Question

what is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient temperature is $t_2 = 22 c$? the person has a normal skin temperature of $t_1 = 33 c$ and surface area of about $a = 1.5 m^2$. the emissivity of skin is $epsilon=0.97$ in the infrared part of the spectrum where radiation takes place.
$p=sigmaepsilon a(t_1^{4}-t_2^{4})$
$t_2 = 22^{circ}c$
$t_1>t_2$ $p>0$
$t_1 = 33^{circ}c$
$t_1$p = 5.67\times10^{-8}\times0.97\times1.5\times(273 + 33)^{4}-(273+22)^{4}$
$p = ?$ $j/s$
$t(k)=t(^{circ}c)+273$

Explanation:

Step1: Convert temperatures to Kelvin

$T_1 = 33 + 273=306\ K$
$T_2 = 22+ 273 = 295\ K$

Step2: Identify the Stefan - Boltzmann constant and emissivity

The Stefan - Boltzmann constant $\sigma=5.67\times 10^{-8}\ W/(m^{2}\cdot K^{4})$, and the emissivity $\epsilon = 0.97$, the surface area $A = 1.5\ m^{2}$

Step3: Calculate the power of heat transfer by radiation

$P=\sigma\epsilon A(T_1^{4}-T_2^{4})$
$P = 5.67\times 10^{-8}\times0.97\times1.5\times(306^{4}-295^{4})$
First, calculate $306^{4}=8.58734\times10^{9}$ and $295^{4}=7.53571\times 10^{9}$
Then $306^{4}-295^{4}=8.58734\times 10^{9}-7.53571\times 10^{9}=1.05163\times 10^{9}$
$P = 5.67\times 10^{-8}\times0.97\times1.5\times1.05163\times 10^{9}$
$P=5.67\times0.97\times1.5\times10.5163$
$P\approx87.6\ W$

Answer:

$87.6\ W$