Question

The balanced combustion reaction for C 6 H 6 C6H6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 5.500 g C 6 H 6 5.500 g C6H6 is burned and the heat produced from the burning is added to 5691 g 5691 g of water at 21 ∘ 21 ∘ C, what is the final temperature of the water?

Answer 1

Explanation:

First, we will calculate the molar mass of $C_{6}H_{6}$ as follows.

Molar mass of $C_{6}H_{6}$ = $6 \times 12 + 6 \times 1$

                                   = 78 g/mol

So, when 2 mol of $C_{2}H{6}$ burns, then heat produced = 6542 KJ

Hence, this means that 2 molecules of $C_{6}H{6}$ are equal to $78 \times 2 = 156 g$ of $C_{6}H_{6}$ burns, heat produced = 6542 KJ

Therefore, heat produced by burning 5.5 g of $C_{6}H{6}$ =                  

       $6542 kJ \times \frac{5.5 g}{156 g}$

            = 228.97 kJ

            = 228970 J           (as 1 kJ = 1000 J)

It if given that for water, m = 5691 g

And, we know that specific heat capacity of water is 4.186 $J/g^{o}C$ .

As,             Q = $m \times C \times (T_{f} - T_{i})$

          228970 J = $5691 g \times 4.184 J/g^{o}C \times (T_{f} - 21) ^{o}C$

                $T_{f} - 21^{o}C = 9.616^{o}C$

                $T_{f} = 30.6^{o}C$

Thus, we can conclude that the final temperature of the water is $30.6^{o}C$.