Practice question
Question
A 0.3kg lead block at 350∘C is placed in 0.6kg water at 25∘C. Find the final temperature. (Specific heat of lead = 127.7J kg−1K−1, water = 4186J kg−1K−1)
Explanation
0.3×127.7×(350−T) = 0.6×4186×(T−25). 13408.5−38.31T = 2511.6T−62790. 13408.5+62790 = 2511.6T+38.31T. 76198.5 = 2549.91T⇒T≈29.89∘C≈29.9∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 29.9°C. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.