Consider a person whose exposed surface area is \(1.7 \mathrm{~m}^{2}\), emissivity is \(0.5\), and surface temperature is \(32^{\circ} \mathrm{C}\). Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of \((a) 300 \mathrm{~K}\) and (b) \(280 \mathrm{~K}\). Answers: (a) \(26.7 \mathrm{~W}\), (b) \(121 \mathrm{~W}\)

We are given the surface temperature \(32^{\circ} \mathrm{C}\). First, convert this temperature to Kelvin by adding \(273.15\): \(T_{surface} = 32 + 273.15 = 305.15 \mathrm{~K}\)

Stefan-Boltzmann Law states that the rate of heat transfer by radiation is: \(Q = e \cdot s \cdot A (T_{surface}^4 - T_{room}^4)\) where \(e\) is the emissivity, \(s\) is the Stefan-Boltzmann constant (\(5.67 \cdot 10^{-8} \mathrm{~W/m^2K^4}\)), \(A\) is the person's exposed surface area, \(T_{surface}\) is the surface temperature in Kelvin, and \(T_{room}\) is the room walls' temperature in Kelvin. Now, we will apply this formula to calculate the rate of heat loss for both cases: (a) For a room with walls' temperature \(300 \mathrm{~K}\): \(Q_{a} = (0.5) \cdot (5.67 \cdot 10^{-8}) \cdot (1.7) \cdot (305.15^4 - 300^4)\) \(Q_{a} \approx 26.7 \mathrm{~W}\) (b) For a room with walls' temperature \(280 \mathrm{~K}\): \(Q_{b} = (0.5) \cdot (5.67 \cdot 10^{-8}) \cdot (1.7) \cdot (305.15^4 - 280^4)\) \(Q_{b} \approx 121 \mathrm{~W}\) Therefore, the rate of heat loss from the person by radiation in a large room having walls at a temperature of (a) \(300 \mathrm{~K}\) is \(26.7 \mathrm{~W}\), and (b) \(280 \mathrm{~K}\) is \(121 \mathrm{~W}\).