The radiative heat transfer rate per unit area ($W/m²$) between two plane parallel grey surfaces (emissivity 0.9) maintained at 400 K and 300 K is (Stefan Boltzmann constant $\sigma = 5.67 \times 10^{-8} W/m^2K^4$)
- A. 1020
- B. 464
- C. 812
- D. 567
Answer: C. 812
Explanation: $q = \frac{\sigma(T_1^4 - T_2^4)}{1/\epsilon_1 + 1/\epsilon_2 - 1}$. Substituting values gives $q \approx 812 W/m²$.