A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727 and power radiated by the wire is 94.2 W. Its emissivity is x8 where x = . (Given = 6.0 10-8 W m-2 K-4, = 3.14 and assume that the emissivity of wire material is same at all wavelengths.)
Answer
5
Solution
Stefan–Boltzmann law: P = e A T4
T = 1727 + 273 = 2000 K
Radiating (curved) surface area of the wire:
A = d L = 3.14 0.5 10-3 0.1 = 1.57 10-4 m2
T4 = 6.0 10-8 (2000)4 = 6.0 10-8 1.6 1013 = 9.6 105 W m-2
⇒ 94.2 = e 1.57 10-4 9.6 105
⇒ 94.2 = e 150.72
⇒ e = 0.625 = 58
∴ x = 5
Answer: 5