Question

The amount of heat radiated per second by a body of surface area $20c{m}^2$ kept at temperature $27{}^{o}C$ is H and wavelength having maximum share of emission is $\lambda ,$ find H and $\lambda$ approximately
(e = emissivity = 0.6, $\sigma =6\times {10}^{-6}\frac{watt}{m^2{K}^4}$
Wein's constant $=0.288cm-K$)

• A. $H=58watt,\lambda =0.44\times {10}^{-3}cm$
• B. $H=68watt,\lambda =0.96\times {10}^{-3}cm$
• C. $H=58watt,\lambda =0.96\times {10}^{-3}cm$
• D. $H=68watt,\lambda =0.44\times {10}^{-3}cm$

Answer 1

The correct option is C $H=58watt,\lambda =0.96\times {10}^{-3}cm$

$H=\sigma eA{T}^4$
$=6\times {10}^{-6}\times 0.6\times 20\times {10}^{-4}\times (300{)}^4$
$=6\times {10}^{-6}\times 12\times {10}^{-4}\times 81\times {10}^8$
$=81\times 6\times 12\times {10}^{-2}\approx 58watt$
Also,
${\lambda }_m.T=0.288cm-K$
${\lambda }_m=\frac{.288}{300}cm=0.96\times {10}^{-3}cm$