If you were to walk briefly in space without a spacesuit while far from the Sun (as an astronaut does in the movie 2001, A Space Odyssey), you would feel the cold of space—while you radiated energy, you would absorb almost none from your environment. (a) At what rate would you lose energy? (b) How much energy would you lose in $\mathbf{30}\mathbf{}\mathit{s}$? Assume that your emissivity is $0.90$, and estimate other data needed in the calculations.

1. Surface area of average human body, $A=2{\text{\hspace{0.17em}m}}^{\text{2}}$
2. Skin temperature of human, $T=310\text{\hspace{0.17em}K}$
3. Emissivity of human, $\in =0.90$

Thermal radiation is the process of transferring heat through electromagnetic radiation that is produced by the thermal motion of matter particles. We use the concept of radiation to calculate the rate of energy loss. Using the equation for the rate at which the object loses energy, we can find the energy lost by the person.

Formulae:

The rate at which the sphere emits thermal radiation, $P_{\text{rad}}=\sigma ϵA{T}^4$ …(i)

Where,is the Stefan–Boltzmann constant and is equal to$(5.67\times {10}^{-8}{\text{W/m}}^{\text{2}}{\text{K}}^{\text{4}})$

$ϵ$is the emissivity, A is the area and T is the temperature

The energy lost by the body due to radiation, $\Delta E=P_{\text{rad}}\Delta t$ …(ii)

Where,$\Delta t$ is the time

Using equation (i), the rate of the energy lost by the person can be given as:

$\begin{array}{c}{P}_{rad}=(5.67\times {10}^{-8}{\text{\hspace{0.17em}W/m}}^{\text{2}}{\text{.K}}^{\text{4}})\left(0.9\right)\left(2.0{\text{\hspace{0.17em}m}}^{\text{2}}\right){\left(300\text{\hspace{0.17em}K}\right)}^4\\ =8.26\times {10}^2\text{W}\end{array}$

Hence, the rate at which energy is lost is $8.26\times {10}^2\text{W}$

Energy that the person would lose incan be given using equation (ii) as:

$\begin{array}{c}\Delta E=(8\times {10}^2\text{W})\left(30\text{s}\right)\\ =2.47\times {10}^4\text{J}\end{array}$

Hence, the energy lost by the person is $2.47\times {10}^4\text{J}$