Topic: Heat Transfer
Surface A in the figure is coated with white paint and maintained at temperature of $27^{\circ}C$. It is located directly opposite to surface B which can be considered a black body and maintained at temperature of $727^{\circ}C$. Calculate the amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature (consider emissivity of the white paint = 1 and view factor = 1, $\sigma =$ Stefan - Boltzmann constant)
- A. $9919\times 10^{8}\sigma~W/m^{2}$
- B. $9999\times 10^{8}\sigma~W/m^{2}$
- C. $9990\times 10^{8}\sigma~W/m^{2}$
- D. $9909\times 10^{8}\sigma~W/m^{2}$
Answer: A. $9919\times 10^{8}\sigma~W/m^{2}$
Explanation: The absolute temperatures are $T_{1} = 27 + 273 = 300\text{ K}$ for surface A, and $T_{2} = 727 + 273 = 1000\text{ K}$ for surface B. The heat transfer per unit area is given by $Q/A = \sigma \epsilon F_{12} (T_{2}^{4} - T_{1}^{4})$. With $\epsilon = 1$ and $F_{12} = 1$, the heat flux is $\sigma (1000^{4} - 300^{4}) = \sigma (10^{12} - 81\times 10^{8})$. Factoring out $10^{8}$, we get $\sigma \times 10^{8}(10000 - 81) = 9919 \times 10^{8} \sigma\text{ W/m}^{2}$.