Answer:
[tex]5.63498482\ ^{\circ}C[/tex]
Explanation:
[tex]\sigma[/tex] = Stefan-Boltzmann constant = [tex]5.67\times 10^{-8}\ W/m^2K^4[/tex]
I = Intensity = [tex]1370\ W/m^2[/tex]
r = Radius
T = Temperature
The power going in is equal to the power going out
[tex]I\epsilon\pi r^2=4\pi r^2\epsilon\sigma T^4\\\Rightarrow T=(\dfrac{I}{4\sigma})^{\dfrac{1}{4}}\\\Rightarrow T=(\dfrac{1370}{4\times 5.67\times 10^{-8}})^{\dfrac{1}{4}}\\\Rightarrow T=278.78498482\ K[/tex]
Converting to Celcius
[tex]278.78498482-273.15=5.63498482\ ^{\circ}C[/tex]
So, yes it is chilly