Question

An indirectly heated filament is radiating maximum energy of wavelength `2.16xx10^-5cm`. Find the net amount of heat energy lost per second per unit area, the temperature of the surrounding air is `13^@C`. Given `b=0.288cm-K.sigma=5.77xx10^-5erg//s-cm^(2)-K^(4)`).

Answer 1

If T is the surface temperature of the filament then
`lamda_mT=b`
or `T=(b)/(lamda_m)=(0.288)/(2.16xx10^-5)=13333.3K`
The temperature of surrounding air, `T_0=13+273=286K`
The net amount of energy radiated per unit area per second
`E=sigma(T^4-T_^4)`
`=5.77xx10^-5[(13333.3)^4-(286)^4]`
`=1.824xx10^12ergs^-1cm^-2`