Question

A hot filament radiates 1.50 W of light. The filament has a surface area of 0.250 mm2

and an emissivity of 0.950. what is the temperature (in kelvin) of the filament?

Answer 1

Solution;

Given;

Area,A=$0.025mm^2=0.25×10^{-6}m^2$

Emmisivity of the filament ,e=0.95

Power radiated,P=1.50W

Stefan Boltzmann's constant,$\sigma=5.6696×10^{-8}W/m^2K^4$

We know that;

$P=\sigma eAT^4$

By direct substitution;

$T^4=\frac{P}{\sigma e A}=\frac{1.5}{5.6696×10^{-8}×0.95×0.25×10^{-6}}=1.114$

$T=(1.114)^{\frac14}=3248.77K$