Question

A log of burning wood in the fireplace has a surface temperature of 450°C. Assume that the emissivity is 1 (a perfect black body), calculate the radiant emission of energy per unit surface area.

Answer 1

It radiates at a rate of
`15506.98Wm^(-2)`

Step-by-step explanation:

We shall use Stefan–Boltzmann law to calculate radiant emission of energy

According to Stefan–Boltzmann law we have

`P=e\sigma AT^(4)`

For a perfect black body we have 'e' =1

It is given surface temperature =
`450^(o)C=723.15 Kelvins`

Thus applying values in the above equation we have

energy per unit surface area=

`(P)/(A)=e\sigma T^(4)\\\\\therefore (P)/(A)=1* 5.67* 10^(-8)* 723.15^(4)\\\\(P)/(A)=15506.98Wm^(-2)`