Find the temperature of an oven if it radiates 8.28 cal per second through an opening, whose area is6.1cm^(2). Assume that the radiation is close to that of a black body.

Find the temperature of an oven if it radiates 8.28 cal per second thr

1.	Find the temperature of an oven if it radiates 8.28 cal per second through an opening, whose area is6.1cm^(2). Assume that the radiation is close to that of a black body.
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The emittance of the oven (the <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> radiated in <a href="https://interviewquestions.tuteehub.com/tag/one-585732" style="font-weight:bold;" target="_blank" title="Click to know more about ONE">ONE</a> second by unit <a href="https://interviewquestions.tuteehub.com/tag/surface-1235573" style="font-weight:bold;" target="_blank" title="Click to know more about SURFACE">SURFACE</a> area) is <br/> `E=(8.28cal//s)/(6.1cm^(2))` <br/> = `(8.28xx4.2J//s)/(6.1xx10^(-4)m^(2))=5.7xx10^(4)"watt"//m^(2)` <br/> From stefan - <a href="https://interviewquestions.tuteehub.com/tag/boltzmann-400317" style="font-weight:bold;" target="_blank" title="Click to know more about BOLTZMANN">BOLTZMANN</a> formula `E=sigma-T^(4)`, where `sigma=5.67xx10^(-8)W//m^(2).K^(4)` <br/> `T^(4)=E/sigma=(5.7xx10^(4))/(5.67xx10^(-8))=1xx10^(12)K^(4)` <br/> `thereforeT=10^(3)K=1000K`.</body></html>