A child has a temperature of 101°F. If her total cross-sectional area is 2 m², find the energy lost each second (in W) due to radiation, assuming the emissivity is 1. (Assume the room temperature is 70°F.) A) 217 B) 180 C) 90 D) 68 E) 850

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Quizplus · Accepted Answer

The energy lost per second due to radiation can be calculated using the Stefan-Boltzmann law, which is given by $P = \epsilon \sigma A (T^4 - T_0^4)$, where $P$ is the power or energy lost per second in watts, $\epsilon$ is the emissivity, $\sigma$ is the Stefan-Boltzmann constant ($5.67 \times 10^{-8} \, \text{Wm}^{-2}\text{K}^{-4}$), $A$ is the surface area in square meters, $T$ is the temperature of the object in Kelvin, and $T_0$ is the temperature of the surroundings in Kelvin.First, convert the temperatures from Fahrenheit to Kelvin:- Child's temperature: $101°F = (101 - 32) \times \frac{5}{9} + 273.15 = 38.33 \times \frac{5}{9} + 273.15 \approx 310.93 \, K$- Room temperature: $70°F = (70 - 32) \times \frac{5}{9} + 273.15 = 21.11 \times \frac{5}{9} + 273.15 \approx 294.26 \, K$Given that the emissivity $\epsilon = 1$, the total cross-sectional area $A = 2 \, \text{m}^2$, and using the Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8} \, \text{Wm}^{-2}\text{K}^{-4}$, we can calculate the energy lost per second as follows:$$P = 1 \times 5.67 \times 10^{-8} \times 2 \times (310.93^4 - 294.26^4) \, \text{W}$$$$P \approx 217 \, \text{W}$$Therefore, the correct answer is 217 W, which corresponds to choice (A).

A Child Has a Temperature of 101°F