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Calorimetry

200 ml of 1...

Question

$200$ ml of $1$ M $H C l$ is mixed with $400$ ml of $0.5$ M $N a O H$ . The temperature rise in the calorimeter was found to be ${4.0}^{o} C .$ Water equivalent of calorimeter is $25$ g and the specific heat of the solution is $1$ cal/mL/degree. If the theoritical heat of neutralization of a strong acid and strong base is $- 13.5$ kcal, then the percentage error (as nearest integer) incurred in this experiment while calculating the heat of neutralization is

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Solution

Total volume $200 + 400 = 600 m l$
200 ml of 1 M acid $=$ 400 ml of 0.5 M $N a O H$
200 m. eq. of acid reacts with 200 m.eq. of base $= Δ H$
$∴$ Heat of neutralization $= 5 \times Δ H$
Heat produced during neutralization $=$ heat taken up by calorimeter + solution
$= m_{1} s_{1} Δ T + m_{2} s_{2} Δ T$
$= (25 \times 1 \times 4) + (600 \times 1 \times 4) = 2500 c a l$
$∴$ Heat of neutralization $= - 5 \times 2500$ $= - 12500 c a l = - 12.5 k c a l$
% error $= (\frac{13.5 - 12.5}{13.5}) \times 100 = 7.4$

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