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# A recruit can join the semi-secret “300 F” club at the Amundsen-Scott South Pole Station only when the outside temperature is below $-{{70}^{0}}C$. On such a day, the recruit first basks in a hot sauna and then runs outside wearing only shoes. (This is against the constant danger of the cold). Assuming that upon stepping out of the sauna, the recruit’s skin temperature is ${{102}^{0}}F$ and the walls, ceiling and the floor of the sauna room have the temperature of ${{30}^{0}}C$. Estimate the recruit’s surface area and take the skin emissivity to be 0.80. What is the approximate net rate of power 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 of $-{{25}^{0}}C$ and the other half exchanges thermal radiation with the snow and ground at a temperature of $-{{80}^{0}}C$. What is the approximate net rate at which the recruit loses energy via thermal radiation exchanges?

Last updated date: 13th Jul 2024
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Hint: We need to relate the given temperatures of the recruit’s body, the sauna room, the outside sky and the snow to find the thermal radiation exchange rates. We need to use Stefan-Boltzmann's law to find the solution for this problem.

We know from Stefan-Boltzmann's law of thermal radiation that the rate of emission of thermal radiation is proportional to the area of the radiating surface, the fourth power of the temperature difference between the body and the surrounding and the emissivity of the body. This can be mathematically given as –
$P=e\sigma A({{T}^{4}}-{{T}_{S}}^{4})$
Where, P is the thermal radiation rate,
e is the emissivity of the object,
$\sigma$ is the Stefan’s constant,
T is the temperature of the object and
${{T}_{S}}$ is the temperature of the surrounding.
Now, let us consider the situation given to us. We have men being recruited to the club. We can estimate the height of the man to be almost 2 meters. We can consider his body as a cylinder (body) of height 1.8 m and radius 0.15 m and a sphere (head) of radius 0.10 m. We can find the surface area of his entire body including head as –
\begin{align} & A={{A}_{cyl}}+{{A}_{sph}} \\ & \Rightarrow A=2\pi rh+\pi {{r}^{2}}+4\pi r{{'}^{2}} \\ & \Rightarrow A=\pi \{(2\times 0.15\times 1.8)+({{0.15}^{2}})+4({{0.10}^{2}})\} \\ & \therefore A=1.8{{m}^{2}} \\ \end{align}
Now, lets us consider each of the cases given in the question one-by-one.
Case I: The thermal radiation exchange between the body at ${{102}^{0}}F$and the sauna bath at ${{30}^{0}}C$.
Let us convert both the temperatures into Kelvin scale as –
\begin{align} & \Rightarrow {{102}^{0}}F=\dfrac{5}{9}{{(102-32)}^{0}}C={{38.89}^{0}}C \\ & \therefore {{102}^{0}}F=(273+38.89)K=311.89K \\ & \Rightarrow {{30}^{0}}C=(273+30)K=303K \\ \end{align}
The thermal radiation exchange rate using Stefan-Boltzmann’s law as –
\begin{align} & P=e\sigma A({{T}^{4}}-{{T}_{S}}^{4}) \\ & \Rightarrow P=e\sigma A({{T}_{body}}^{4}-{{T}_{sauna}}^{4}) \\ & \Rightarrow P=0.80\times 5.6703\times {{10}^{-8}}\times 1.8({{311.89}^{4}}-{{303}^{4}}) \\ & \therefore P=84.4W \\ \end{align}
The power radiated is 84.4W from the recruit’s body to the sauna bath.
Case II: The thermal radiation exchange between the half the area of the body at 311.89K and the sky temperature of $-{{25}^{0}}C$.
The sky temperature of $-{{25}^{0}}C$ can be converted to Kelvin scale as –
$\Rightarrow -{{25}^{0}}C=(273-25)K=248K$
The thermal radiation exchange rate using the Stefan-Boltzmann’s law is given as –
\begin{align} & P=e\sigma A({{T}^{4}}-{{T}_{S}}^{4}) \\ & \Rightarrow P=e\sigma {{A}_{half}}({{T}_{body}}^{4}-{{T}_{sky}}^{4}) \\ & \Rightarrow P=0.80\times 5.6703\times {{10}^{-8}}\times 0.9({{311.89}^{4}}-{{248}^{4}}) \\ & \therefore P=231.8W \\ \end{align}
The power radiated is 231.8W from the recruit’s body to the sky.
Case III: The thermal radiation exchange rate from the half area of the recruit’s body at 311.89K to the snow and ground at temperature $-{{80}^{0}}C$.
The ground temperature can be converted to the Kelvin scale as –
$\Rightarrow -{{80}^{0}}C=(273-80)K=193K$
The thermal radiation exchange rate using the formula is given as –
\begin{align} & P=e\sigma A({{T}^{4}}-{{T}_{S}}^{4}) \\ & \Rightarrow P=e\sigma {{A}_{half}}({{T}_{body}}^{4}-{{T}_{snow}}^{4}) \\ & \Rightarrow P=0.80\times 5.6703\times {{10}^{-8}}\times 0.9({{311.89}^{4}}-{{193}^{4}}) \\ & \therefore P=329.6W \\ \end{align}
The power radiated from the recruit to the ground is about 330W.
So, we get the thermal radiation exchange rate for the three situations as –
(i)From body to sauna as 84.4W
(ii) From body to sky as 231.8W
(iii) From body to ground as 330W.
These are the required solutions.

Note:
The quantity we have found is the net power radiated from the recruit’s body in the form of heat. We know that the power radiated is the rate of change of energy radiated. So, it is the thermal radiation exchange rate that we have found as required in the problem.