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Practice question

MCQ

Question

A silver wire of length 1.2m at 25∘C is cooled until its length decreases by 0.0228cm. What is the final temperature? (αl\=1.9×10−5K−1)

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Explanation

Given: L0 = 1.2m = 120cm, ΔL = −0.0228cm, αl = 1.9×10−5K−1, T1 = 25∘C. ΔL = L0αlΔT⇒−0.0228 = 120×1.9×10−5×ΔT. ΔT = −0.0228120×1.9×10−5 = −0.02282.28×10−3 = −10K. T2 = 25−10 = 15∘C.

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