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Practice question

MCQ

Question

A steel cylinder has a volume of 1.5L at 55∘C. What temperature must it be heated to for the volume to increase by 0.0054L? (αl\=1.2×10−5K−1)

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Explanation

Given: V0 = 1.5L = 1500cm3, ΔV = 0.0054L = 5.4cm3, αl = 1.2×10−5K−1, T1 = 55∘C. αv = 3αl = 3×1.2×10−5 = 3.6×10−5K−1. ΔV = V0αvΔT⇒5.4 = 1500×3.6×10−5×ΔT. ΔT = 5.41500×3.6×10−5 = 5.40.054 = 100K. T2 = 55+100 = 155∘C.

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