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Given: L0 = 20cm, V0 = 203 = 8000cm3, ΔT = 125−25 = 100∘C, αl = 1.2×10−5K−1. αv = 3αl = 3×1.2×10−5 = 3.6×10−5K−1. ΔV = V0αvΔT = 8000×3.6×10−5×100 = 28.8cm3.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
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