The tungsten filament of a lightbulb has an operating temperature of about 2800
ID: 1632166 • Letter: T
Question
The tungsten filament of a lightbulb has an operating temperature of about 2800 degree C when the room temperature is 20 degree C. If the emitting area of the filament is 1.2 cm^2, and its emissivity is 0.4, what is the power output of the lightbulb? (Stephan-Boltzmann Constant = 5.67 times 10^-8 W/m^2 K^4) 25.76 Watts 84 Watts 24.27 Watts 11.2 Watts 312.48 Watts What is the entropy change to turn 12 grams of ice (C_ice = 53 cal/gr degree C) at -15 degree to 10 grams of steam? 10.44 cal./K 30.6 cal./K 19.89 cal./K 31.32 cal./K 22.8 cal./KExplanation / Answer
Filament of bulb
Area, A = 1.2 cm2 = 1.2x10-4 m2
Temperature, T = 28000 C = 2800 + 273 = 3073 K
Emissivity, = 0.4
Temperature of surrounding, T0 = 200 C = 20 + 273 = 293 K
As per Stefan's Law, the energy emitted by a black body per second per unit area is given by,
E = T4
Where
(Stefan’s constant) = 5.67x10-8 W/m2 K4
Energy emitted by any arbitrary body per second per unit area is given by,
E = T4
Energy emitted by any arbitrary body per second or power emitted is given by,
Pe = AT4
Energy absorbed from surrounding by any arbitrary body per second or power absorbed is given by,
Po = A(T0)4
Hence net power emitted by the body, P = Pe – Po
P = A{T4 – (T0)4}
P = (5.67x10-8)(0.4)(1.2x10-4){(3073)4 – (293)4}
P = 242.683 = 242.7 watt
So the answer is 242.7 watt.
As option C has 24.27 watt in spite of 242.7 watt as the answer. So in case there is a typing error, correct option is C.
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