Answer:
The final temperature of water = 35.2 °C
Explanation:
Step 1: Data given
Mass of C6H6 = 8.100 grams
Mass of water = 5691 grams
Temperature = 21 °C
Step 2: The balanced equation
2C6H6(l) + 15O2(g) ⟶ 12 CO2 (g) + 6H2O (l) + 6542 kJ
Step 3:
Q = m*c*ΔT.
⇒with Q = the heat released during this reaction (this depends on the amount of reactants used)
⇒ with m= the mass of the water
⇒with c = the "specific heat" of water = how much energy it takes to raise the temp of 1g of water by 1°C
⇒with ΔT = the change in the temperature of the water
For every 2 moles of C6H6 consumed, 6542 kJ of heat is released.
Step 4: Calculate moles for 8.100 grams
8.100grams / 78.11 g/mol= 0.1037 mol es
So, according to the equation, the amount of heat released is:
(0.1037 moles / 2 moles)* (6542 kJ) = 339.2 kJ
Step 5: Calculate the final temperature
Q = mcΔT
ΔT = Q / (m*c)
T2- T1 = Q / (m*c)
T2 = [Q / (m*c)] + T1 = [(339.2 kJ) / (5691g)(0.004186 kJ/g°C)] + 21°C = 35.2°C
The final temperature of water = 35.2 °C
