If 6.200 g of 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

The balanced equation for complete combustion of C6H6 is

2C6H6 + 15O2 ==> 12CO2 + 6H2O

From the literature, the ∆Hcombustion for C6H6 (benzene) =  -3270 kJ/mole

moles C6H6 present = 6.200 g x 1 mole/78.11 g = 0.07938 moles

Heat produced from this = 0.07938 moles x (- 3270 kJ/mole) = -250.6 kJ

q = mC∆T where q = heat; m=mass; C=sp.heat; ∆T = change in temperature

250.6 kJ = (5.691kg)(4.184 kJ/kg/deg)(∆T)

Solving for ∆T = 250.6/23.81 = 10.52 degrees

Final temperature of water = 21º + 10.5º = 31.5ºC

Answer 2

6.200 g C6H6 x (1 mol C6H6/78.11 g C6H6) = .07938 mol C6H6

2C6H6 + 15O2 → 6H2O + 12CO2

This is the balanced eq. But, the H for one mole of benzene is –3273 kJ/mol. I got this off the Internet.

So, .09738 mol C6H6 x -3273 kJ/mol = -259.8 kJ this will be absorbed by the water. (Change sign)

5691 g H2O x 1 mol H2O/18.02 g H2O = 315.8 mol H2O

cp = q/mΔT

Solve for ΔT

ΔT = q/m cp

ΔT = 259.8 x 10^3 J/(5691 g H2O)(4.18 J/gC) = 10.92 C

Begin at 21, add 10.92, end at 31.92C