Under steady-state operation the surface temperature of a small 20-w incandescent light bulb is 125c when the temperature of the room air and walls is 25c. Approximating the bulb as a sphere 40 mm in diameter with a surface emissivity of 0.8, what is the rate of heat transfer from the surface of the bulb to the surroundings?
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Answer:
stefan Boltzmann constant sigma =
5.67*10^-8 watt/kelvin*m^2
emissivity = 0.8
bulb temperature = 125 ℃ = 398 °k
room temperature = 25 ℃ = 298 °k
area of bulb = 4π*0.02^2 = 0.005 m^2
Explanation:
rate of heat transfer
= (5.67*10^-8)* {(398^4) - (298^4)}*0.8*0.005
= (2.28 *10^-10)*(1.72*10^10) = 3.92 Watt
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