Practice question
Question
A 0.3kg lead block at 400∘C is placed in 0.7kg water at 15∘C in a 0.1kg aluminium calorimeter at 15∘C. Find the final temperature. (Specific heat of lead = 127.7J kg−1K−1, water = 4186J kg−1K−1, aluminium = 900Jkg−1K−1)
Explanation
0.3×127.7×(400−T) = (0.7×4186+0.1×900)×(T−15). 15324−38.31T = (2930.2+90)×(T−15) = 3020.2T−45303. 15324+45303 = 3020.2T+38.31T. 60627 = 3058.51T⇒T≈19.82∘C≈19.8∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 19.8°C. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.