Q.A 60 watt filament lamp loses all its energy by radiation from its surface. The emissivity of the surface is 0.5. The area of the surface is 5 × 10⁻⁵ m². Find the temperature of the filament (σ = 5.67 × 10⁻⁸ J m⁻² s⁻¹ K⁻⁴).

Stefan's law gives the total radiant energy from a black body as $\sigma A T^4$ per second; multiply by time for the total energy.

By Stefan's law, the power radiated by a perfectly black body of surface area $A$ at absolute temperature $T$ is

$P = \sigma A T^{4}$

Given $A = 200\text{ cm}^2 = 200\times10^{-4}\text{ m}^2 = 0.02\text{ m}^2$, $T = 127^\circ\text{C} = 400\text{ K}$, and $\sigma = 5.67\times10^{-8}\text{ W m}^{-2}\text{K}^{-4}$:

$T^4 = (400)^4 = 2.56\times10^{10}\text{ K}^4$

$P = 5.67\times10^{-8}\times 0.02\times 2.56\times10^{10} = 5.67\times10^{-8}\times5.12\times10^{8} \approx 29.03\text{ W}$

Energy radiated in one minute ($t = 60\text{ s}$):

$E = P\times t = 29.03\times 60 \approx 1741.8\text{ J}$

Rate of loss of heat by radiation follows Stefan's law with a surrounding-temperature correction.

By Stefan's law (with surroundings at temperature $T_0$), the net rate of loss of heat is $R \propto (T^4-T_0^4)$.

Convert to kelvin: $T_1=827+273=1100$ K, $T_2=427+273=700$ K, $T_0=27+273=300$ K.

$\frac{R_1}{R_2} = \frac{T_1^4-T_0^4}{T_2^4-T_0^4} = \frac{(1100)^4-(300)^4}{(700)^4-(300)^4}$

Computing: $1100^4 = 1.4641\times10^{12}$, $700^4=2.401\times10^{11}$, $300^4=8.1\times10^{9}$.

$\frac{R_1}{R_2} = \frac{1.4641\times10^{12}-0.0081\times10^{12}}{0.2401\times10^{12}-0.0081\times10^{12}} = \frac{1.4560\times10^{12}}{0.2320\times10^{12}} \approx 6.28$