A ring hole diameter is 1.0000 cm at 0°C. A sphere diameter is 1.0015 cm at 10 0 ∘ C . α r = 1 0 − 5 , α s = 2 × 1 0 − 5 per °C. Equal constant specific heats, no external heat transfer. Using ref

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A ring hole diameter is 1.0000 cm at 0°C. A sphere diameter is 1.0015 cm at 10 0 ∘ C . α r ​ = 1 0 − 5 , α s ​ = 2 × 1 0 − 5 per °C. Equal constant specific heats, no external heat transfer. Using reference-length linear laws, choose (equilibrium T, ring/sphere mass ratio) for just passing.

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Diameter equality 1 + 1 0 − 5 T = 1.0015 [ 1 + 2 × 1 0 − 5 ( T − 100 ) ] gives T = 0.00001003 0.000503 ​ = 50.1495 5 ∘ C . Equal specific heats give mass ratio T ( 100 − T ) ​ = 0.994036 . A.

Physics - Heat & Thermo Question with Solution | TestHub

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