Question

1.000 g sample of octane (c8h18) is burned in a bomb calorimeter containing 1200 grams of water at an initial temperature of 25.00c. After the reaction, the final temperature of the water is 33.20c. The heat capacity of the calorimeter (also known as the calorimeter constant) is 837 j/c. The specific heat of water is 4.184 j/g c. Calculate the heat of combustion of octane in kj/mol.

Answer 1

Answer : The heat of combustion of octane is, $-5475.42KJ/mole$

Solution :

First we have to calculate the heat absorbed by water.

Formula used :

$q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})$

where,

$q_w$ = heat absorbed = ?

$m_w$ = mass of water = 1200 g

$c_w$ = specific heat of water= $4.184J/g^oC$

$T_{final}$ = final temperature = $33.2^oC$

$T_{initial}$ = initial temperature = $25^oC$

Now put all the given values in the above formula, we get

$q_w=1200g\times 4.184J/g^oC\times (33.2-25)^oC$

$q_w=41170.56J$

Now we have to calculate the heat of calorimeter.

$q_b=c_b\times (T_{final}-T_{initial})=837J/^oC\times (33.2-25)^oC=6863.4J$

Now we have to calculate the heat released in combustion.

$q_{comb}=-(q_w+q_b)\\\\q_{comb}=-(41170.56+6863.4)\\\\q_{comb}=-48033.96J=-48.03KJ$

Now we have to calculate the heat of combustion of octane.

As, 1 gram of octane released heat of combustion -48.03 KJ

So, 114 g/mole of octane released heat of combustion $-48.03\times 114=-5475.42KJ/mole$

Therefore, the heat of combustion of octane is, $-5475.42KJ/mole$