Question

The following equation is the balanced combustion reaction for C6H6:

2C6H6 (l) + 15O2 (g) --> 12CO2(g) + 6H2O(l)+6542KJ

If 8.800 g of C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 Celcius, what is the final temp of the water? (in celcius)

Answer 1

The balanced combustion reaction for C₆H₆ is---

2C₆H₆ (l) + 15O₂ (g) --> 12CO₂(g) + 6H₂O(l)

Now, C₆H₆ =( 12 x 6 ) +(1 x 6) = 72 + 6 = 78 g

The above equation indicates that 2x78 = 156 g C₆H₆ will produce 6542 KJ
Now, by using proportions we can then calculate that ---

x/6542kJ = 8.800 g / 156g
=> x = 6542 x 8.800 g / 156 g

=> x =57569.6 /156 KJ

=> x = 369.035 KJ

=> x = 369035 J (heat)
We have the relation,

heat(Q) = mass(m) *specific heat(c) * T  

=>T = 369035 J / (5691g x 4.18J/g°C)

=> T= 15.5⁰C

Therefore, final temperature = (21 + 15.5)° C= 36.5°C