Question

The total surface area of the human body is 1.20 m2 and the surface temperature is 30∘C=303∘K. If the surroundings are at a temperature of 4.0 ∘C , what is the net rate of heat loss from the body by radiation? The emissivity of the body is very close to unity, irrespective of skin pigmentation. Express your answer using two significant figures.

Answer 1

To solve this problem we must consider the expressions of Stefan Boltzmann's law for which the rate of change of the radiation of energy H from a surface must be

$H = Ae\sigma T^4$

Where

A = Surface area

e = Emissivity that characterizes the emitting properties of the surface

$\sigma$ = Universal constant called the Stefan-Boltzmann constant $(5.67*10^{-8}m^{-2}K^{-4})$

T = Absolute temperature

The total heat loss would be then

$Q = H_2 -H_1$

$Q =Ae\sigma T_2^4-Ae\sigma T_1^4$

$Q = Ae\sigma (T_2^4-T_1^4)$

$Q = (1.2)(1)(5.67*10^{-8})(303^4-280^4)$

$Q = 155.29J$

Therefore the net rate of heat loss from the body by radiation is 155.29J