Question

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?

Answer 1

The final temperature of the water : 30.506 °C

Further explanation  

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released  

Qin = Qout  

Heat can be calculated using the formula:  

Q = mc∆T  

m = mass, g  

∆T = temperature difference, °C / K  

From reaction:  

2C₆H₆ (l) + 15O₂ (g) ⟶12CO₂ (g) + 6H₂O (l) +6542 kJ, heat released by +6542 kJ to burn 2 moles of C₆H₆

If there are 5.400 g of C₆H₆ then the number of moles:  

mol = mass: molar mass C₆H₆

mol = 5.4 : 78  

mol C₆H₆ = 0.0692

so the heat released in combustion 0.0692 mol C₆H₆:  

$\rm Q=heat=\dfrac{0.0692}{2}\times 6542\:kJ\\\\Q=226.353\:kJ$

the heat produced from the burning is added to 5691 g of water at 21 C

So :

Q = m . c . ∆T (specific heat of water = 4,186 joules / gram ° C)  

226353 = 5691 . 4.186.∆T  

$\rm \Delta T=\dfrac{226353}{5691\times 4.186}\\\\\Delta T=9.506\\\\\Delta T=T(final)-Ti(initial)\\\\9.506=T_f-21\\\\T_f=30.506\:C$  

Learn more  

the difference between temperature and heat  

brainly.com/question/3821712  

Specific heat  

brainly.com/question/9525136  

relationships among temperature, heat, and thermal energy.  

brainly.com/question/224374  

When heat is added to a substance  

brainly.com/question/3417580  

Answer 2

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