Practice question
Question
A 0.4kg aluminium block at 160∘C is placed in 1.2kg water at 28∘C in a 0.2kg lead calorimeter at 28∘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
0.4×900×(160−T) = (1.2×4186+0.2×127.7)×(T−28). 57600−360T = (5023.2+25.54)×(T−28) = 5048.74T−141364.72. 57600+141364.72 = 5048.74T+360T. 198964.72 = 5408.74T⇒T≈36.78∘C≈36.8∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 36.8°C. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.