Respuesta :
The correct answer of this question: 18789.79 watt.
EXPLANATION:
As per the question, the surface area of the log is given as A = [tex]0.25\ m^2[/tex]
The temperature of the of the surface of the log t = 800 [tex]^0C[/tex]
We know that [tex]t^0C\ =(273+t)\ K[/tex]
⇒ 800 degree celsius= 800+273 K
= 1073 K.
Here, K stands for kelvin scale.
We are asked to calculate the thermal power radiation .
The power is defined as the amount of energy released per unit time.
Hence, power is calculated as-
Power P = [tex]\sigma AT^4[/tex]
= [tex]5.67\times 10^{-8} \times 0.25\times (1073)^4\ watt[/tex]
= 18789.79 watt. [ans]
Here, [tex]\sigma[/tex] is the Stefan-Boltzmann constant and T is the absolute temperature of the log.
The power emitted as thermal radiation by the burning log is 18.79 Kilowatt.
Given the data in the question;
- Surface area; [tex]A = 0.25m^2[/tex]
- Temperature; [tex]T = 800 ^oC = [ 800 + 273]K = 1073K[/tex]
Power; [tex]P = \ ?[/tex]
To determine how much power it emit as thermal radiation, we use Stefan's law of black body radiation:
[tex]Power = A*\alpha * T^4[/tex]
Where A is area, T is the temperature and α is the Stefan-Boltzmann constant ([tex]5.67 * 10^{-8} W/m^2 K^4[/tex])
We substitute our values into the equation
[tex]Power = 0.25m^2\ *\ (5.67 * 10^{-8} W/m^2 K^4) * (1073K)^4\\\\Power = 0.25m^2\ *\ (5.67 * 10^{-8} W/m^2 K^4) * (1.3255*10^{12} K^4)\\\\Power = 18789.8W\\\\Power = 18.79 kW[/tex]
Therefore, the power emitted as thermal radiation by the burning log is 18.79 Kilowatt.
Learn more: https://brainly.com/question/10731660
