Question 11.12: Polar Bear Club Goal Apply Stefan’s law. Problem A member of...
Polar Bear Club
Goal Apply Stefan’s law.
Problem A member of the Polar Bear Club, dressed only in bathing trunks of negligible size, prepares to plunge into the Baltic Sea from the beach in St. Petersburg, Russia. The air is calm, with a temperature of 5^{\circ} \mathrm{C}. If the swimmer’s surface body temperature is 25^{\circ} \mathrm{C}, compute the net rate of energy loss from his skin due to radiation. How much energy is lost in 10.0 min? Assume his emissivity is 0.900 , and his surface area is 1.50 \mathrm{~m}^{2}.
Strategy Use Equation 11.11,
\mathscr{P}_{\mathrm{net}}=\sigma A e(T^{4}-T_{0}^{4}) (11.11)
the thermal radiation equation, substituting the given information. Remember to convert temperatures to Kelvin by adding 273 to each value in degrees Celsius!
Convert temperatures from Celsius to Kelvin:
\begin{aligned} T_{5^{\circ} \mathrm{C}} & =T_{C}+273=5+273=278 \mathrm{~K} \\ T_{25^{\circ} \mathrm{C}} & =T_{C}+273=25+273=298 \mathrm{~K} \end{aligned}
Compute the net rate of energy loss, using Equation 11.11:
\begin{aligned} \mathscr{P}_{\text {net }}= & \sigma A e\left(T^{4}-T_{0}^{4}\right) \\ = & \left(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)\left(1.50 \mathrm{~m}^{2}\right) \\ & \times(0.90)\left[(298 \mathrm{~K})^{4}-(278 \mathrm{~K})^{4}\right] \\ \mathscr{P}_{\text {net }}= & 146 \mathrm{~W} \end{aligned}
Multiply the preceding result by the time, 10 minutes, to get the energy lost in that time due to radiation:
Q=\mathscr{P}_{\text {net }} \times \Delta t=(146)\left(6.00 \times 10^{2} \mathrm{~s}\right)=8.76 \times 10^{4} \mathrm{~J}
Remarks Energy is also lost from the body through convection and conduction. Clothing traps layers of air next to the skin, which are warmed by radiation and conduction. In still air these warm layers are more readily retained. Even a Polar Bear Club member enjoys some benefit from the still air, better retaining a stagnant air layer next to the surface of his skin.