Physics Thermodynamics

Question

Find irradiative heat loss for a person with 𝐴𝑟=1.5m2, if skin temperature is 𝑇𝑠= 32◦C and

temperature of the nearby radiating surface 𝑇𝑟= 25◦C per hour. The emissivity of the skin in this

wavelength range is nearly unity 𝑒=1, independent of the skin pigmentation.

Answer 1

The equation for gray body radiative heat transfer is Q = A*σ*e*(T₂⁴-T₁⁴)

We're assuming that the emissivity of the person (e) is equal to the emissivity of the room, otherwise we'd have to use a different equation.

It's important to note that the radiative heat transfer equation expects the temperature to be in K, so we'll convert from C. Our knowns are unknowns are:

A = 1.5 m²

e = 1

Ts = 32C = 305.15 K

T_r = 25C = 298.15 K

σ = 5.67x10^-8 W/m²/K⁴ (Stefan-Boltzmann constant)

Plugging in gives

Q = 1.5 * 5.67x10^-8 * 1 * (305.15⁴ - 298.15⁴)

Q = 65.37 W

It's important to note which way the heat is flowing. In this example the person is hotter than the room, therefore the heat flow Q is heat transferring from the person to the room.