If 5.400 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?

Actually, we will use this formula:

Q = m. C.ΔT

when Q is the heat released during the reaction.

and m is the mass of water.

and C is the specific heat of water

and ΔT is the change of the temperature of the water.

So according to the balanced equation for this reaction:

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

So when 2 moles of C6H6 released 6542 KJ of heat, So how many moles of the 5.4 g

no of moles of C6H6 = 5.4 / molar mass of C6H6

= 5.4g/78.11 g/mol = 0.069 moles

so 2 moles of C6H6 released → 6542

∴ 0.069 moles of C6H6 will release → X

∴Q = (0.069 * 6542) / 2 = 226 KJ

So by substitution in Q = m. c.ΔT to get ΔT

∴Δ T = Q / (m. c) = 226 / (5691*0.004186) = 9.496

when we have Ti

ΔT=Tf-Ti

∴Tf = 9.496 + 21 = 30.5 °C