Question

Consider a person floating in space. Calculate the blackbody luminosity (energy radiated per second) of the person and also the rate of energy absorbed from CMB photons (T 2.73 K). At what rate does the person cool off, i.e. dT/dt? For simplicity, assume a spherical person with 1 m diameter, temperature of 310 K, and who absorbs and radiates as a perfect blackbody. To calculate temperature change, assume that this person has the same heat capacity as water, C 4200.Jkg K-1

Answer 1

The blackbody luminosity (energy radiated per second) of a person which will be given by -

L = (4$\pi$R²) $\sigma$ T⁴

where, R = radius of a spherical person = 0.5 m

$\sigma$ = Stefan-Boltzmann constant = 5.67 x 10^-8 W/m².K⁴

T = temperature of spherical blackbody = 310 K

then, we get

L = [4 (3.14) (0.5 m)² (5.67 x 10^-8 W/m².K⁴) (310 K)⁴]

L = 1644.2 W