Question

Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?

Answer 1

$493 \; \text{W}\cdot \text{m}^{-2}$.

Explanation

The Stefan-Boltzmann Law gives the energy radiation per unit area of a black body:

$\dfrac{P}{A} = \sigma \cdot T^{4}$

where,

• $P$ the total power emitted,
• $A$ the surface area of the body,
• $\sigma$ the Stefan-Boltzmann Constant, and
• $T$ the temperature of the body in degrees Kelvins.

$\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}$.

$T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}$.

$\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}$.

Keep as many significant figures in $T$ as possible. The error will be large when $T$ is raised to the power of four. Also, the real value will be much smaller than $493\; \text{W}\cdot \text{m}^{-2}$ since the emittance of a human body is much smaller than assumed.