Given, cross sectional area of nozzle
$$
A=\frac{5}{\sqrt{21}} \times 10^{-3} \mathrm{m}^{2}
$$
and rate of heat transfer $Q=\frac{1}{10}=10^{-1} \mathrm{m}^{3} / \mathrm{s}$
Heat transfer $Q=A V$ $\therefore V=\frac{Q}{A}=\frac{10^{-1} \mathrm{m}^{3} / \mathrm{s}}{\frac{5}{\sqrt{21}} \times 10^{-3} \mathrm{m}^{2}}=2 \sqrt{21} \times 10 \mathrm{m} / \mathrm{s}$
When it hits a rigid wall then maximum possible increase in temperature of water can be expressed as
$$
\frac{1}{2} m v^{2}=m s \Delta T
$$
where, $m=$ mass, $s=$ specific heat of water
$$
=1 \mathrm{cal} / \mathrm{g}=4.2 \times 10^{3} \mathrm{J} / \mathrm{kg}
$$
and $\Delta T=$ increase in temperature
$$
\begin{aligned}
\text { From Eq. }(i), \Delta T=& \frac{1}{2} \frac{v^{2}}{s} \\
&=\frac{\left(2 \sqrt{21} \times 10^{1}\right)^{2}}{2 \times 4.2 \times 10^{3}} \\
&=\frac{84 \times 10^{2}}{8.4 \times 10^{3}} \\
&=1^{\circ} \mathrm{C}
\end{aligned}
$$