Question

Your roof radiates heat away at night. At what net rate does the heat radiate from a 275 m² black roof on a night when the roof temperature is 33°C in the surrounding temperature is 14°C? The emissivity of the roof is 0.900.

Answer 1

We are asked to determine the net rate of radiation of a roof given its area and its emisivity. To do that we will use the following formula:

$P_{net}=\sigma eA(T_0^4-T^4)$

Where:

$\begin{gathered} P_{net}=\text{ net rate of radiation} \\ \sigma=\text{ Stefan-Boltzmann's constant} \\ e=\text{ emisivity} \\ A=\text{ area} \\ T_0=\text{ surrounding temperature} \\ T=\text{ temperature of the roof} \end{gathered}$

The Stefan-Boltzmann's constant is given by:

$\sigma=5.67\times10^{-8}\frac{W}{m^2K^4}$

Now, we need to convert the temperature to Kelvin. To do that we use the following:

$T_K=T_c+273.15$

Where:

$\begin{gathered} T_K=\text{ temperature in Kelvin} \\ T_C=\text{ temperature in Celsius} \end{gathered}$

For the 33°C we have:

$T_k(33)=33+273.15=306.15$

For the 14°C:

$T_K(14)=14+273.15=287.15$

Now, we substitute the values:

$P_{net}=(5.67\times10^{-8}\frac{W}{m^2K^4})(0.9)(275m^2)((287.15K)^4-(306.15K)^4)$

Solving the operation:

$P_{net}=-27870.84W$

Therefore, the net rate of radiation is -27870.84 Watts.