Respuesta :
Answer:
31.9178 °C is the final temperature of the water
Explanation:
[tex]2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ[/tex]
Mass of benzene burned = 6.200 g
Moles of benzene burned = [tex]\frac{6.200 g}{78 g/mol}=0.0794 mol[/tex]
According to reaction , 2 moles of benzene gives 6542 kJ of energy on combustion.
Then 0.0794 mol of benzene on combustion will give:
[tex]\frac{6542 kJ}{2}\times 0.0794 kJ=259.7174 kJ=Q[/tex]
Mass of water in which Q heat is added = m = 5691 g
Initial temperature = [tex]T_i=21^oC[/tex]
Final temperature = [tex]T_f[/tex]
Specific heat of water = c = 4.18 J/g°C
Change in temperature of water = [tex]T_f-T_i[/tex]
[tex]Q=mc\Delta t=mc(T_f-T_i)[/tex]
[tex]259,717.4 J=5691 g\times 4.18 J/g^oC\times (T_f-21^oC)[/tex]
[tex]T_f=31.91 ^oC[/tex]
31.9178 °C is the final temperature of the water
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:
[tex]\rm Q=heat=\dfrac{0.0795}{2}\times 6542\:kJ\\\\Q=260.0445\: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)
260044.5 = 5691 . 4.186.∆T
[tex]\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[/tex]
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