## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$\lambda = 2.42~\mu m$
We can find the temperature of the cube: $P = e~\sigma~A~T^4$ $T^4 = \frac{P}{e~\sigma~A}$ $T = \sqrt[4] {\frac{P}{e~\sigma~A}}$ $T = \sqrt[4] {\frac{630~W}{(1)~(5.67\times 10^{-8}~W~m^{-2}~K^{-4})~(6)~(3.0\times 10^{-2}~m)^2}}$ $T = \sqrt[4] {2.057613169\times 10^{12}~K^4}$ $T = 1197.7~K$ We can use Wien's law to find the peak wavelength: $\lambda = \frac{2.9\times 10^6~nm~K}{T}$ $\lambda = \frac{2.9\times 10^6~nm~K}{1197.7~K}$ $\lambda = 2.42\times 10^{3}~nm$ $\lambda = 2.42\times 10^{-6}~m$ $\lambda = 2.42~\mu m$