Question

How much energy does a person with an emissivity of 0.85 and a surface area of `"1.42 m"^2` lose per minute when the external temperature is 20 °C?

Answer 1

The person loses 7.7 kJ/min.

The relationship governing the net radiation from hot objects is called the Stefan-Boltzmann Law:

`P = εσA(T^4-T_C^4)`, where

`P = "net radiated [power](http://socratic.org/physics/work-and-energy/power)"`
`ε = "emissivity"`
`σ = "Stefan's constant" = 5.670 ×10^(-8) "W·m"^(-2)"K"^(-4)`
`T = "temperature of radiator"`
`T_c = "temperature of surroundings"`

`P = 0.85 × 5.670 ×10^(-8) "W"·cancel("m⁻²K⁻⁴") × 1.42 cancel("m²") × ((310 cancel("K"))^4 – (293 cancel("K"))^4) = 127.6 cancel("W") × (1 "J·"cancel("s⁻¹"))/(1 cancel("W")) × (60 cancel("s"))/"1 min" = "7700 J/min" = "7.7 kJ/min"`