organisms, or processes from one another. The feature described by Adiabatic is a defining property that arises from its unique molecular structure, evolutionary history, or physiological role. The other options (Isothermal, Isobaric, and Isochoric) describe characteristics of different entities, represent incorrect properties, or apply to the subject under different conditions.

Ref: Lehninger Principles of Biochemistry, Nelson & Cox, 8th Ed., Ch. 1

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