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Given: L0 = 80cm, ΔL = −0.0136cm, αl = 1.7×10−5K−1, T1 = 40∘C. ΔL = L0αlΔT⇒−0.0136 = 80×1.7×10−5×ΔT. ΔT = −0.013680×1.7×10−5 = −0.01361.36×10−3 = −10K. T2 = 40−10 = 30∘C.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
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