1.

Two black bodies at temperature 327^(@)C and 427^(@)C are kept in an evacuated chamber at 27^(@)C. The ratio fo their rates of loss of heat is

Answer»

`((6)/(7))`
`((6)/(7))^(2)`
`((6)/(7))^(3)`
`(243)/(464)`

Solution :Rate of loss of heat by a black BODY through radiation is where, `E= sigma A (T^(4) - T_(S)^(4))`
`:. (E_(1))/(E_(2)) = (T_(1)^(4) - T_(S)^(4))/(T_(2)^(4) - T_(S)^(4)) = ((327 + 273)^(4)- (27+ 273)^(4))/((427 +273)^(4)- (27 + 273)^(4)) `
`=((600)^(4) - (300)^(4))/((700)^(4)-(300)^(4)) = (10^(8) (6^(4)-3^(4)))/(10^(8) (7^(4)-3^(4)))= (((6^(2))^(2) - (3^(2))^(2)))/(((7^(2))^(2)- (3^(2))^(2)))`
`= ((6^(2) -3^(2)) (6^(2) + 3^(2)))/((7^(2) - 3^(2)) (7^(2) + 3^(2))) [ :. (a^(2)-b^(2)) = (a +b) (a-b)]`
`=((36-9) (36+ 9))/((49-9) (49+ 9)) = ((27) (45))/((40) (58)) = (243)/(464)`


Discussion

No Comment Found