A solid sphere has a temperature of 556 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

Cube temperature = 526.83 K

Explanation:

Volume of the cube and sphere will be the same.

Now, volume of cube = a³

And ,volume of sphere = (4/3)πr³

Thus;

a³ = (4/3)πr³

a³ = 4.1187r³

Taking cube root of both sides gives;

a = 1.6119r

Formula for surface area of sphere is;

As = 4πr²

Also,formula for surface area of cube is; Ac = 6a²

Thus, since a = 1.6119r,

Then, Ac = 6(1.6119r)²

Ac = 15.5893r²

The formula for radiant power is;

Q' = eσT⁴A

Where;

e is emissivity

σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k

T is temperate in kelvin

A is Area

So, for the cube;

(Qc)' = eσ(Tc)⁴(Ac)

For the sphere;

(Qs)' = eσ(Ts)⁴(As)

We are told (Qc)' = (Qs)'

Thus;

eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)

eσ will cancel out to give;

(Tc)⁴(Ac) = (Ts)⁴(As)

Since we want to find the cube's temperature Tc,

(Tc)⁴ = [(Ts)⁴(As)]/Ac

Plugging in relevant figures, we have;

(Tc)⁴ = [556⁴ × 4πr²]/15.5893r²

r² will cancel out to give;

(Tc)⁴ = [556⁴ × 4π]/15.5893

Tc = ∜([556⁴ × 4π]/15.5893)

Tc = 526.83 K

b would be the answer

i don't know that i agree with that but what's the question ?