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

0.3659

Step-by-step 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

`(A_1)/(A_2) =(T_(2)^(4))/(T_(1)^(4))`

substituting the values in the above question we get

`(A_1)/(A_2) =(2100_(2)^(4))/(2700_(1)^(4))`

or

`(A_1)/(A_2) }`=0.3659