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?

1 Answer
Ernest Z.
May 23, 2015

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"#
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"

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