In an electric furnace used for refining steel,the temperature is monitored by measuring the radiant power emitted through a small hole in the wall of the furnace,of area 0.5 cm².This hole acts like a perfect blackbody radiator having the same temperature as the interior of the furnace.If the temperature of the furnace (and therefore of the hole)is to be maintained at 1650°C,how much power will the hole radiate?

Question

Quizplus · Accepted Answer

The power radiated from a black body is given by the Stefan-Boltzmann law:
P = σAT^4
where P is the radiant power, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the area of the hole (0.5 cm^2 = 5 x 10^-6 m^2), and T is the temperature in Kelvin (1650 + 273 = 1923 K).

Substituting the values, we get:
P = (5.67 x 10^-8)(5 x 10^-6)(1923^4)
P ≈ 40 W

Therefore, the power radiated from the hole is approximately 40 W. So, the correct choice is C) 40 W.

In an Electric Furnace Used for Refining Steel,the Temperature Is