Eureka!

This problem involves the usage of several basic laws from the field of radiation and thermodynamics.

First, we make use of Wien's Displacement Law, assuming that the filament can be taken to be blackbody $\lambda_{max}T=\beta$ (where $\beta\approx2.898\times10^{-3}m K$ is Wien's constant), which relates the peak wavelength $\lambda_{max}$ of an object's blackbody radiation spectrum to its temperature in Kelvin $T$ . From this, we find that the incandescent light bulb's filament is at a temperature of $T=\frac{2.898\times10^{-3}}{10^{-6}}=2898K$

Next, we find the power input to the light bulb from the wall outlet. The power $P$ from the wall is simply given by the input voltage $V$ times the current $I$ , giving us $P=VI=120\times0.5=60W$

Lastly, we assume that all the input power from the wall outlet has been converted into the blackbody radiation that the light bulb emits, and that the filament is a perfect blackbody. The latter assumption allows us to use the Stefan Boltzmann Law, which relates the power emitted by a blackbody to its area $A$ and temperature $T$ . The law states: $P=A\sigma T^4$ where $\sigma$ is the Stefan Boltzmann constant with a value of $5.67\times10^{-8}Wm^{-2}K^{-4}$

Substituting in all our previous values for $P$ and $T$ , we find that the area $A$ of the filament is:

$A=\frac{P}{\sigma T^4}=\frac{60}{(5.67\times10^{-8})(2898)^4}=1.50\times10^{-5}m^2$

By wien's law , $\lambda_{peak}T = 2.9 \times 10^{-3}, \Rightarrow T = 2900K$ , Also , $T_{surr} = 300K$

Heat dissipation = $iV = 60W$

Power radiated through radiation is given by stephan's law, and it must be equal to heat dissipation , Therefore

$60 = A \sigma (T^4 - T_{surr}^{4})$

Plugging values, we get $A = \boxed{1.5 \times 10^{-5} m^2}$

We can assume that filament acts as a black body.

Using Wien's displacement law , the temperature of of the filament is:

$T=\frac{b}{\lambda_{max}}$

Then using Stefan-Boltzmann law , we get,

$j^{\ast}A=IV$

where $j^{\ast}$ is the power emitted per unit area of the surface of filament and A is the area of the filament. Then solving for $A$ , we get,

$\boxed{A=1.5 \times 10^{-5} m^{2}}$

Let the temperature of the surroundings be \­( T_{o}\­) and the temperature of the object at maximum wavelength \­( λ\­) be \­( T\­).

The rate of power radiated is described by

\­( P) = \(A e σ ( T^{4} - T_{o}^{4}\­) (1)

assuming that the bulb is a blackbody (e = 1) and σ is the Stefan-Boltzmann constant.

The rate of power radiated is described by

\­( P\) = \(A e σ ( T^{4} - T_{o}^{4}\­) (1)

and the temperature of the object at maximum wavelength \­( λ\­) can be determined using Wien's displacement law

\­( T\­) = (\frac{2.898 mm. K}{λ}\) (2)

and the power delivered by the bulb is

\­( P\­) = \­(IV\­) (3)