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*10^(-8)(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*10^(-8)(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.