A super-insulated house is at a temperature of 20$\degree$C.The temperature outside is 0$\degree$C.The surface area of the house is 200 m²,and the emissivity is 1.Approximately how much energy is radiated (in W)per second? A)20 000 B)2 000 C)200 D)2 E)0.2

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Explanation:The Stefan-Boltzmann law states that the power radiated per unit area of a blackbody is proportional to the fourth power of its absolute temperature. The emissivity of a surface is a measure of how closely it approximates a blackbody, with an emissivity of 1 indicating a perfect blackbody. In this case, the surface area of the house is 200 m^2, the emissivity is 1, and the temperature difference between the inside and outside of the house is 20°C. Using the Stefan-Boltzmann law, we can calculate the power radiated per second as follows:P = σ * A * e * (T1^4 - T2^4)where:σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)A is the surface area of the house (200 m^2)e is the emissivity of the surface (1)T1 is the temperature of the inside of the house (293 K)T2 is the temperature of the outside of the house (273 K)Plugging in these values, we get:P = 5.67 x 10^-8 W/m^2K^4 * 200 m^2 * 1 * (293 K^4 - 273 K^4)P = 20,000 WTherefore, the correct answer is A) 20,000.

A Super-Insulated House Is at a Temperature of 20 °\degree° CThe Temperature Outside Is 0

A Super-Insulated House Is at a Temperature of 20 $\degree$ CThe Temperature Outside Is 0