Question

A body at a temperature of ${727}^{o}C$ and having surface area 5 $c{m}^2$ , radiates 300 J of energy each minutes.The emissivity is

• A. 0.18
• B. 0.02
• C. 0.2
• D. 0.15

Answer 1

The correct option is A 0.18

Given,

Temperature, $T={728}^{0}C=728+273=1001K$

Surface Area, $A=5\times {10}^{-4}{m}^2$

Energy radiated, $Q=300J$

Time, $t=60s$

By Stephan-Boltzmann equation,

$Q=e\sigma T^{4}At$

where,

$Q$ is radiate energy

$\sigma$ is Stephan-Boltzmann Constant

$T$ is absolute temperature in Kelvin

$e$ is emissivity of the material

$A$ is the area of the emitting body

So,

Emissivity, $e=\frac{Q}{\sigma T^{4}At}$

$=\frac{300}{(5.67\times {10}^{-8})\times (1001{)}^4\times (5\times {10}^4)\times 60}=0.18$