Question

When $600mL$ of $0.2MHN{O}_3$ is mixed with $400mL$ of $0.1MNaOH$ solution in a flask, the rise in temperature of the flask is ___$\times 10^{-2}{}^{0}C$.
(Enthalpy of neutralisation $=57kJmo{l}^{-1}$ and Specific heat of water $=4.2J{K}^{-1}{g}^{-1}$ )
(Neglect heat capacity of flask)

Answer 1

Answer: 54

${\Delta }_{r}H=40mmol\times \left(57\times 10^3\right)\frac{J}{mol}$
$=40\times 10^{-3}mol\times 57\times 10^3\frac{J}{mol}$
$=2280J$
$mS\Delta T=2280$
$\Rightarrow 1000mL\times \frac{1gm}{mL}\times 4,2\times \Delta T=2280<br/>$
$\Delta T=\frac{2280}{4.2}\times 10^{-3}$
$=\frac{22800}{42}\times 10^{-3}$
$=542.86\times 10^{-3}$
$\Delta T=54.286\times 10^{-2}K$
$\Delta T=54.286\times 10^{-20}C$
Ans. $54.286$
Answer mentioned as 54 (Closest integer)