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Practice question

MCQ

Question

A steel rod of length 50cm at 20∘C is heated until its length increases by 0.03cm. What is the final temperature? (αl\=1.2×10−5K−1)

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Explanation

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.

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