When 600 mL of 0.2  HNO₃ is mixed with 400 mL of 0.1 M NaOH solution in a flask, rise in temperature of the flask is ___________ $\times {10}^{-2}{}^{0}C$

(Enthalpy of neutralization = $57kJmo{l}^{-1}$ and specific heat of water  $4.2J{K}^{-1}{g}^{-1}$ ]

(Neglect heat capacity of flask)

$\begin{array}{l}4.2J{K}^{-1}{g}^{-1}HN{O}_{3}NaOH\\ 600mL\times 0.2M400mL\times 0.1M\\ =120mmol=40mmol\end{array}$

$HN{O}_3+NaOH\to NaN{O}_3+H_{2}O$

Bef:     120                             40

Aft:      80                                0                                  40 m mol

${\Delta }_{r}H=40mmol\times (57\times {10}^3)\frac{J}{mol}$

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

= 2280 J

$mS\Delta T=2280$

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

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

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

$\Delta T=54.26\times {10}^{-2}K$

$\Delta T=54.26\times {10}^{-2 0}C$

Answer mentioned as 54 (Closest integer)