1. Write the reaction

$\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$

For a strong acid–strong base reaction, the molar heat of neutralization is given as:

$\Delta H = -57.1\,\text{kJ mol}^{-1}$

This is per mole of water formed (or per mole of acid/base neutralized).

2. Calculate moles of HCl and NaOH

For HCl:

$n_{\mathrm{HCl}} = 0.2\times 0.200 = 0.040\,\text{mol}$

For NaOH:

$n_{\mathrm{NaOH}} = 0.1\times 0.300 = 0.030\,\text{mol}$

So, NaOH is the limiting reagent.

Therefore, moles neutralized:

$n = 0.030\,\text{mol}$

3. Heat released during neutralization

$q = n\times 57.1\,\text{kJ} = 0.030\times 57.1 = 1.713\,\text{kJ}$

Convert to joules:

$q = 1713\,\text{J}$

4. Mass of the resulting solution

Total volume mixed:

$200\,\text{mL} + 300\,\text{mL} = 500\,\text{mL}$

Given density $=1.00\,\text{g cm}^{-3}$, mass is:

$m = 500\,\text{g}$

5. Use heat formula

Assuming all released heat raises the temperature of the solution,

$q = mc\Delta T$

So,

$\Delta T = \frac{q}{mc} = \frac{1713}{500\times 4.18}$

$\Delta T = \frac{1713}{2090} \approx 0.8196^\circ\text{C}$

Thus,

$\Delta T \approx 0.82^\circ\text{C}$

6. Match with the form given

Given:

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

Since

$0.82 = 82\times 10^{-2}$

we get

$x = 82$

7. Comparison with stored answer

Stored correct answer = 82, which matches our result.