$600 \;mL \times 0.2\; M \quad 400 \;mL \times 0.1 \;M$

$=120\; m \;mol \quad=40 \;m\; mol$

$HNO _{3}+ NaOH \rightarrow NaNO _{3}+ H _{2} O$

$\begin{array}{lccc}\text { Bef. } & 120 & 40 \\ \text { Aft. } & 80 & 0 & 40\; m\; mol\end{array}$

$\Delta_{ r } H =40 \;m \;mol \times\left(57 \times 10^{3}\right) \frac{ J }{ mol }$

$=40 \times 10^{-3} mol \times 57 \times 10^{3} \frac{ J }{ mol }$

$=2280\; J$

$m\; S \Delta T =2280$

$\Rightarrow 1000\; mL \times \frac{1 gm }{ mL } \times 4,2 \times \Delta T =2280$

$\Delta T =\frac{2280}{4.2} \times 10^{-3}$

$=\frac{22800}{42} \times 10^{-3}$

$=542.86 \times 10^{-3}$

$\Delta T =54.286 \times 10^{-2}\; K$

$\Delta T =54.286 \times 10^{-20} \;C$

Ans $54.286$