Question

A recruit can join the semi-secret “$\mathbf{300}\mathit{F}$” club at the Amundsen–Scott South Pole Station only when the outside temperature is below $\mathbf{-}{\mathbf{70}}^{\mathbf{0}}\mathbf{}\mathit{C}$. On such a day, the recruit first basks in a hot sauna and then runs outside wearing only shoes. (This is, of course, extremely dangerous, but the rite is effectively a protest against the constant danger of the cold). Assume that upon stepping out of the sauna, the recruit’s skin temperature is ${\mathbf{102}}^{\mathbf{0}}\mathbf{}\mathit{F}$and the walls, ceiling, and floor of the sauna room have a temperature of ${\mathbf{30}}^{\mathbf{0}}\mathit{C}$. Estimate the recruit’s surface area, and take the skin emissivity to be 0.80. (a) What is the approximate net rate P_net at which the recruit loses energy via thermal radiation exchanges with the room? Next, assume that when outdoors, half the recruit’s surface area exchanges thermal radiation with the sky at a temperature ofand the other half exchanges thermal radiation with the snow and ground at a temperature of.What is the approximate net rate at which the recruit loses energy via thermal radiation exchanges with (b) the sky and (c) the snow and ground?

Short answer

a. The approximate net rate $p_{net}$at which the recruit loses energy via thermal radiation exchanges with the room is $-86\text{W}$.

b. The approximate net rate $p_{net}$at which the recruit loses energy via thermal radiation exchanges with the sky is $2.3\times {10}^2\text{W}$.

c. The approximate net rate $p_{net}$at which the recruit loses energy via thermal radiation exchanges with the snow and ground is $3.3\times {10}^2\text{W}$.

Step-by-step solution

Stating the given data

i. Temperature of recruit’s skin $T={102}^{\text{0}}\text{F or}312 \text{K}$.

ii. Temperature of environment $T_{env}=303\text{K}$.

iii. Emissivity $\in =0.80$.

Understanding the concept of energy via thermal radiation

Thermal radiation is the process of transferring heat through electromagnetic radiation that is produced by the thermal motion of matter particles. We use the formula of the net rate at which energy exchanges via thermal radiation. Using this formula, we can find the approximate net rate at which the recruit loses energy via thermal radiation exchanges with the room, sky, snow, and ground.

Formula:

Net rate of radiation by the body, $P_{net}=\sigma \in A\left(T_{env}^4-T^4\right)$ …(i)

where$\sigma$ is the Stefan–Boltzmann constant and is equal to $(5.67\times {10}^{-8} {\text{W/m}}^{\text{2}}{\text{K}}^{\text{4}})$.

Calculation of net rate of radiation when exchange is with the room

Assume the surface area of the body to comprise the surface areas of a cylinder $A_c$, and of a sphere, $A_s$.

The total surface area of the sphere is

$$\begin{gathered} A_s=4\pi R^2 \\ =4\times 3.14\times {\left(0.10m\right)}^2 \\ =0.1256m^2 \end{gathered}$$

The lateral surface area of the cylinder is

$$\begin{gathered} A_c=2\pi rh \\ =2\pi \left(0.15m\right)\left(1.8m\right) \\ =1.6964m^2 \end{gathered}$$

Therefore, the total area of the room is given by

$$\begin{gathered} A=A_s+A_c \\ =0.1256m^2+1.6964m^2 \\ =1.8220m^2 \end{gathered}$$

Thus, using the formula of equation (i), the net rate of radiation while the exchange is within the room is given by

$$\begin{gathered} P_{net}=\left(5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}} {\text{K}}^{\text{4}}}\right)\left(0.80\right)\left(1.8220{\text{m}}^{\text{2}}\right)\left[{\left(303\text{K}\right)}^4-{\left(312 \text{K}\right)}^4\right] \\ =-86\text{W} \end{gathered}$$.

Therefore, the approximate net rate at which the recruit loses energy via thermal radiation exchanges with the room is$-86W$.

Calculation of approximate net rate of radiation when exchange is with the sky

b.

For the half surface, the area is given as:

$$\begin{gathered} A=\frac{1.8220{\text{m}}^{\text{2}}}{2} \\ =0.91m^2 \end{gathered}$$

and the temperature of the environment, $T_{\text{env}}=248\text{K}$.

Thus, using the formula of equation (i), the net rate of radiation while the exchange is within the sky is given by

$$\begin{gathered} P_{net}=\left(5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}}\right)\left(0.80\right)\left(0.91 {\text{m}}^{\text{2}}\right)\left[{\left(248\text{K}\right)}^4-{\left(312 \text{K}\right)}^4\right] \\ =-2.3\times {10}^2\text{W} \\ \left|P_{net}\right|=\left|-2.3\times {10}^2\text{W}\right| \\ =2.3\times {10}^{2}W \end{gathered}$$

Therefore, the approximate net rate$p_{net}$ at which the recruit loses energy via thermal radiation exchanges with the sky is$2.3\times {10}^2\text{W}.$

Calculation of net rate of radiation when exchange is with the snow and the ground

c.

The half of the total area is given by $A=0.91 {\text{m}}^{\text{2}}$.

The temperature of the environment, $T_{env}=193\text{K}$.

Thus, using the formula of equation (i), the net rate of radiation while the exchange is within the snow and the ground is given by

$$\begin{gathered} P_{net}=\left(5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}}\right)\left(0.80\right)\left(0.91 {\text{m}}^{\text{2}}\right)\left[{\left(193\text{K}\right)}^4-{\left(312 \text{K}\right)}^4\right] \\ =-3.3\times {10}^2\text{W} \\ \left|P_{net}\right|=\left|-3.3\times {10}^2\text{W}\right| \\ =3.3\times {10}^2\text{W} \end{gathered}$$

Therefore, the approximate net rate $p_{net}$ at which the recruit loses energy via thermal radiation exchanges with the snow and ground is $3.3\times {10}^2\text{W}.$