Answer : The final temperature of the water is, [tex]36.4^oC[/tex]
Explanation :
First we have to calculate the moles of benzene.
Mass of [tex]C_6H_6[/tex] = 8.700 g
Molar mass of [tex]C_6H_6[/tex] = 78 g/mol
[tex]\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[/tex]
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 = [tex]\frac{0.112}{2}\times 6542=366.4kJ[/tex]
Now we have to calculate the final temperature of the water.
[tex]q=m\times c\times (T_2-T_1)[/tex]
where,
q = heat produced = 366.4 kJ = 366400 J
m = mass of water = 5691 g
c = specific heat capacity of water = [tex]4.18J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]21^oC[/tex]
[tex]T_2[/tex] = final temperature = ?
Now put all the given values in the above formula, we get:
[tex]366400J=5691g\times 4.18J/g^oC\times (T_2-21.0)[/tex]
[tex]T_2=36.4^oC[/tex]
Thus, the final temperature of the water is, [tex]36.4^oC[/tex]
