1. Write the reaction

$HNO_3 + NaOH \rightarrow NaNO_3 + H_2O$

For strong acid–strong base neutralisation, heat released is:

$\Delta H_{\text{neut}} = 57\,\text{kJ mol}^{-1}$

This is per mole of water formed (or per mole of acid-base neutralised).

2. Calculate moles of acid and base

For $HNO_3$:

$n_{HNO_3} = 0.2 \times 0.6 = 0.12\,\text{mol}$

For $NaOH$:

$n_{NaOH} = 0.1 \times 0.4 = 0.04\,\text{mol}$

Since the reaction is $1:1$, the limiting reagent is $NaOH$.

So, moles neutralised:

$n = 0.04\,\text{mol}$

3. Heat evolved

$q = n \times 57\,\text{kJ} = 0.04 \times 57 = 2.28\,\text{kJ}$

$q = 2280\,\text{J}$

4. Mass of solution

Total volume mixed:

$600\,\text{mL} + 400\,\text{mL} = 1000\,\text{mL}$

Assuming density of solution $\approx 1\,\text{g mL}^{-1}$,

$m = 1000\,\text{g}$

5. Use heat relation

$q = mc\Delta T$

Given:

$c = 4.2\,\text{J g}^{-1}\text{K}^{-1}$

So,

$\Delta T = \frac{q}{mc} = \frac{2280}{1000 \times 4.2}$

$\Delta T = \frac{2280}{4200} = 0.542857\,^\circ\text{C}$

$\Delta T \approx 0.54\,^\circ\text{C} = 54 \times 10^{-2}\,^\circ\text{C}$

6. Final integer answer

The blank is:

$\boxed{54}$

This means the rise in temperature is

$54 \times 10^{-2}\,^\circ\text{C}$