Question

The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJIf 8.700 g C6H6is 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 : The final temperature of the water is, $36.4^oC$

Explanation :

First we have to calculate the moles of benzene.

Mass of $C_6H_6$ = 8.700 g

Molar mass of $C_6H_6$ = 78 g/mol

$\text{Moles of }C_6H_6=\frac{\text{Mass of }C_6H_6}{\text{Molar mass of }C_6H_6}=\frac{8.700g}{78g/mole}=0.112mole$

Now we have to calculate the heat produced by the reaction.

As, 2 moles of benzene burns to give heat = 6542 kJ

So, 0.112 moles of benzene burns to give heat = $\frac{0.112}{2}\times 6542=366.4kJ$

Now we have to calculate the final temperature of the water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 366.4 kJ = 366400 J

m = mass of water = 5691 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $21^oC$

$T_2$ = final temperature = ?

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

$366400J=5691g\times 4.18J/g^oC\times (T_2-21.0)$

$T_2=36.4^oC$

Thus, the final temperature of the water is, $36.4^oC$