A solution is 500 mL of 2M $KOH$ is added to 500 mL of 2M $HCl$ and the mixture is well shaken. The rise in temperature $T_{1}$ is noted. The experiment is then repeated using 250 mL of each solution and rise in temperature $T_{2}$ is again noted. Assume all heat is taken by the solution :

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$T_{1} = T_{2}$

$T_{2}$ is twice of $T_{1}$

$T_{1}$ is 2 times as large as $T_{2}$

$T_{1}$ is 4 times as large as $T_{2}$

answer is A.

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Detailed Solution

In the reaction,

$KOH + HCl \to KCl + H_{2} O$

$Q_{t} = Total weight \times Raise in T$

$\begin{array}{l} Q_{1} = 1000 \times T_{1} \\ Q_{2} = 500 \times T_{2} \\ Q_{1} = 2 Q_{2} \\ 1000 T_{1} = 2 \times 500 T_{2} \\ \Rightarrow T_{1} = T_{2} \end{array}$

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