When 600mL of 0 . 2M HNO _ 3 is mixed with 400mL of 0 . 1MNaOH...

Number of moles of given acid and base are

$$\begin{array}{ll}{HNO}_{3} = \dfrac{600}{1000} \times 0 . 2M = 0 . 12moles\end{array}$$NaOH = \dfrac{400}{1000} \times 0 . 1M = 0 . 04moles$</p><p>Now neutralisation reaction between them will be</p><p>${HNO}_{3} + NaOH \to {NaNO}_{3} + {H}_{2}O0 . 12mole0 . 04mole0 . 08mole0mole0 . 04mole$</p><p>Heat relesed during this reaction will be </p><p>${\Delta}_{r}H = 0 . 04 \times \left(57 \times {10}^{3}\right)J$</p><p>$= 2280J$</p><p>Now temperature increased can be calculated by</p><p>$mS \Delta T = 2280J$</p><p>$⇒ 1000mL \times \dfrac{1gm}{mL} \times 4 . 2 \times \Delta T = 2280$</p><p>$\Delta T = \dfrac{2280}{4 . 2} \times {10}^{- 3}$</p><p>$= 542 . 86 \times {10}^{- 3}$</p><p>$\Delta T = 54 . 286 \times {10}^{- 2} ° C$