Respuesta :
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
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
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:
[tex]\rm Q=heat=\dfrac{0.0692}{2}\times 6542\:kJ\\\\Q=226.353\:kJ[/tex]
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
[tex]\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[/tex]
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