Final answer:
The net rate at which heat radiates from a 275-m² black roof at a temperature difference using radiative heat transfer is calculated using the Stefan-Boltzmann law. The temperatures are converted to Kelvin, and values are inserted into the equation. The resulting net rate of heat radiation is approximately 41.25 kW.
Step-by-step explanation:
The student is asking about the net rate at which heat radiates from a black roof at a certain temperature difference with its surroundings. The calculation for this scenario uses the Stefan-Boltzmann law for radiative heat transfer, which can be expressed as:
\(P = e \cdot A \cdot \sigma \cdot (T_{roof}^4 - T_{surroundings}^4)\)
where:
- \(P\) is the net power of heat radiation,
- \(e\) is the emissivity of the roof (0.900),
- \(A\) is the area of the roof (275 m²),
- \(\sigma\) is the Stefan-Boltzmann constant \(\approx 5.67 \times 10^{-8} W/m^2K^4\),
- \(T_{roof}\) is the roof's temperature in Kelvin,
- \(T_{surroundings}\) is the surrounding temperature in Kelvin.
First, convert the temperatures from Celsius to Kelvin by adding 273.15:
- \(T_{roof} = 30.0^\circ C + 273.15 = 303.15 K\),
- \(T_{surroundings} = 15.0^\circ C + 273.15 = 288.15 K\).
Then plug the values into the equation:
\(P = 0.900 \cdot 275 m^2 \cdot 5.67 \times 10^{-8} W/m^2K^4 \cdot ((303.15 K)^4 - (288.15 K)^4)\)
After calculating, we find that the net rate of heat radiation is approximately 41.25 kW, which is option (a).