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A 40 W lamp gives out all its power as radiation from a tungsten filament at 2500 K, whose…

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A $40\ \text{W}$ lamp gives out all its power as radiation from a tungsten filament at $2500\ \text{K}$ , whose emissivity is $0.35$ . What is the surface area of the filament? ( $\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$ .)

Need a reminder? The key ideaShow

Radiation: $P = e\sigma AT^4$ , net $e\sigma A(T^4 - T_0^4)$ , with $\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$ and $T$ in kelvin. Wien: $\lambda_mT = 2.90 \times 10^{-3}\ \text{m K}$ .

The worked solution

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