A tall unclothed athlete with a skin temperature of 34oC stands in a dark room with a temperature o 23oC. The athlete’s skin has a surface area of 1.7m2. What is the rate of heat transfer via radiation from the athlete? (eskin = 0.97)
Q(t)=εσAT4Q(t)=ε σAT^4Q(t)=εσAT4
εskin=0.97ε_{skin}=0.97εskin=0.97
σ=5.67×10−5σ=5.67×10^{-5}σ=5.67×10−5
A=1.7m2A=1.7m^2A=1.7m2
T=307KT=307KT=307K
Q(t)=0.97×5.67×10−8×1.7×3074Q(t)=0.97×5.67×10^{-8}×1.7×307^4Q(t)=0.97×5.67×10−8×1.7×3074
Q(t)=488.55WattQ(t)=488.55WattQ(t)=488.55Watt
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