Question

A 567 W bulb has a tungsten filament of length 40 cm and radius $\dfrac{2}{\pi}$ mm. If the radiation of the filament is 81% of that of a perfect black body, then the temperature of the filament is
(Stefan's constant, $\sigma = 5.67 \times 10^{-8}$ W m$^{-2}$ K$^{-4}$)

• A. 2500 K
• B. 1666.7 K
• C. 1333.3 K
• D. 999.6 K

Hint:

Use $P = e\sigma A T^4$ for real objects; make sure to convert mm to meters when calculating area.

Answer 1

The Correct Option is B

Given: Power $P = 567$ W, emissivity $e = 0.81$, $\sigma = 5.67 \times 10^{-8}$ W m$^{-2}$ K$^{-4}$
Length $l = 0.4$ m, radius $r = \dfrac{2}{\pi} \times 10^{-3}$ m
Surface area $A = 2\pi r l = 2\pi \cdot \dfrac{2}{\pi} \cdot 10^{-3} \cdot 0.4 = 1.6 \times 10^{-3}$ m$^2$
Using Stefan-Boltzmann law: $P = e\sigma A T^4$
$\Rightarrow 567 = 0.81 \cdot 5.67 \times 10^{-8} \cdot 1.6 \times 10^{-3} \cdot T^4$
Solving: $T^4 = \dfrac{567}{0.81 \cdot 5.67 \times 10^{-8} \cdot 1.6 \times 10^{-3}} \approx 7.74 \times 10^{11}$
$T \approx \sqrt[4]{7.74 \times 10^{11}} = 1666.7$ K