Practice question
Question
A 0.25kg iron block at 150∘C is dropped into 1.5kg of water at 30∘C. What is the final temperature? (Specific heat of iron = 450J kg−1K−1, water = 4186J kg−1K−1)
Explanation
Heat lost = Heat gained. 0.25×450×(150−T) = 1.5×4186×(T−30). 16875−112.5T = 6279T−188370. 16875+188370 = 6279T+112.5T. 205245 = 6391.5T⇒T≈32.11∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 32.11°C. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.
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