A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727° C and power radiated by the wire is 94.2 W . Its emissivity is x/8 where x= _________.
(Given σ=6.0 × 10⁻⁸ W m⁻² K⁻⁴, π=3.14 and assume that the emissivity of wire material is same at all wavelength.)

Solution

Physics sir · 4 steps

The wire radiates from its curved side, so use π d L as the area and put T in kelvin.

1. Radiating area. A thin wire radiates from its curved surface. The two tiny end faces are negligible, so A = π d L = 3.14 × (0.5 × 10⁻³) × 0.1 = 1.57 × 10⁻⁴ m².
2. Absolute temperature. T = 1727 + 273 = 2000 K, so T⁴ = 16 × 10¹² = 1.6 × 10¹³ K⁴.

   P = e σ A T⁴ ⇒ e = P/(σ A T⁴)

3. Put in numbers. σ A T⁴ = 6.0 × 10⁻⁸ × 1.57 × 10⁻⁴ × 1.6 × 10¹³ = 150.72 W. So e = 94.2/150.72 = 0.625.
4. So. e = 0.625 = 5/8, which gives x = 5.

The common slip: 2.5. Taking 0.5 mm as the radius instead of the diameter doubles the area and gives e = 5/16, so x = 2.5.

The idea to keep. In Stefan-Boltzmann problems, always convert to kelvin before raising to the fourth power, and take the area that actually faces the surroundings: π d L for a wire.

Answer from the NTA final answer key. Solution by PrepWiser.