Question

An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a wire that heats due to resistance to electrical current. The ends of the wire are not exposed. The total filament emissivity is 0.300 and its temperature is 3068 K. Calculate the power emitted by the lamp in watts. Note: required precision is 1%.

Answer 1

power emitted is 1.75 W

Step-by-step explanation:

given data

length l = 5 cm = 5 ×
`10^(-2)` m

diameter d = 0.074 cm = 74 ×
`10^(-5)` m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 ×
`10^(-5)`×5 ×
`10^(-2)`

area = 1162.3892 ×
`10^(-5)` m²

so here power emitted is express as

power emitted = E × σ × area × (temperature)^4

put here all value

power emitted = 0.300× 5.67 × 1162.3892 ×
`10^(-5)` × (3068)^4

power emitted = 1.75 W