A wire of length 10 cm and diameter 0.5 mm radiates 94.2 W at temperature 1727°C. Its emissivity is x/8. Given σ = 6.0×10⁻⁸ W m⁻² K⁻⁴ and π = 3.14, find x.

Question

G1 Physics · Accepted Answer

Convert temperature: T = 1727°C + 273 = 2000 K. For a thin cylindrical wire, radiating area is approximately its curved surface area: A = πdL. Given, d = 0.5 mm = 0.5×10⁻³ m, L = 10 cm = 0.10 m. Stefan-Boltzmann law: P = εσAT⁴. Therefore, 94.2 = ε(6×10⁻⁸)[3.14(0.5×10⁻³)(0.10)](2000)⁴. The numerical factor evaluates to 150.72 W for ε=1. Thus, ε = 94.2/150.72 = 0.625 = 5/8. Given ε = x/8, x = 5.