Practice question
Question
A 0.1kg aluminium block at 80∘C is placed in 0.5kg water at 25∘C. Calculate the final temperature. (Specific heat of aluminium = 900J kg−1K−1, water = 4186J kg−1K−1)
Explanation
Heat lost = Heat gained. 0.1×900×(80−T) = 0.5×4186×(T−25). 7200−90T = 2093T−52325. 7200+52325 = 2093T+90T. 59525 = 2183T⇒T≈27.27∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 27.3°C. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
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