When $600 \mathrm{mL}$ of $0.2\mathrm{M} {\mathrm{HNO}}_3$ is mixed with $400 \mathrm{mL}$ of $0.1\mathrm{M} \mathrm{NaOH}$ solution in a flask, the rise in temperature of the flask is_____$\times {10}^{-2} °\mathrm{C}$
(Enthalpy of neutralisation $=57 \mathrm{kJ}{ \mathrm{mol}}^{-1}$ and Specific heat of water $=4.2{\mathrm{JK}}^{-1}{ \mathrm{g}}^{-1}$) (Neglect heat capacity of flask)

Number of moles of given acid and base are

$$\begin{gathered} \begin{array}{cc}{\mathrm{HNO}}_3 = \frac{600}{1000}\times 0.2 \mathrm{M}= 0.12 \mathrm{moles} & \end{array} \\ \mathrm{NaOH} = \frac{400}{1000}\times 0.1 \mathrm{M} = 0.04 \mathrm{moles} \end{gathered}$$

Now neutralisation reaction between them will be

$$\begin{gathered} {\mathrm{HNO}}_3 + \mathrm{NaOH} \to {\mathrm{NaNO}}_3+{\mathrm{H}}_2\mathrm{O} \\ 0.12 \mathrm{mole} 0.04 \mathrm{mole} \\ 0.08 \mathrm{mole} 0 \mathrm{mole} 0.04 \mathrm{mole} \end{gathered}$$

Heat relesed during this reaction will be 

${\mathrm{\Delta }}_{\mathrm{r}}\mathrm{H}=0.04\times \left(57\times {10}^3\right) \mathrm{J}$

$=2280 \mathrm{J}$

Now temperature increased can be calculated by

$\mathrm{mS\Delta T}=2280 \mathrm{J}$

$\Rightarrow 1000 \mathrm{mL}\times \frac{1\mathrm{gm}}{\mathrm{mL}}\times 4.2\times \mathrm{\Delta T}=2280$

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

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

$\mathrm{\Delta T}=54.286\times {10}^{-2} °\mathrm{C}$