$\left.\frac{d Q}{A d t}=e \sigma\left( T _0+\Delta T \right)^4- T _0^4\right)=\sigma T _0^4\left[\left(1+\frac{\Delta T }{ T _0}\right)^4-1\right]$

$=e \sigma T _0^4\left[\left(1+4 \frac{\Delta T }{ T _0}\right)-1\right]$

$\frac{ dQ }{ Adt }=\sigma e T _0^3 \cdot 4 \Delta T$    $. . . . .(ii)$

Now from equ. $(i)$

$ms \frac{ dT }{ dt }=\sigma e T \left( T ^4- T _0^4\right)$

$\frac{ dT }{ dt }=\frac{\sigma e A }{ ms }\left[\left( T _0+\Delta T \right)^4- T _0^4\right]$

$=\frac{\sigma e e }{ ms } T _0^4 \times\left[\left(1+\frac{\Delta T }{ T _0}\right)^4-1\right]$

$\frac{ dT }{ dt }=\frac{\sigma e A }{ ms } T _0^4-4 \Delta T$

$\frac{ dT }{ dt }=e \Delta T ;\left( K =\frac{4 \sigma e A T_0^s}{ ms }\right)$

$\Rightarrow 4 \sigma e A T_0^3=\frac{ K }{ A }( ms )$

from equ. $(1)$

$\frac{d Q}{A d t}=e \sigma I _0^3 \cdot 4 \Delta T$

$700=( K / A )( ms ) \Delta T$

$\therefore \Delta T =\frac{700 \times 5 \times 10^{-2}}{10^{-5} \times 4200}=\frac{50}{6}=\frac{25}{3}$

$\Delta T =8.33$