200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is -57.1 kJ. The increase in temperature in °C of the system on mixing is $$x \times 10^{-2}$$. The value of x is _________. (Nearest integer)
[Given: Specific heat of water = 4.18 J g$$^{-1}$$ K$$^{-1}$$
Density of water = 1.00 g cm$$^{-3}$$]
(Assume no volume change on mixing)

We have an acid-base neutralisation: $$\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$$.

First we calculate the number of moles present in each solution.

The given data are:

Volume of HCl = 200 mL = 0.200 L,   Molarity of HCl = 0.2 M

Volume of NaOH = 300 mL = 0.300 L,   Molarity of NaOH = 0.1 M

Using $$n = M \times V$$ (moles = molarity × volume in litres):

$$n_{\text{HCl}} = 0.2 \, \text{mol L}^{-1} \times 0.200 \, \text{L} = 0.040 \, \text{mol}$$

$$n_{\text{NaOH}} = 0.1 \, \text{mol L}^{-1} \times 0.300 \, \text{L} = 0.030 \, \text{mol}$$

Comparing the two, NaOH has fewer moles, so NaOH will be the limiting reagent and only $$0.030 \, \text{mol}$$ of each reactant will actually neutralise.

The molar heat of neutralisation (enthalpy change) is given as $$\Delta H = -57.1 \, \text{kJ mol}^{-1}$$. The negative sign denotes that heat is released.

The total heat released is found from $$q = n \times \Delta H$$:

$$q = 0.030 \, \text{mol} \times 57.1 \, \text{kJ mol}^{-1} = 1.713 \, \text{kJ}$$

We convert this to joules because the specific heat capacity is in J g−1 K−1:

$$1.713 \, \text{kJ} = 1.713 \times 10^{3} \, \text{J} = 1713 \, \text{J}$$

Now we determine the temperature rise of the entire solution that absorbs this heat. The total volume after mixing is

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

Since the density of water is $$1.00 \, \text{g cm}^{-3}$$ (1 g mL−1), the mass of the solution is

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

The specific heat capacity of water is $$c = 4.18 \, \text{J g}^{-1} \text{K}^{-1}$$.

Using the relation $$q = m \, c \, \Delta T$$ and solving for $$\Delta T$$, we get

$$\Delta T = \dfrac{q}{m \, c} = \dfrac{1713 \, \text{J}}{500 \, \text{g} \times 4.18 \, \text{J g}^{-1} \text{K}^{-1}}$$

$$\Delta T = \dfrac{1713}{2090} \, \text{K} = 0.8196 \, \text{K}$$

For aqueous solutions the numerical value in Kelvin and in degrees Celsius is the same, so the temperature rise is $$0.8196^{\circ}\text{C}$$.

The problem states that this rise can be written as $$x \times 10^{-2}$$ °C. Hence,

$$x = 0.8196 \times 10^{2} = 81.96$$

Rounding to the nearest integer, $$x = 82$$.

Hence, the correct answer is Option 82.