Answer:581.87 K
Explanation:
Given
Sphere is melted to form a square
Let the radius of sphere be r and square has a side a
Therefore
[tex]\frac{4\pi}{3}r^3=a^3[/tex]
Surface area of sphere [tex]A_s=4\pi r^2[/tex]
Surface area of cube [tex]A_c=6a^2[/tex]
Total emmisive remains same
Thus [tex]P=A\epsilon \sigma T^4[/tex]
[tex]A_sT_s^4=A_cT_c^4[/tex]
[tex]\frac{T_c^4}{T_s^4}=\frac{A_s}{A_c}[/tex]
[tex]\frac{T_c^4}{T_s^4}=\frac{1}{2}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{3}}[/tex]
[tex]\frac{T_c}{T_s}=\frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}[/tex]
[tex]T_c=T_s\times \frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}[/tex]
[tex]T_c=614\times \frac{1.12679}{1.189}[/tex]
[tex]T_c=581.87 K[/tex]
