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 7.300 g C 6 H 6 7.300 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?

Respuesta :

Answer: The final temperature of the water is [tex]33.85^{o}C[/tex].

Explanation:

We know that molar mass of [tex]C_{6}H_{6}[/tex] is 78 g/mol. And, the amount of heat produced when 2 mol of [tex]C_{2}H_{6}[/tex] burns is 6542 KJ.

This means that,

   [tex]78 \times 2[/tex] = 156 g of [tex]C_{2}H_{6}[/tex] burns, heat produced is 6542 kJ.

Therefore, heat produced (Q) by burning  7.3 g of [tex]C_{6}H_{6}[/tex] is as follows.

               [tex]\frac{6542 \times 7.3 g}{156 g}[/tex]

              = 306.13 kJ

or,           = 306130 J      (as 1 KJ = 1000 J)

For water, mass is given as 5691 g and specific heat capacity of water is 4.186 [tex]J/g^{o}C[/tex].

So, we will calculate the value of final temperature as follows.

            Q = [tex]m \times C \times (T_{f} - T_{i})[/tex]

  306130 J = [tex]5691 g \times 4.186 J/g^{o}C \times (T_{f} - 21)^{o}C[/tex]

       [tex](T_{f} - 21)^{o}C = \frac{306130 J}{23822.53 J/^{o}C}[/tex]

          [tex]T_{f}[/tex] = 12.85 + 21

                      = [tex]33.85^{o}C[/tex]

Thus, we can conclude that the final temperature of the water is [tex]33.85^{o}C[/tex].