A brick wall with a thermal emissivity of 0.8 is heated by the sun all day and is at a temperature of 25 °C in early evening. A pedestrian walking past the wall feels its warmth radiating in to the cool evening air. The wall has surface area of 5m^2. What is the total out put power of the thermal radiation emitted by the wall? Ignore radiation absorbed by the wall from surrounding bodies.
(Area)*(sigma)*(0.8)*(298)^4 = ?
sigma is the Stefan-Bolzmann constant, which you should look up.
Note the T^4 term, where the temeprature is in Kelvn.
THANK YOU
To calculate the total output power of the thermal radiation emitted by the wall, we can use the Stefan-Boltzmann Law. This law states that the power radiated by an object is proportional to the fourth power of its temperature.
The equation is given by:
P = εσAT^4
where P is the power emitted, ε is the emissivity of the object, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area of the object, and T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
T = 25 + 273.15 = 298.15 K
Now, we can substitute the values into the equation:
P = (0.8)(5)(5.67 x 10^-8)(298.15^4)
Calculating this expression will give us the total output power of the thermal radiation emitted by the wall.