Question

The balanced combustion reaction for C 6 H 6 C6H6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 5.500 g C 6 H 6 5.500 g C6H6 is burned and the heat produced from the burning is added to 5691 g 5691 g of water at 21 ∘ 21 ∘ C, what is the final temperature of the water?

Answer 1

Step-by-step explanation:

First, we will calculate the molar mass of
`C_(6)H_(6)` as follows.

Molar mass of
`C_(6)H_(6)` =
`6 * 12 + 6 * 1`

= 78 g/mol

So, when 2 mol of
`C_(2)H{6}` burns, then heat produced = 6542 KJ

Hence, this means that 2 molecules of
`C_(6)H{6}` are equal to
`78 * 2 = 156 g` of
`C_(6)H_(6)` burns, heat produced = 6542 KJ

Therefore, heat produced by burning 5.5 g of
`C_(6)H{6}` =

`6542 kJ * (5.5 g)/(156 g)`

= 228.97 kJ

= 228970 J (as 1 kJ = 1000 J)

It if given that for water, m = 5691 g

And, we know that specific heat capacity of water is 4.186
`J/g^(o)C` .

As, Q =
`m * C * (T_(f) - T_(i))`

228970 J =
`5691 g * 4.184 J/g^(o)C * (T_(f) - 21) ^(o)C`

`T_(f) - 21^(o)C = 9.616^(o)C`

`T_(f) = 30.6^(o)C`

Thus, we can conclude that the final temperature of the water is
`30.6^(o)C`.