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)`.

Step-by-step explanation

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

`(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 * 10^(-8) \;\text{W}\cdot \text{m}^(-2) \cdot \text{K}^(-4)`.

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

`(P)/(A) = \sigma \cdot T^(4) = 5.67 * 10^(-8) * 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.