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 > 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)

Millimoles of HCl = 200 * 0.2 = 40

Millimoles of NaOH = 300 * 0.1 = 30

Heat released =  $\frac{30}{1000}*57.1*1000J=1713J$

$\left[\begin{array}{l}\rho =\frac{m}{v}\\ m=\rho v\end{array}\right]$

Mass of solution = 500 * 1 g

= 500 g

Specific heat of water = 4.18 Jg^-1 K^-1

$\Delta T=\frac{q}{mc}$

$=\frac{1713}{500*4.18}°C$  

$=81.96*10^{-2}°C\approx 82*10^{-2}°C$

Ans. = 82