Question

The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 11.9 W. HooRU's surface area is 1.59 m2 and the emissivity of its surface is 0.217. Ignore the radiation HooRU absorbs from the cold universe. What is HooRU's temperature?

Answer 1

According to the Stefan–Boltzmann Law, how much power (energy per time) is radiated per surface area at a given absolute temperature by a perfect blackbody (emissivity = 1):

P = σT⁴A
where
σ = 2π⁵k⁴/15c²h³ = 5.67·10⁻⁸ W/(m²·K⁴)
A = surface area = 1.59 m^2 (given)

When the emissivity, ε, is < 1, it multiplies the perfect blackbody amount:
P = εσT⁴A

In our case, this is,

ε = 0.217

Put all these variables in the above expression -

11.9 = 0.217 x 5.67 x 10^-8 x T^4 x 1.59

=> T^4 = 11.9 / (0.217 x 5.67 x 10^-8 x 1.59) = 6.083 x 10^8

=> T = (6.083 x 10^8)^1/4 = 1.5704 x 10^2 = 157.04 K

So, the temperature of HooRU is 157.04 K.