Question

200 mL of $0.2MHCl$ is mixed with 300 mL of$0.1MNaOH$. The molar heat of neutralization of this reaction is $-57.1kJ$. The increase in temperature in ${}^{o}C$ of the system on mixing is $x\times {10}^{-2}$. 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 $c{m}^{-3}$]
[Assume no volume change on mixing]

Answer 1

$HCl+NaOH\to NaCl+H_{2}O$
Moles 0.04 0.0. -- --

0.01 -- 0.03 0.03

$Q,\text{Heat released}=0.03\times 57.1kJ=1.713kJ$

$Q=m\times s\times \Delta T$
$s\text{is specific heat}$

$\Delta T=\frac{1.713\times 1000}{500\times 4.18}=81.96\times {10}^{-2}\approx 82\times {10}^{-2}$
$x=82$