Question

The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJIf 8.700 g C6H6is 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 : The final temperature of the water is,
`36.4^oC`

Explanation :

First we have to calculate the moles of benzene.

Mass of
`C_6H_6` = 8.700 g

Molar mass of
`C_6H_6` = 78 g/mol

`\text{Moles of }C_6H_6=\frac{\text{Mass of }C_6H_6}{\text{Molar mass of }C_6H_6}=(8.700g)/(78g/mole)=0.112mole`

Now we have to calculate the heat produced by the reaction.

As, 2 moles of benzene burns to give heat = 6542 kJ

So, 0.112 moles of benzene burns to give heat =
`(0.112)/(2)* 6542=366.4kJ`

Now we have to calculate the final temperature of the water.

`q=m* c* (T_2-T_1)`

where,

q = heat produced = 366.4 kJ = 366400 J

m = mass of water = 5691 g

c = specific heat capacity of water =
`4.18J/g^oC`

`T_1` = initial temperature =
`21^oC`

`T_2` = final temperature = ?

Now put all the given values in the above formula, we get:

`366400J=5691g* 4.18J/g^oC* (T_2-21.0)`

`T_2=36.4^oC`

Thus, the final temperature of the water is,
`36.4^oC`