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#copper block

7 public questions tagged with this topic.

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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Temperature Scales and Triple Point

A 0.15kg copper block at 200∘C is dropped into 0.9kg water at 15∘C in a calorimeter of mass 0.1kg (specific heat = 386J

Heat lost = Heat gained. 0.15×386×(200−T) = (0.9×4186+0.1×386)×(T−15). 11580−57.9T = (3767.4+38.6)×(T−15) = 3806T−57090. 11580+57090 = 3806T+57.9T. 68670 = 3863.9T⇒T≈17.77∘C≈17.8∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 17.8°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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Shear Modulus and Rigidity

A copper block of dimensions 0.5m×0.3m×0.1m is subjected to a shearing force of 6×104N. If the shear modulus of copper i

Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.5×0.3 = 0.15m2, L = 0.1m. Substitute: Δx = 6×104×0.10.15×4.2×1010 = 60006.3×109≈9.52×10−7m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.52×10−7m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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