question below in the description

Question

The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 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

2 C₆H₆ + 15 O₂ ==> 12 CO₂ + 6 H₂O + 6542 kJ (exothermic)

calculate heat generated from 8.600 g of C6H6:

molar mass C6H6 = 78.11 g

8.600 g x 1 mol / 78.11 g = 0.1101 mols x 6542 kJ / 2 mols = 360.1 kJ of heat

Next, we use this amount of heat to find the final temperature of the water, as follows:

q = mC∆T

q = heat = 360.1 kJ = 360100 J

m = mass of water = 5691 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature of the water = ?

Solving for ∆T, we have...

∆T = q / (m)(C) = 360100 J / (5691 g)(4.184 J/gº)

∆T = 15.1º

Final temperature of the water = 21º + 15.1º = 36º (2 sig.figs.