You were lost while hiking outside wearing only a bathing suit. a) Calculate the power radiated from your body, assuming that your body's surface area is about \(2.00 \mathrm{~m}^{2}\) and your skin temperature is about \(33.0^{\circ} \mathrm{C} .\) Also, assume that your body has an emissivity of 1.00 . b) Calculate the net radiated power from your body when you were inside a shelter at \(20.0^{\circ} \mathrm{C}\). c) Calculate the net radiated power from your body when your skin temperature dropped to \(27.0^{\circ} \mathrm{C}\).

To work with the Stefan-Boltzmann law, we must convert the given temperatures from Celsius to Kelvin. To do this, add 273.15 to the Celsius temperature: Skin Temperature: \(33.0^{\circ} \mathrm{C} + 273.15 = 306.15 \mathrm{K}\) Shelter Temperature: \(20.0^{\circ} \mathrm{C} + 273.15 = 293.15 \mathrm{K}\) Lower Skin Temperature: \(27.0^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{K}\)

Using the Stefan-Boltzmann law, we can calculate the radiated power \(P\): \(P = e \cdot A \cdot \sigma \cdot T^{4}\) Where: \(e = 1.00\) (emissivity) \(A = 2.00 \mathrm{~m}^{2}\) (surface area) \(\sigma = 5.67 \times 10^{-8} \mathrm{W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-4}\) (Stefan-Boltzmann constant) \(T = 306.15 \mathrm{K}\) (skin temperature) Using these values: \(P = 1.00 \cdot 2.00 \cdot 5.67 \times 10^{-8} \cdot (306.15)^{4}\) \(P \approx 1159.4 \mathrm{W}\) The power radiated from the body when lost outside is approximately \(1159.4 \mathrm{W}\).

Using the Stefan-Boltzmann law, we can calculate the radiated power from the body when inside the shelter. This time, we must subtract the power absorbed from the shelter: \(P_{\text{net}} = P_{\text{body}} - P_{\text{shelter}}\) Calculate the power radiated by the shelter: \(P_{\text{shelter}} = e \cdot A \cdot \sigma \cdot T_{\text{shelter}}^{4}\) With \(T_{\text{shelter}} = 293.15 \mathrm{K}\): \(P_{\text{shelter}} = 1.00 \cdot 2.00 \cdot 5.67 \times 10^{-8} \cdot (293.15)^{4}\) \(P_{\text{shelter}} \approx 892.3 \mathrm{W}\) Now calculate the net power radiated: \(P_{\text{net}} = 1159.4 \mathrm{W} - 892.3 \mathrm{W} \approx 267.1 \mathrm{W}\) The net radiated power from the body when inside the shelter is approximately \(267.1 \mathrm{W}\).

When the skin temperature drops to \(27.0^{\circ} \mathrm{C}\) or \(300.15 \mathrm{K}\), we need to recalculate the power radiated from the body: \(P_{\text{new}} = e \cdot A \cdot \sigma \cdot (300.15)^{4} \approx 860.1 \mathrm{W}\) Calculating the new net radiated power: \(P_{\text{net, new}} = 860.1 \mathrm{W} - 892.3 \mathrm{W} \approx -32.2 \mathrm{W}\) The net radiated power from the body when the skin temperature is lowered to \(27.0^{\circ} \mathrm{C}\) is approximately \(-32.2 \mathrm{W}\), which means the body is absorbing power at this temperature.