Q 11-10-030NEETJEE MainEAMCET 2014 (Engineering)Medium
A very small hole in an electric furnace is used for heating metals. The hole nearly acts as a black body. The area of the hole is $200\ \text{mm}^2$. To keep a metal at $727°$C, the heat energy flowing through this hole per second, in joules, is ($\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$)
Answer: (D) $11.34$
Concept. The hole radiates as a black body: Stefan's law $P = \sigma AT^4$.
Step 1. Units. $T = 727 + 273 = 1000$ K; $A = 200\ \text{mm}^2 = 2 \times 10^{-4}\ \text{m}^2$.
Step 2. Power.
$$P = 5.67 \times 10^{-8} \times 2 \times 10^{-4} \times 10^{12} = 11.34\ \text{W}$$
Answer. $11.34$ (Option D).
Solution by Sreeraj P, M.Sc Physics