Question

In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area of a human body may be considered to be that of the sides of a cylinder of length L=2.0m and circumference C=0.8m.For the Stefan-Boltzmann constant use σ=5.67×10−8W/m2/K4.If the surface temperature of the skin is taken to be Tbody=30∘C, how much thermal power Prb does the body described in the introduction radiate?Take the emissivity to be e=0.6.Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.

Answer 1

Answer:$P=14.6 W$

Explanation:

According to the Stefan-Boltzmann law for real radiating bodies:

$P=\sigma A \epsilon T^{4}$ (1)

Where:

$P$ is the energy radiated (in Watts)

$\sigma=5.67(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.  

$A$ is the Surface area of the body  

$T=30\°C + 273.15= 303.15 K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin

$\epsilon=0.6$ is the body's emissivity

On the other hand, we are told the human body is roughly approximated to a cylinder of length $L=2.0m$ and circumference $C=0.8m$.

The circumference of a circle is:$C=0.8m=2 \pi r$ where $r$ is the radius. Hence $r=\frac{0.8m}{2 \pi}=0.1273 m$.

Now we have to input this value for $r$  in the Area of a cylinder formula:

$A=\pi r^{2}L$

$A=\pi (0.1273 m)^{2}(2 m)$

$A=0.0509 m^{2}$ (2)

Substituting (2) in (1):

$P=(5.67(10)^{-8}\frac{W}{m^{2} K^{4}}) (0.0509 m^{2}) (0.6) (303.15 K)^{4}$ (3)

Finally:

$P=14.62 W \approx 14.6 W$