Question

Consider a surface at \(0^{\circ} \mathrm{C}\) that may be assumed to be a blackbody in an environment at \(25^{\circ} \mathrm{C}\). If $300 \mathrm{~W} / \mathrm{m}^{2}$ of radiation is incident on the surface, the radiosity of this black surface is (a) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(15 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(132 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(300 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(315 \mathrm{~W} / \mathrm{m}^{2}\)

Short answer

Answer: (e) 315 W/m².

Step-by-step solution

Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power per unit area radiated (P/A) from an objects is proportional to the fourth power of its temperature (T) in Kelvin (K). $$ \frac{P}{A} = σ \times ε \times T^4 $$ where σ is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W} / \mathrm{m^2 K^4}\)), and ε is the emissivity of the object. For a black body, ε = 1.

Calculate Temperatures in Kelvin

Before applying the Stefan-Boltzmann Law, we need to convert the given temperatures from Celsius to Kelvin. $$ \mathrm{Surface \ Temperature} = 0^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 273.15 \mathrm{K} $$ $$ \mathrm{Environment \ Temperature} = 25^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 298.15 \mathrm{K} $$

Find Emissive Power of Black Surface

Now we have the temperature of the black surface in Kelvin and can calculate the emissive power per unit area radiated by the surface using the Stefan-Boltzmann Law. For a black body, using ε=1: $$ \frac{P}{A} = σ \times T^4 = (5.67 \times 10^{-8} \mathrm{W} / \mathrm{m^2 K^4}) \times (273.15 \mathrm{K})^4 $$ $$ \frac{P}{A} = 315 \mathrm{W} / \mathrm{m^2} $$

Calculate Radiosity of the Black Surface

Radiosity of the black surface is the sum of the power reflected and the power emitted by the surface. Since the surface is a black body and absorbs all incident radiations, its reflectivity is 0. Hence, the power reflected will also be 0. The power emitted by the surface is the emissive power we calculated in the previous step: $$ \mathrm{Radiosity} = \mathrm{Power \ Reflected} + \mathrm{Power \ Emitted} $$ $$ \mathrm{Radiosity} = 0 \mathrm{W} / \mathrm{m^2} + 315 \mathrm{W} / \mathrm{m^2} $$ $$ \mathrm{Radiosity} = 315 \mathrm{W} / \mathrm{m^2} $$ The correct answer is: (e) \(315 \mathrm{W} / \mathrm{m}^{2}\).