CGPET 2019 Solved Paper

Given,
Rate of radiation of black body at 0°C,

dθ1

dt

=E
Now, Stefan's law, rate at which heat is emitted is given as

dθ

dt

=σεAT4
where, σ= Stefan constant and A= area of surface.
∴

dθ1

dt

=(σεA)T14
or E=(σεA)(0+273)4 .......(i)
Then, the rate of radiation of this black body at 273°C,
∴

dθ2

dt

=(σεA)T24
Let,

dθ2

dt

=X
or X=(σεA)T24(∵T2=273°C)
or X=(σεA)(273°+273°) ........(ii)
Dividing Eqs. (i) by (ii), we get
∴

E

X

=

(0°+273)4

(273+273)4

=

(273)4

(546)4

=

(1)4

(2)4

=

1

24

=

1

16

or X=16E
Therefore, the rate of radiation of this black body at 273°C will be 16E.