Question

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.

Answer 1

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

$\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}$

substituting the values in the above question we get

$\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}$

or

$\frac{A_1}{A_2} }$=0.3659