Question

The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 6.200 g 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

Answer:

31.9178 °C is the final temperature of the water

Explanation:

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ$

Mass of benzene burned = 6.200 g

Moles of benzene burned = $\frac{6.200 g}{78 g/mol}=0.0794 mol$

According to reaction , 2 moles of benzene gives 6542 kJ of energy on combustion.

Then 0.0794 mol of benzene on combustion will give:

$\frac{6542 kJ}{2}\times 0.0794 kJ=259.7174 kJ=Q$

Mass of water in which Q heat is added = m = 5691 g

Initial temperature = $T_i=21^oC$

Final temperature = $T_f$

Specific heat of water = c = 4.18 J/g°C

Change in temperature of water = $T_f-T_i$

$Q=mc\Delta t=mc(T_f-T_i)$

$259,717.4 J=5691 g\times 4.18 J/g^oC\times (T_f-21^oC)$

$T_f=31.91 ^oC$

31.9178 °C is the final temperature of the water

Answer 2

The final temperature of the water : 31.916 °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

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆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 6,200 g of C₆H₆ then the number of moles:

mol = mass: molar mass C₆H₆

mol = 6.2: 78

mol C6H6 = 0.0795

so the heat released in combustion 0.0795 mol C6H6:

$\rm Q=heat=\dfrac{0.0795}{2}\times 6542\:kJ\\\\Q=260.0445\:kJ$

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)

260044.5 = 5691 . 4.186.∆T

$\rm \Delta T=\dfrac{260044.5}{5691\times 4.186}\\\\\Delta T=10.916\\\\\Delta T=T(final)-Ti(initial)\\\\10.916=T-21\\\\T=31.916\:C$

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