Transmission Of Heat Question 324
Question: The emissivity and the surface area of filament of electric bulb are 0.7 and $ 5\times {{10}^{-5}}m^{2} $ . The operating temperature of the filament is 3000 K. The power of the bulb is approximately $ [\sigma =5.67\times {{10}^{-8}}watt/m^{2}-K^{4}] $
Options:
A) 230 W
B) 160 W
C) 9000 W
D) 4150 W
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The rate at which the filament of the bulb is radiating the energy is equal to the rate at which it is consuming the electrical energy (no energy loss has been considered). So power of bulb is $ P=\sigma eAT^{4} $
$ =5.67\times {{10}^{-8}}\times 0.7\times 0.5\times {{10}^{-5}}\times {{(3000)}^{4}}\approx 160W $
BETA
