Question

A solid sphere has a temperature of 674 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in kelvins?

Answer 1

The total radiant power is given Stefan-Boltzmann law:
P = A·ε·σ·T^4
(A surface area, ε emissivity, σ Stefan-Boltzmann constant, T absolute surface temperature)

Sphere and cube emit the same radiant power, i.e.
A_sphere · ε · σ · (T_sphere)^4 = A_cube · ε · σ · (T_cube)^4
<=>
A_sphere · (T_sphere)^4 = A_cube · (T_cube)^4
<=>
T_cube = T_sphere · (A_sphere/A_cube)^(1/4)

surface area of a cube of edge length a is
A_cube = 6·a^2
surface area of a sphere of edge length r is
A_cube = 4·π·r^2

Hence:
T_cube = T_sphere · ( 4·π·r^2 / (6·a^2) )^(1/4)
= T_sphere · ( (2/3)·π·(r/a)^2 )^(1/4)

Both objects are made from the same mass. Assuming constant density the volumes are the same, too, i.e.
V_sphere = V_cube
<=>
(4/3)·π·r^3 = a^3
<=>
r/a = ( 3/(4·π) )^(1/3)

Therefore:
T_cube = T_sphere · ( (2/3)·π·(r/a)^2 )^(1/4)
= T_sphere · ( (2/3) · π · ( 3/(4·π) )^(2/3) )^(1/4)
= T_sphere · ( (π/6)^(1/3) )^(1/4)
= T_sphere · (π/6)^(1/12)
= 674K · (π/6)^(1/12)
= 638.6K