Practice question
Question
A 0.25kg silver block at 150∘C is placed in 1kg of water at 25∘C in a 0.2kg aluminium calorimeter at 25∘C. What is the final temperature? (Specific heat of silver = 236J kg−1K−1, water = 4186J kg−1K−1, aluminium = 900Jkg−1K−1)
Explanation
Heat lost = Heat gained. 0.25×236×(150−T) = (1×4186+0.2×900)×(T−25). 8850−59T = (4186+180)×(T−25) = 4366T−109150. 8850+109150 = 4366T+59T. 118000 = 4425T⇒T≈26.67∘C≈26.7∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 26.7°C. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
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