Question

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?

Answer 1

Answer: 3.7×10¹²watts

Explanation:

Radiation is one of the mode of heat transfer and modes differs from each other based on their medium of heat transfer. Radiation is a process of transferring heat energy from one point to another without heating the intervening medium (no material medium is required).

According to Stefan's law of radiation, the rate of emission of radiant energy is directly proportional to the fourth power of its absolute temperature.

Mathematically, R = eAT⁴

e is constant of proportionality called emissivity. Emissivity varies depending on the type of body being considered.

For the question, we are considering black body and emissivity of black body is 1 being a perfect body.

A is the area of the body

T is the absolute temperature

e = 1

A = 0.5cm²

T = 1650°C

Rate of radiation = 1×0.5×1650⁴

= 3.7×10¹²watts.

The hole will therefore radiate 3.7×10¹²watts