A) What area of a black body kept at 300°C radiates 100W?
B) If a 25m2 roof has a U-value of 0.4Wm^-2 °C^-1, how much heat, expressed in MJ, would escape through it over a day if the temperature inside the building was maintained at 7°C above the outside temperature? can anyone please just tell me what Fourmile shall I use?
A) To determine the area of a black body that radiates 100W when kept at 300°C, we can use the Stefan-Boltzmann law. This law states that the total power radiated by a black body is proportional to the fourth power of its temperature. The formula is:
P = σ * A * T^4
Where:
P is the power radiated (in watts),
σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 Wm^-2K^-4),
A is the area of the black body (in square meters),
T is the temperature (in Kelvin).
We know the power radiated (P = 100W) and the temperature (T = 300°C = 573K). Rearranging the formula, we can solve for A:
A = P / (σ * T^4)
Substituting the values, we get:
A = 100 / (5.67 x 10^-8 * 573^4)
Calculating this expression will give you the required area of the black body.
B) To determine the amount of heat lost through a roof with a given U-value over a day, you can use the formula:
Heat loss = U * A * ΔT * time
Where:
U is the U-value of the roof (in Wm^-2°C^-1),
A is the area of the roof (in square meters),
ΔT is the temperature difference between inside and outside (in degrees Celsius),
time is the duration (in seconds or hours).
In this case, you mentioned that the roof has a U-value of 0.4Wm^-2°C^-1, the area is 25m^2, and the temperature inside is maintained 7°C above the outside temperature.
To convert the heat loss from watts to megajoules (MJ), you need to divide the result by 3.6 x 10^6 (since 1 watt-hour equals 3.6 x 10^6 joules).
Regarding your mention of "Fourmile shall I use," it seems you might be referring to a specific calculation or parameter. However, with the provided information, the formulas mentioned above should be sufficient for your calculations.