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When $9.2g$ of Toluene $C_{7}H_{8}$ undergoes combustion in a bomb calorimeter whose heat capacity is $4.15\, KJ/^{o}C$ , the temperature rises from $20^{o}C$ to $45^{o}C$ . Calculate the change in internal energy (in J/mol) for the combustion of toluene.
Option: 1
1576

Option: 2
103.75

Option: 3
157.6

Option: 4
1037.5

Answers (1)

best_answer

As we have learned,

Internal energy in a bomb calorimeter is calculated as $\Delta E=\frac{Z*M*\Delta T}{w}$

Here,
$Z=$ heat capacity of bomb calorimeter
$M=$ the molecular mass of substrate
$W=$ weight of substrate
$T=$ temperature
Given,
$Z = 4.15\, KJ/^{o}C$
$M = 92 g$
$W = 9.2g$
Change in Temperature $=45-20=25^{o}C$

Putting values in the equation,

$\Delta E=\frac{4.15*92*25}{9.2}$

$\Delta E=1037.5\, K\! J/mol$

Posted by

Kuldeep Maurya

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