A super-insulated house is at a temperature of 20°C. The temperature outside is 0°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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Quizplus · Accepted Answer

The rate of heat transfer by radiation is given by the Stefan-Boltzmann law:$$P = \epsilon \sigma A T^4$$where:* P is the power (in watts)* $\epsilon$ is the emissivity (a dimensionless number between 0 and 1)* $\sigma$ is the Stefan-Boltzmann constant ($5.67 imes 10^{-8} ext{ W/m}^2 ext{K}^4$)* A is the surface area (in square meters)* T is the temperature (in Kelvin)In this problem, we have:* $\epsilon = 1$* $\sigma = 5.67 imes 10^{-8} ext{ W/m}^2 ext{K}^4$* A = 200 m²* $T_1 = 20°C = 293 K$* $T_2 = 0°C = 273 K$So, the power radiated by the house is:$$P = \epsilon \sigma A (T_1^4 - T_2^4)$$$$P = (1)(5.67 imes 10^{-8} ext{ W/m}^2 ext{K}^4)(200 ext{ m}^2) (293 K^4 - 273 K^4)$$$$P = 20,000 ext{ W}$$Therefore, the correct answer is A.

A Super-Insulated House Is at a Temperature of 20°C