Question

Radiation from the head is a major source of heat loss from the human body. Model a head as a 20-cm-diameter, 20-cm-tall cylinder with a flat top. If the body’s surface temperature is 35°C, what is the net rate of heat loss on a chilly 5°C day? All skin, regardless of color, is effectively black in the infrared where the radiation occurs, so use an emissivity of 0.95.

Short answer

The accurate answer is 25.6 W.

Step-by-step solution

Given Information

Given:

Diameter of the head, $\mathrm{d}=20\mathrm{cm}$

Therefore, radius, $r=10\mathrm{cm}=0.1\mathrm{m}$

Height of the head, $h=20\mathrm{cm}=0.2\mathrm{m}$

Emissivity, $\mathrm{e}=0.95$

Surface temperature of the body, $\mathrm{T}=35°\mathrm{C}=(35+273)\mathrm{K}=308\mathrm{K}$

Ambient temperature,

$\mathrm{T}0=5°\mathrm{C}=(5+273)\mathrm{K}=278\mathrm{K}$

Calculation

On the basis of temperature $T_0$

$P=e\delta A(T^4-T_0^4)$

On the basis of provided information:

$A=\mathrm{\pi Dh}+·{\mathrm{\pi D}}^2/4=\mathrm{\pi D}(\mathrm{h}+\frac{\mathrm{D}}{2})$

On the basis of substituting numerical case

$\begin{array}{rcl}P& =& e\delta \mathrm{\pi D}(\mathrm{h}+\frac{\mathrm{D}}{4})({\mathrm{T}}^4-{\mathrm{T}}_0^4)\\ & =& 0.95·5.67·{10}^{-8}·\mathrm{\pi }·0.2·(0.2+0.2/4)·({308}^4-{278}^4)\\ & =& 25.6\mathrm{W}\end{array}$

Therefore, the accurate answer is 25.6 W.