Question

An electric heater of surface area $200c{m}^2$ emits radiant energy of $60kJ$ at time interval of $1min$. If its emissivity be $0.45$, the radiant energy emitted by a black body in one hour, identical to the electrical heater in all respects in kJ is $x\times 4000$.Find the value of x.

Answer 1

Given $\Delta Q=60kJ$; $A=200\times {10}^{-4}{m}^2$ and $\Delta t=60s$
Using, emissive power $E=\left(\frac{\Delta Q}{\Delta t}\right)\frac{1}{A}$, we have
$E=\frac{60kJ}{200\times {10}^{-4}{m}^2\times 60s}=50kW/m^2$
$\therefore$ Emissivity (of a surface)
$e=\frac{emissivepower\left(ofthatsurface\right)}{emissivepower\left(ofablackbody\right)}$
$\therefore$ Emissive power of a black body
$=\frac{50}{0.45}kW/m^2=111.11kW/m^2$
$\therefore$ Heat radiated $\Delta Q=EA\Delta t$
$=\left(111.11kW/m^2\right)(200\times {10}^{-4}{m}^2)(60\times 60s)=8000kJ$