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 $0.5 \mathrm {~cm} ^ { 2 }$ 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 $1650 ^ { \circ } \mathrm { C } \text {, }$ how much power will the hole radiate?

Question

Quizplus · Accepted Answer

The power radiated by a blackbody is given by the Stefan-Boltzmann law, $P = \sigma A T^4$, where $\sigma = 5.67 \times 10^{-8} \, \text{W m}^{-2} \text{K}^{-4}$ is the Stefan-Boltzmann constant, $A$ is the area in square meters, and $T$ is the temperature in Kelvin. To find the power radiated through the hole, convert the area to square meters ($0.5 \, \text{cm}^2 = 0.5 \times 10^{-4} \, \text{m}^2$) and the temperature to Kelvin ($1650^\circ \text{C} = 1650 + 273 = 1923 \, \text{K}$). Plugging these values into the formula gives $P = 5.67 \times 10^{-8} \times 0.5 \times 10^{-4} \times (1923)^4 \approx 40 \, \text{W}$.

In an Electric Furnace Used for Refining Steel, the Temperature 0.5 cm20.5 \mathrm {~cm} ^ { 2 }0.5 cm2

In an Electric Furnace Used for Refining Steel, the Temperature $0.5 \mathrm {~cm} ^ { 2 }$