Practice question
Question
A 0.15kg aluminium block at 280∘C is placed in 0.7kg water at 22∘C in a 0.05kg lead calorimeter at 22∘C. What is the final temperature? (Specific heat of aluminium = 900J kg−1K−1, water = 4186J kg−1K−1, lead = 127.7Jkg−1K−1)
Explanation
0.15×900×(280−T) = (0.7×4186+0.05×127.7)×(T−22). 37800−135T = (2930.2+6.385)×(T−22) = 2936.585T−64599.87. 37800+64599.87 = 2936.585T+135T. 102399.87 = 3071.585T⇒T≈33.34∘C≈33.3∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 33.3°C. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
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