If 5.700 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?

Question

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

Answer 1

First, write the balanced equation for the combustion of C₆H₆:

2C₆H₆ + 15O₂ ==> 12CO₂ + 6H₂O ... balanced equation

Next, we find the moles of C₆H₆:

5.700 g x 1 mol C₆H₆/78 g = 0.07308 moles

Now we can determine the heat of this reaction:

0.07308 moles x 6542 kJ/2 mol = 239.0 kJ

To find the change in temperature (and the final temperature) we use the following:

q = mC∆T

q = heat = 239.0 kJ

m = mass of water = 5691 g = 5.691 kg

C = specific heat water = 4.184 J/gº (look it up if you don't know it) = 4.184 kJ/kgº

∆T = change in temperature = ?

Solving for ∆T we have...

∆T = q/mC = 239.0 kJ/(5.691 kg)(4.184 kJ/kgº)

∆T = 10.04º = 10.º

Since the reaction is exothermic (heat generated), the temperature will increase. Thus, final temperature of the water is 21º + 10.º = 31ºC