A black body radiates heat energy at the rate of 2 xx 10^(5)J//s m^(2) at the temperature of 127^(@)C. Temperature of the black body at which rate of heat radiation 32 xx 10^(5) J//s m^(2), is

A black body radiates heat energy at the rate of 2 xx 10^(5)J//s m^(2)

1.	A black body radiates heat energy at the rate of 2 xx 10^(5)J//s m^(2) at the temperature of 127^(@)C. Temperature of the black body at which rate of heat radiation 32 xx 10^(5) J//s m^(2), is
Answer» <html><body><p>400K<br/>600K<br/>800K<br/>200K</p>Solution :Here, `E_(1) = 2 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)J//s m^(2), T_(1) = 127^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>= 400K and E_(2) = 32 xx 10^(5) J//s m^(2)` <br/> By Stefan.s <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a>, the rate of emission of radiated energy per unit area per unit time is <br/> `E= sigma T^(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` <br/> `:. (E_(2))/(E_(1)) = (T_(2)^(4))/(T_(1)^(4)) or (T_(2))/(T_(1)) = ((E_(2))/(E_(1)))^(1//4) = ((32 xx 10^(5))/(2 xx 10^(5)))^(1//4) = 2` <br/> or `T_(2) = 2 xxT_(1) =2 xx 400= 800K`</body></html>