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