A 0.05kg gold block at 1200∘C is placed in 0.5kg water at 20∘C. What is the final temperature? (Specific heat of gold =
0.05×134×(1200−T) = 0.5×4186×(T−20). 8040−6.7T = 2093T−41860. 8040+41860 = 2093T+6.7T. 49900 = 2099.7T⇒T≈23.77∘C≈23.8∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 23.8°C. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.