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Estimate the rate of heat loss by radiation of an athlete, sitting unclothed in a locker room, whose dark walls (emissivity 0.90) are at a temperature of \SI{15}{\degreeCelsius}. Assume the skin temperature to be \SI{34}{\degreeCelsius} and its emissivity to be $0.70$. The surface area of the body not in contact with the chair is assumed to be $\SI{1.5}{\meter\squared}$.
The heat loss of the athlete is the difference between the from the athlete emitted heat and his absorbed heat, i.e.: \begin{align} P &= P_{\text{\scriptsize emitted}}  P_{\text{\scriptsize absorbed}}\\ &= \epsilon_1\cdot \sigma\cdot A \cdot T_1^4  \epsilon_0\cdot \sigma\cdot A \cdot T_0^4\\ &= \left(\epsilon_1 \cdot T_1^4  \epsilon_1 \epsilon_0 \cdot T_0^4\right)\cdot\sigma A\\ &= \left(0.70\cdot (\pq{307}{K})^4  0.90 \cdot 0.70 \cdot (\SI{288}{K})^4\right)\cdot \SI{5.67e8}{\watt\per\kelvin\tothe 4 \per\meter\squared} \cdot \SI{1.5}{\meter\squared}\\ &= \SI{139}{W} \end{align}
14:08, 12. June 2017  si  Urs Zellweger (urs)  Current Version 