Question

The balanced combustion reaction for C 6 H 6 is 2 C 6 H 6 ( l ) 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) 6 H 2 O ( l ) 6542 kJ If 7.700 g C 6 H 6 is burned and the heat produced from the burning is added to 5691 g of water at 21 ∘ C, what is the final temperature of the water

Answer 1

Answer: $156.4^0C$

Explanation:

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542kJ$ $\Delta H=-6542kJ$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.700g}{78.11g/mol}=0.9858moles$

According to stoichiometry :

2 moles of $C_6H_6$ releases = 6542 kJ of heat

0.9858 moles of $C_6H_6$  release =$\frac{6542}{2}\times 0.9858=3224kJ$ of heat

Thus heat given off by burning 7.700 g of $C_6H_6$  will be absorbed by water.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= 3224kJ = 3224000J   (1kJ=1000J)

m= mass of water = 5691 g

c = specific heat capacity = $4.184J/g^0C$

Initial temperature of the water = $T_i$ = 21.0°C

Final temperature of the water = $T_f$  = ?

Putting in the values, we get:

$3224000J =5691g\times 4.184J/g^0C\times (T_f-21)$

$T_f=156.4^0C$

The final temperature of the water is $156.4^0C$