Explanation

(d) From Stefan’s law of radiation,
$E \propto T^{4}\ \Rightarrow \ \frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
Given, $T_{1} = T,\ \ \ T_{2} = 2T$
$\therefore$ $\frac{E_{1}}{E_{2}} = \frac{(T)^{4}}{(2T)^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$
$E_{2} = 16\ E_{1}$
Heat taken by water from radiation
$E = mc\mathrm{\Delta}\theta$
Where c is specific heat, $\mathrm{\Delta}\theta$ the change in temperature and $m$the mass.
$\therefore$ $E = m \times 1 \times (20.5 - 20)$
$E = m \times 0.5$ …(i)
When energy supplied is 16 times the previous one, then let temperature rise to $\theta'$
$\therefore\ \ \ \ \ \ \ \ 16E = m \times 1 \times (\theta^{'} - 20)$ …(ii)
Dividing Eq. (i) by (ii), we get
$\frac{1}{16} = \frac{0.5}{\theta^{'} - 20}$
$\Rightarrow \ \ \ \ \ \theta^{'} - 20 = 16 \times 0.5 = 8$
$\Rightarrow \ \ \ \ \ \theta^{'} = 20 + 8 = 28℃$