The temperature of body is increased from `T_(1)=127^(@)C" to "T_(2)=227^(@)C`. The ambient temperature is `27^(@)`C. The energies emitted per second by the body at `T_(1)andT_(2)` are `E_(1)andE_(2)` respectively. The ratio of `(E_(2))/(E_(1))` is

Options

A. 1.8
B. 2.7
C. 3.1
D. 4.3

Explanation

Given, temperature of bohy increases from `T_(0)=127^(@)C" to "T_(2)=227^(@)C`
and ambient temperature,
`T_(0)=27^(@)`
By Stefan-Boltzman's law energy emitted per second is given by
`E=epsisigma(T^(4)-T_(0)^(1))`
i.e., `Eprop[T^(4)-T_(0)^(4)]`
`(E_(2))/(E_(1))=((T_(2)^(4)-T_(0)^(4)))/((T_(1)^(2)-T_(0)^(4)))`
Putting the given values, we get
`=((227)^(2)-(27)^(4))/((127)^(4)-(27)^(4))=(2654706400)/(259613200)`
Hence, the ratio `(E_(2))/(E_(1))` is 10.22. So, no option is correct.