Question

An electric heater of surface area $200$ cm${}^2$ emits radiant energy of $60kJ$ at a time interval of $1$ min. Determine its emissive power.

• A. 50000 W/m${}^2$
• B. 40000 W/m${}^2$
• C. 20000 W/m${}^2$
• D. 60000 W/m${}^2$

Answer 1

Answer: (A)

50000 W/m${}^2$

Given,

$\Delta Q=60kJ$

$A=200\ast {10}^{-4}{m}^2$

$\Delta T=60s$

Emissive power $E=\frac{\Delta Q}{\Delta T}\frac{1}{A}=\frac{60}{200\ast {10}^{-4}\ast 60}=50kW{m}^{-2}$