The balanced combustion reaction for C6 H6 is

Question

The balanced combustion reaction for C₆ H₆ is

2 C₆ H₆ (l) + 15 O₂ (g)---> 12 CO₂ (g) + 6 H₂O(l) + 6542 kJ

If 7.100 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?

final temperature: ______∘C

Answer 1

2 C₆ H₆ (l) + 15 O₂ (g)---> 12 CO₂ (g) + 6 H₂O(l) + 6542 kJ

q = mC∆T

q = heat = see below

m = mass of water = 5691 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = ?

To find q (heat), we use the information from the combustion reaction:

7.100 g C6H6 x 1 mol C6H6 / 78.11 g x 6542 kJ / 2 mol C6H6 = 297.3 kJ of heat generated by the combustion

This heat is absorbed by the water to raise the temperature of the water.

q = mC∆T

297300 J = 5691 g x 4.184 J/gº x ∆T

∆T = 12.49º

Final temperature of the water = 21º + 12.49º = 33.49ºC = 33ºC (2 sig.figs.)

Answer 2

7.1 g / 78 g/mol = 0.091026 mol

0.091026 / 2 = 0.045513

0.045513 x 6542 KJ = 297.7 KJ released

q = m x c x (change in temp)

c is heat capacity. The heat capacity of water is 4.184 J/gC

297.7 KJ x (1000 J / 1 KJ) = 5691 g x 4.184 J/gC x (temp change)

297700 / 5691 / 4.184 = 12.5 C

12.5 + 21 = 33.5 C