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Given: L0 = 50cm, ΔL = 0.03cm, αl = 1.2×10−5K−1, T1 = 20∘C. ΔL = L0αlΔT⇒0.03 = 50×1.2×10−5×ΔT. ΔT = 0.0350×1.2×10−5 = 500K. T2 = T1+ΔT = 20+500 = 520∘C.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
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