Define emissive power.

The filament of a light bulb has surface area $64 {\mathrm{mm}}^2.$ The filament can be considered as a black body at temperature $2500 \mathrm{K}$ emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of $100 \mathrm{m}$. Assume the pupil of the eyes of the observer to be circular with radius $3 \mathrm{mm}$. Then:

(Take Stefan-Boltzmann constant $=5.67\times {10}^{-8} \mathrm{W} {\mathrm{m}}^{-2} {\mathrm{K}}^{-4},$ Wiens' displacement constant $=2.90\times {10}^{-3} \mathrm{m} \mathrm{K}$, Planck's constant $=6.60\times {10}^{-34} \mathrm{J} \mathrm{s}$, speed of light in vacuum $=3.00\times {10}^8 \mathrm{m}{\mathrm{s}}^{-1}$)

A container with $1 \mathrm{kg}$ of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is $700 \mathrm{W} {\mathrm{m}}^{-2}$ and it is absorbed by the water over an effective area of $0.05 {\mathrm{m}}^2.$ Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in ${}^{\circ }\mathrm{C}$) in the temperature of water and the surroundings after a long time will be_________. (Ignore effect of the container, and take constant for Newton's law of cooling $=0.001 {\mathrm{s}}^{-1},$ Heat capacity of water $=4200 \mathrm{J} {\mathrm{kg}}^{-1} {\mathrm{K}}^{-1}$)

(Approximate the answer to two decimal place)