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

Question

If  6.200g C6H6 is burned and the heat produced from the burning is added to 5691g of water at 21C, what is the final temperature of the water?

Answer 1

6.200 g C6H6 x 1 mol C6H6 / 78.11 g = 0.07938 mols

0.07938 mols x 6542 kJ / 2 mols = 259.6 kJ

q = mC∆T

q = heat = 259.6 kJ

m = mass of water = 5691 g

C = specific heat of water = 4.184 J/gº = 0.004184 kJ/gº

∆T = change in temperature

Solving for ∆T we have...

∆T = 259.6 kJ / (5691 g)(0.004184 kJ/gº)

∆T = 10.89 degrees

Final temperature = 21º + 10.89º = 31.89º = 32ºC