Question

A body surface area $5c{m}^2$ and temperature ${727}^{o}C$ emits 300J of energy per minute.Find its emissivity

• A. e=0.02813
• B. e=0.01858
• C. e=0.18
• D. e=0.09838

Answer 1

The correct option is C e=0.18

For black body radiations power emitted by the body is given by:

$P=A\sigma \epsilon (T^4-T_0^4)$

Where A is area of the surface.

$\epsilon$ is the emisity or emission coefficient

T is the temperature of the surface of the body.

$T_0$ is the ambient temperature, since nothing is given then $T_0$ can be ty taken to be zero.

$\frac{300}{6}=5\times {10}^{-4}\times 5.67\times {10}^{-8}\times \epsilon \left[\right(1000{)}^4-0]$

$\epsilon =0.18$.