Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both fila
ments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.
1 answer:
Answer:
0.3659
Explanation:
The power (p) is given as:
P = AeσT⁴
where,
A =Area
e = transmittivity
σ = Stefan-boltzmann constant
T = Temperature
since both the bulbs radiate same power
P₁ = P₂
Where, 1 denotes the bulb 1
2 denotes the bulb 2
thus,
A₁e₁σT₁⁴ = A₂e₂σT₂⁴
Now e₁=e₂
⇒A₁T₁⁴ = A₂T₂⁴
or
substituting the values in the above question we get
or
=0.3659
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